Lesson QUADRATIC EQUATIONS
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This lesson covers an overview of QUADRATIC EQUATIONS This lesson concerns itself with finding the roots of quadratic equations. It does not discuss properties of PARABOLAS that are the result of graphing quadratic equations. As such, the minimum and maximum value of the parabola (or the graph of the quadratic equation depending on what you want to call it) and the axis of symmetry and all other properties of the parabola are not discussed. See the lesson on PARABOLAS to learn more about those properties. REFERENCES <a href = "http://www.themathpage.com/alg/quadratic-equations.htm" target = "_blank">http://www.themathpage.com/alg/quadratic-equations.htm</a> <a href = "http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut17_quad.htm" target = "_blank">www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut17_quad.htm</a> <a href = "http://www.purplemath.com/modules/factquad.htm" target = "_blank">http://www.purplemath.com/modules/factquad.htm</a> <a href = "http://www.mathsisfun.com/algebra/completing-square.html" target = "_blank">http://www.mathsisfun.com/algebra/completing-square.html</a> <a href = "http://www.purplemath.com/modules/quadform.htm" target = "_blank">http://www.purplemath.com/modules/quadform.htm</a> DEFINITION OF QUADRATIC EQUATION A quadratic equation is a polynomial equation where the highest degree is 2. Example: y = x^2 + 2x + 2 STANDARD FORM OF A QUADRATIC EQUATION f(x) = ax^2 + bx + c = 0 where: a = coefficient of x^2 b = coefficient of x c = constant term Please note that x^2 and {{{x^2}}} are the same thing. The first form is just typed on the page as you see it. The second form uses the formula generating functions of algebra.com. Please note that the standard form of the quadratic has already set the equation equal to 0. This sets the coefficients of a, b, and the constant of c to the right value. Example: x^2 - 2x + 1 = 2 is NOT in the correct standard form for finding the roots of the equation. Subtract 2 from both sides of this equation to get: x^2 - 2x - 1 = 0 IS in the correct standard form for finding the roots of the equation. Standard form for the quadratic equation is for is ax^2 + bx + c = 0 This quadratic equation is x^2 - 2x - 1 = 0 where: a = 1 b = -2 c = -1 ROOTS OF A QUADRATIC EQUATION The roots of a quadratic equation are determined by the value of x when the equation crosses the x-axis. The roots are found by setting the equation equal to 0 and solving for x. If you get a real number for an answer, then the equation has roots. If you get an imaginary or complex number for an answer, then the equation has no roots. A quadratic equation can have: 0 roots (none) 1 root 2 roots QUADRATIC EQUATION WITH ONE ROOT Following is the graph of a quadratic equation with one root. {{{graph(300,300,-5,5,-5,5,x^2-4x+4)}}} The quadratic equation used for this graph is: y = x^2 -4x + 4 the root for this equation is x = 2. If you plug x = 2 into the equation of y = x^2 -4x + 4, you will get y = 0. QUADRATIC EQUATION WITH TWO ROOTS Following is the graph of a quadratic equation with two roots. {{{graph(300,300,-5,5,-5,5,x^2-4x+3)}}} The quadratic equation used for this graph is: y = x^2 -4x + 3 The roots for this equation are x = 1 and x = 3. If you plug x = 1 into the equation of y = x^2 -4x + 4, you will get y = 0 If you plug x = 3 into the equation of y = x^2 -4x + 4, you will also get y = 0. QUADRATIC EQUATION WITH 0 ROOTS Following is the graph of a quadratic equation with 0 roots. {{{graph(300,300,-5,5,-5,5,x^2-4x+5)}}} The quadratic equation used for this graph is: y = x^2 -4x + 5 The values of x for this equation when y = 0 are x = 2 + sqrt(-1) and x = 2 - sqrt(-12). If you plug x = 2 + sqrt(-1) into the equation of y = x^2 -4x + 4, you will get y = 0 If you plug x = 2 - sqrt(-1) into the equation of y = x^2 -4x + 4, you will also get y = 0, but these are not roots of the equation because the value of x is not real. The value of x is not real because it contains an imaginary part. The imaginary part is sqrt(-1) which is the square root of a negative number. The graph of this equation does not cross the x-axis, as can be seen. If you solve for the roots of a quadratic equation and you get an imaginary number or complex number for an answer, then you can assume that the equation has no roots. Graphing the equation will show you that the equation does not cross the x-axis. More information regarding imaginary or complex numbers can be found at the following websites. <a href = "http://www.purplemath.com/modules/complex.htm" target = "_blank">http://www.purplemath.com/modules/complex.htm</a> <a href = "http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut12_complexnum.htm" target = "_blank">http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut12_complexnum.htm</a> <a href = "http://tutorial.math.lamar.edu/Extras/ComplexPrimer/ComplexNumbers.aspx" target = "_blank">http://tutorial.math.lamar.edu/Extras/ComplexPrimer/ComplexNumbers.aspx</a> FINDING THE ROOTS OF A QUADRATIC EQUATION You can find the roots of a quadratic equation in several ways. Those are: 1: Factoring the Equation to get the roots. 2: Use the Quadratic Formula to get the roots. 3: Use the Complete the Squares method to get the roots. 4: Plot the equation on a graph and see where the graph of the equation crosses the x-axis. Selection 1 (Factoring the Quadratic Equation) can find the roots if they are integers, but this can be difficult and time consuming where they are not readily apparent. The method is very quick, however, for those cases where the roots are easy to find. It is recommended to try this method first. If it looks like the roots can't be found easily, then revert to either the Quadratic Formula or the Completing the Squares Method. Selection 2 (Quadratic Formula) will always find the roots, whether they are real or imaginary. It is recommended you commit this formula to memory as it is your fall back position if all else fails. Selection 3 (Completing the Square) should also always find the roots, whether they are real or imaginary. This method can be used to prove the quadratic formula works. It is not always, however, easier to use than the quadratic equation. It requires some additional logic versus just plugging in the numbers when using the quadratic formula. Use it if you like, but the fall back position should still be the quadratic formula, unless you are more comfortable using the completing the squares method. Selection 4 (Plot the Equation on a graph) can find the roots in all cases where they are real, although it can be difficult to pinpoint the exact value. It cannot, however, find the roots if they are imaginary. It is not recommended to use this method to find the roots. This method is, however, very useful to see if the roots are where you expect them to be. It can also show you why your equation does not have any real roots. This will usually be because the graph is above or below the x-axis and never crosses it. Easy roots can be spotted by factoring. If an examination of the equation does not lead to them right away, then use the quadratic formula or complete the squares method to get the roots. Because explaining how to find the roots by factoring takes so long (many examples), I'll give you the quadratic formula first, then follow that with the completing the square method, then follow that with finding the roots by factoring. QUADRATIC FORMULA The quadratic formula is given by the equation: {{{x = ((-b +- sqrt(b^2-4ac))/(2a))}}} where: a is the coefficient of the x^2 term. b is the coefficient of the x term. c is the constant factor. The quadratic equation must be in standard form of {{{ax^2 + bx + c = 0}}} before using the quadratic formula to find the roots. It is important that the equation is set to 0. Having the equation set to something other than 0 is not the standard form of the quadratic equation. Example: 5x^2 + 6x + 7 = 8 This is not in standard form. Subtract 8 from both sides of this to get: 5x^2 + 6x + 7 - 8 = 0 Combine like terms to get: 5x^2 + 6x - 1 = 0 The equation is now in standard form of: ax^2 + bx + c where: a = 5, b = 6, and c = -1 QUADRATIC FORMULA EXAMPLE 1 Find the roots of y = x^2 - 4x + 4 We set y = 0 to get: x^2 - 4x + 4 = 0 We determine that: a = 1 (coefficient of x^2 term) b = -4 (coefficient of x term) c = 4 (constant factor) We plug these values into the quadratic formula of {{{x = ((-b +- sqrt(b^2-4ac))/(2a))}}} to get: {{{x = ((-(-4) +- sqrt((-4)^2-4*(1)*(4)))/(2*(1)))}}} We simplify this equation to get: {{{x = (4 +- sqrt(16-16)/2)}}} This becomes: {{{x = ((4 +- 0)/2)}}} which becomes: {{{x = 2}}} Note that the expression within the square root sign became 0. this is an indication that this quadratic equation will have only one root. A graph of this equation looks like: {{{graph(200,200,-5,5,-5,5,x^2-4x+4)}}} Note that the graph visually confirms that the single root of x = 2 for this equation is correct. QUADRATIC FORMULA EXAMPLE 2 Find the roots of y = x^2 - 4x + 3 We set y = 0 to get: x^2 - 4x + 3 = 0 We determine that: a = 1 (coefficient of x^2 term) b = -4 (coefficient of x term) c = 3 (constant factor) We plug these values into the quadratic formula of {{{x = ((-b +- sqrt(b^2-4ac))/(2a))}}} to get: {{{x = ((-(-4) +- sqrt((-4)^2-4*(1)*(3)))/(2*(1)))}}} We simplify this equation to get: {{{x = (4 +- sqrt(16-12)/2)}}} This becomes: {{{x = ((4 +- sqrt(4))/2)}}} which becomes: {{{x = ((4 +- 2)/2)}}} which becomes: {{{x = 1}}} or {{{x = 3}}} Note that the expression within the square root sign became positive and was a perfect square (the square root of the expression is an integer). The fact that the expression within the square root sign was positive is an indication that you will have two real roots. The fact that the expression within the square root sign was a perfect square is an indication that those roots will be integers. A graph of this equation is shown below: {{{graph(200,200,-5,5,-5,5,x^2 - 4x + 3)}}} Note that the graph visually confirms that the roots of x = 1 or x = 3 are correct. QUADRATIC FORMULA EXAMPLE 3 Find the roots of y = x^2 - 4x + 5 We set y = 0 to get: x^2 - 4x + 5 = 0 We determine that: a = 1 (coefficient of x^2 term) b = -4 (coefficient of x term) c = 5 (constant factor) We plug these values into the quadratic formula of {{{x = ((-b +- sqrt(b^2-4ac))/(2a))}}} to get: {{{x = ((-(-4) +- sqrt((-4)^2-4*(1)*(5)))/(2*(1)))}}} We simplify this equation to get: {{{x = (4 +- sqrt(16-20)/2)}}} This becomes: {{{x = ((4 +- sqrt(-4))/2)}}} which becomes: {{{x = ((4 +- 2i)/2)}}} which becomes: {{{x = (2+i)}}} or {{{x = (2-i)}}} Note that the expression within the square root sign became negative and was a perfect square (the square root of the expression is an integer). The fact that the expression within the square root sign was negative is an indication that you will have two complex roots. Complex roots have a real part and an imaginary part. The fact that the expression within the square root sign was a perfect square is an indication that those roots will be integers. These roots are composed of integers (they do not contain a fractional part). They are not, however, real. The solution to the roots of this equation is that there are no roots because the graph of the quadratic equation never crosses the x-axis and the roots that have been calculated are not real. A graph of this equation is shown below: {{{graph(200,200,-5,5,-5,5,x^2 - 4x + 5)}}} Note that the graph confirms that this equation has no roots since the graph of the equation never crosses the x-axis. The roots of {{{x = (2+i)}}} or {{{x = (2-i)}}} that were calculated for this equation when we set the equation equal to 0 are not real. They are complex (contain a real part and an imaginary part). This is your indication that the equation has no real roots. COMPLETING THE SQUARE Completing the square is an alternate method for finding the roots of a quadratic equation. In some cases it's easier to complete the square. In other cases it's easier to use the quadratic formula. Most people use the quadratic formula. It is, however, both instructive and useful to learn how to complete the square as well. In order to complete the square, you need to transform your equation into a complete the square form if it is not already in that form. The standard form of the quadratic equation is: {{{ax^2 + bx + c = 0}}} We subtract c from both sides of the equation to get: {{{ax^2 + bx = -c}}} If a is equal to 1, then we are done. If a is not equal to 1, then we have to divide both sides of the equation by a to make the coefficient of x^2 equal to 1. Some references for learning how to find roots by completing the square are shown below: <a href = "http://www.jamesbrennan.org/algebra/quadratics/Completing%20the%20Square.htm" target = "_blank">James Brennan</a> <a href = "http://www.purplemath.com/modules/sqrquad.htm" target = "_blank">Purple Math</a> <a href = "http://www.themathpage.com/aPreCalc/complete-the-square.htm" target = "_blank">The Math Page</a> <a href = "http://www.sosmath.com/algebra/quadraticeq/complsquare/complsquare.html" target = "_blank">SOS Math</a> There are other references on the web that you can find by doing a search on "completing the square". That last reference (sosmath) is useful to show you the very close relationship between the completing the square method and the quadratic formula. COMPLETING THE SQUARE METHOD EXAMPLE 1 Find the roots of {{{y = x^2 - 4x + 5}}} by using the completing the square method. This is the same equation used for quadratic formula example 3. The roots will not be real. The answer should be {{{x = (2-i)}}} or {{{x = (2+i)}}}. We set y = 0 to get: {{{x^2 - 4x + 5}}} = 0 This is the standard form of the quadratic equation. a = coefficient of {{{x^2}}} = 1 b = coefficient of {{{x}}} = -4 c = constant factor = 5 we subtract 5 from both sides of the equation to get: {{{x^2 - 4x = -5}}} This is the standard form of the quadratic equation for finding the roots by using the completing the square method. Since a is already equal to 1, we do not have to divide both sides of the equation by a in order to make a = 1. We divide our b factor by 2 to get: {{{b/2 = -4/2 = -2}}} Our factor becomes (x-2) We square the halved b factor to get: {{{(b/2)^2 = (-4/2)^2 = (-2)^2 = 4}}} Our factored equation is adjusted by subtracting 4 from it. Our factored equation becomes: {{{(x-2)^2 - 4 = 5}}} What happened? Look at our equation of: {{{x^2 - 4x = -5}}} The {{{x^2 - 4x}}} part can be turned into a result of (x-b/2)^2 plus a constant factor. Since b = -4, b/2 = -2 and we derive the factor of {{{(x-2)^2}}}. We square this factor to get: {{{(x-2)^2}}} This is where the term completing the square comes from. When we multiply {{{(x-2)^2}}} however, we get {{{x^2 - 4x + 4}}} This is the same as our original equation of {{{(x^2 - 4x)}}} plus an additional constant factor of 4. To compensate for that, we subtract 4 from {{{(x-2)^2}}} to get {{{x^2 - 4x + 4 - 4}}} which makes it become {{{x^2 - 4x}}}. We have derived our original equation of {{{x^2 - 4x}}} by completing the square and making an adjustment to compensate for the additional constant. We started off with: {{{x^2 - 4x = -5}}} We derived: {{{(x-2)^2 - 4 = -5}}} The equation of {{{x^2 - 4x = -5}}} is equivalent to {{{(x-2)^2 - 4 = -5}}} Now that we derived: {{{(x-2)^2 - 4 = -5}}} We can solve for x. First we add 4 to both sides of this equation to get: {{{(x-2)^2 = -5 + 4}}} which becomes {{{(x-2)^2 = -1}}} We take the square root of both sides of this equation to get: {{{x-2}}} = +/- {{{sqrt (-1)}}} We add 2 to both sides of this equation to get: {{{x}}} = {{{2}}} +/- {{{sqrt(-1)}}} We replace {{{sqrt(-1)}}} with {{{i}}} because i represents {{{sqrt(-1)}}} Our solutions are: {{{x = (2+i)}}} or {{{x = (2-i)}}} This is the same solution as we got from the quadratic formula example 3. You will get the same solution by completing the square as you will get by using the quadratic formula. This is not unexpected since the quadratic formula is derived from the method of completing the square as indicated in at least one of the listed references for completing the square. As indicated previously, there are some complications with completing the square that make using it a bit more complicated than using the quadratic formula, but there are instances where completing the square can be easier than using the quadratic formula. These cases are when the a factor is already 1 and when the quadratic formula is in a form where completing the square can be accomplished with little additional effort. In our example, it was only a three step process before we took the square root of both sides of the equation and solved for x. Our original equation was {{{x^2 - 4x = -5}}}. First step we created our factor of {{{(x-2)}}} and squared it to get {{{(x-2)^2}}} = {{{x^2-4x+4}}}. Second step we subtracted 4 from it to get {{{(x-2)^2 - 4 = x^2-4x}}}. Third step we added 4 to both sides of the equation to get {{{(x-2)^2 = -5 + 4 = -1}}} Check the references for further information involving completing the square method for finding the roots of quadratic equations. MULTIPLYING POLYNOMIALS TOGETHER Each term in one of the factors needs to multiply each term in the other factor once and only once. Here's how that works. You take each term in the first factor and multiply it by each term in the second factor. Work from left to right or right to left is ok but be consistent. Keep each term separate until you're done and then add them together and combine like terms. In this example there will be no like terms because I identified them uniquely to show you the process. (o + p) * (t + u) equals: o*t = ot o*u = ou p*t = pt p*u = pu Add these together and combine like terms to get: ot + ou + pt + pu You'll see this in operation below when we multiply out the factors to confirm that the factors are correct. FOIL If you are familiar with FOIL (First, Outer, Inner, Last), then what I have just described above is simply an alternate, more generalized method, to achieve the same result. Multiply these factors together using FOIL: (o + p) * (t + u) F in FOIL means First. You multiply the first terms in each factor together to get: o*t = ot O in FOIL means Outer. You multiply the outer terms in each factor together to get: o*u = ou I in FOIL means Inner. You multiply the inner terms in each factor together to get: p*t = pt L in FOIL means Last. You multiply the last terms in each factor together to get: p*u = pu You add all these together and combine like terms to get: ot + ou + pt + pu The results from the generalized method above was: ot + ou + pt + pu These resuls are identical. FOIL and the Generalized Method will provide the same results. The difference is that the generalized method can be applied to multiplying expressions that contain more variables within them, such as (m+n) * (o+p+q+r). FOIL doesn't work with these. FOIL is specific to forms like (m+n)*(o+p). FACTORING QUADRATIC EQUATIONS Factoring involves looking at the quadratic equation and figuring out what the roots are without resorting to the quadratic formula or the completing the squares method. This is the quickest method for those quadratic equations that easily lend themselves to this method. It is recommended to try this method first, but if the factors do not magically appear after a short period of investigation, it is then recommended to go to the quadratic formula or the completing the square method. There is no necessity to find the roots of a quadratic equation by factoring unless the exercise specifically calls for that method to be used. You must, at least, remember the quadratic formula. Impress that in your mind until it becomes second nature to recall it. The Standard Form of the Quadratic Equation is: ax^2 + bx + c = 0 When you multiply your factors together, you have to make sure you take one of the factors and multiply each of the terms in the other factor exactly once for each term. This was shown above and will be repeated here. (o + p) * (t + u) equals: o*t = ot (first term in first expression is multiplied by first term in second expression) o*u = ou (first term in first expression is multiplied by second term in second expression) p*t = pt (second term in first expression is multiplied by first term in second expression) p*u = pu (second term in first expression is multiplied by second term in second expression) Add these together and combine like terms to get: ot + ou + pt + pu Of course, you don't work with (o + p) * (t + u) You work with something more like: (5x + 7) * (3x + 2) = 0 5 is a coefficient of x 3 is a coefficient of x 7 and 2 are constants To multiply these factors together, you would follow the rules of multiplying each term in one of the factors by each term in the other factor exactly one time and then adding the results together and combining the results. Following that rule, we would get: (5x + 7) * (3x + 2) equals: 5x*3x = 15x^2 (first term in first expression is multiplied by first term in second expression) 5x*2 = 10x (first term in first expression is multiplied by second term in second expression) 7*3x = 21x (second term in first expression is multiplied by first term in second expression) 7*2 = 14 (second term in first expression is multiplied by second term in second expression) You add these up together and combine like terms to get: 15x^2 + 31x + 14 Set this equation equal to 0 and you get: 15x^2 + 31x + 14 = 0 This equation is now in standard form of: ax^2 + bx + c You can see that: a = 15 which is a product of 5*3 c = 14 which is a product of 7*2 b = 31 which is the sum of 5*2 + 7*3 Look at your quadratic equation and your factors again. Standard form of the quadratic equation equals ax^2 + bx + c Quadratic Equation equals 15x^2 + 31x + 14 Factors equal (5x + 7) * (3x + 2) Your a factor in the quadratic equation equals 15 which is the product of the left coefficient in the first factor multiplied by the left coefficient in the second factor. Your c factor in the quadratic equation equals 14 which is the product of the right constant in the first factor multiplied by the right constant in the second factor. Your b factor in the quadratic equation equals 31 which is the product of the left coefficient in the first factor multiplied by the right constant in the second factor plus the left coefficient in the second factor multiplied by the right constant in the first factor. The standard form for the factors to the quadratic equation is: (dx+e) * (fx+g) = 0 If we multiply these factors out we get: dx*fx = df*x^2 dx*g = dg*x e*fx = ef*x e*g = e*g Add these together and combine like terms to get: df*x^2 + (dg+ef)*x + e*g = 0 Standard form of the quadratic equation is: ax^2 + bx + c = 0 a = df in the factors to this quadratic equation. (df = d*f) b = dg + ef in the factors to this quadratic equation. (dg = d*g; ef = e*f) c = eg in the factors to this quadratic equation. (eg = e*g) In the equation we just factored: Quadratic Equation equals 15x^2 + 31x + 14 = 0 The standard form of the quadratic equation equals ax^2 + bx + c = 0 where: a = 15 b = 31 c = 14 Factors equal (5x + 7) * (3x + 2) = 0 The standard form of the factors to the quadratic equation equals (dx+e) * (fx+g) = 0 where: d = 5 e = 7 f = 3 g = 2 a = 15 = d*f = 5*3 = 15 c = 14 = e*g = 7*2 = 14 b = 31 = d*g + e*f = 5*2 + 7*3 = 10 + 21 = 31 d*f can be shown as df e*g can be shown as eg d*g + e*f can be shown as dg+ef d and f are the two left most terms in (dx+e) * (fx+g) = 0 e and g are the two rightmost terms in (dx+e) * (fx+g) = 0 d*g are the two outer terms in (dx+e) * (fx+g) = 0 e*f are the two inner terms in (dx+e) * (fx+g) = 0 If the equation can be factored, you will be able to find your factors this way eventually. Sometimes it's easy and sometimes it's hard. What can make it frustrating is that you can't always find the answer this way because the quadratic equation can't always be factored. Sometimes you may have trouble finding it even if it can be factored. That's why it is recommended to try factoring first. If the answer doesn't appear right away, use the quadratic formula or the completing the squares method. The only time you have to find the roots by factoring is if the exercise specifically requires that you find them that way. Otherwise, use the method that you are most familiar and comfortable with. Some examples will help clarify the method of finding the roots using the factoring method. I'll work from simple to complex so you can see the process develop as we go along. FACTORING EXAMPLE 1 Find the roots of x^2 + 2x = -1 Before you start, you need to get the quadratic formula into the standard form for finding the roots of ax^2 + bx + c = 0 add 1 to both sides of this equation to get: x^2 + 2x + 1 = 0 It is now in the standard form of ax^2 + bx + c = 0 The equation we want to factor is x^2 + 2x + 1 = 0 where: a = 1 b = 2 c = 1 Standard form of the factors for this quadratic equation is: (dx+e) * (fx+g) = 0 where: a = d*f = 1 b = d*g + e*f = 2 c = e*g = 1 d and f are the two left most terms in (dx+e) * (fx+g) = 0 e and g are the two rightmost terms in (dx+e) * (fx+g) = 0 d*g are the two outer terms in (dx+e) * (fx+g) = 0 e*f are the two inner terms in (dx+e) * (fx+g) = 0 You know that a will be a product of d*f. Possible values for d*f are: 1*1 Minus times a minus to get a plus is not considered for the a term. When d = 1, f = 1 You know that c = e*g = 1 Possible values for e*g are: 1*1 (-1)*(-1) When e = 1, g = 1 When e = -1, g = -1 We plug these values into the general form of the factors to this equation of (dx+e) * (fx+g) = 0 to get possible factors of: (x+1)*(x+1) = 0 (x-1)*(x-1) = 0 We know that b = d*g + e*f = 2 If d and f are both equal to 1, this reduces to g+e = 2 g+e are the two right terms in (dx+e) * (fx+g) = 0 Since the first set of factors will give us e+g = 2, we choose (x+1)*(x+1) = 0 as our possible set of factors for this equation. We multiply (x+1)*(x+1) by each other to get: x*x = x^2 x*1 = x 1*x = x 1*1 = 1 Remember that each term in one factor has to be multiplied by each term in the other factor exactly once. Working from left to right, I took each term in the first factor and multiplied it by each term in the second factor. We add these together and combine like terms to get: x^2 + 2x + 1 = 0 Since our equation to be factored is identical to this, we solve for x and confirm the results and we are done. x+1 = 0 results in x = -1. x+1 = 0 results in x = -1 again. Since the results are the same, they are combined to get one possible root of: x = -1 Substitute x = -1 into the original quadratic equation to get: x^2 + 2x + 1 = (-1)^2 + 2(-1) + 1 = 1 - 2 + 1 = 0 confirming x = -1 is good. Your answer is that the quadratic equation of x^2 + 2x + 1 = 0 has one root and it is x = -1. A graph of this equation looks like: {{{graph(200,200,-5,5,-5,5,x^2+2x+1)}}} The graph confirms that the one root of x = -1 is accurate. That is where the graph of the equation touches or passes through the x-axis. FACTORING EXAMPLE 2 Find the roots of x^2 + x - 6 by factoring. We get this equation into standard form of ax^2 + bx + c = 0 by setting it equal to 0. Equation becomes x^2 + x - 6 = 0 It is now in the standard form of ax^2 + bx + c = 0 a = 1 b = 1 c = -6 Standard form of the factors for this quadratic equation is: (dx+e) * (fx+g) = 0 where: a = d*f = 1 b = d*g + e*f = 1 c = e*g = -6 d and f are the two left most terms in (dx+e) * (fx+g) = 0 e and g are the two rightmost terms in (dx+e) * (fx+g) = 0 d*g are the two outer terms in (dx+e) * (fx+g) = 0 e*f are the two inner terms in (dx+e) * (fx+g) = 0 Possible factors for a = d*f = 1 are: 1*1 Minus times a minus to get a plus is not considered for the a term. When d = 1, f = 1 Possible factors for c = e*g = -6 are: 1*(-6) 2*(-3) 3*(-2) 6*(-1) When e = 1, g = -6 When e = 2, g = -3 When e = 3, g = -2 When e = 6, g = -1 Possible factors for this equation are: (x+1) * (x-6) (x+2) * (x-3) (x+3) * (x-2) (x+6) * (x-1) We know that b = d*g + e*f = 1 We also know that when d and f are both equal to 1 that this reduces to b = e+g We are looking for a set of factors that will make b = e+g = 1 g+e are the two right terms in (dx+e) * (fx+g) = 0 First set of factors yields b = e+g = 1-6 = -5 Second set of factors yields b = e+g = 2-3 = -1 Third set of factors yields b = e+g = 1-2 = 1 Third set of factors is the one that we want. Our possible factors to this equation are: (x+3) * (x-2) = 0 We multiply these factors together to get: x*x = x^2 x*(-2) = -2x 3*x = 3x 3*(-2) = -6 We add these together and combine like terms to get: x^2 + x - 6 = 0 Since our equation to be factored is identical to this, we solve for x and confirm the results and we are done. Solving for x, we get: x+3 = 0 results in x = -3 x-2 = 0 results in x = 2 Our possible roots for this quadratic equation are: x = -3 or x = 2 Plugging -3 into the original equation gets: x^2 + x - 6 = (-3)^2 + (-3) - 6 = 9 - 3 - 6 = 0 confirming that x = -3 is good. Plugging 2 into the original equation gets: x^2 + x -6 = 2^2 + 2 - 6 = 4 + 2 - 6 = 0 confirming that x = 2 is also good. Our answer is that the quadratic equation of x^2 + x - 6 has two roots and they are x = -3 and x = 2 A graph of this equation looks like this: {{{graph(200,200,-5,5,-5,5,x^2+x-6)}}} The graph confirms that the roots of x = -3 and x = 2 are correct. FACTORING EXAMPLE 3 Factor x^2 = 9x - 18 This needs to be put into standard form. Subtract 9x and add 18 to both sides of this equation to get: x^2 - 9x + 18 = 0 The equation is now in standard form of: ax^2 + bx + c = 0 The equation we want to factor is: x^2 - 9x + 18 = 0 where: a = 1 b = -9 c = 18 Standard form of the factors for this quadratic equation is: (dx+e) * (fx+g) = 0 where: a = d*f = 1 b = d*g + e*f = -9 c = e*g = 18 d and f are the two left most terms in (dx+e) * (fx+g) = 0 e and g are the two rightmost terms in (dx+e) * (fx+g) = 0 d*g are the two outer terms in (dx+e) * (fx+g) = 0 e*f are the two inner terms in (dx+e) * (fx+g) = 0 We know that a = d*f = 1 Possible values for d*f are: 1*1 Minus times a minus to get a plus is not considered for the a term. When d = 1, f = 1 We know that c = e*g = 18 We get: e*g can be any of: 1*18 2*9 3*6 6*3 9*2 18*1 (-1)*(-18) (-2)*(-9) (-3)*(-6) (-6)*(-3) (-9)*(-2) (-18)*(-1) When e = 1, g = 18 When e = 2, g = 9 When e = 3, g = 6 When e = 6, g = 3 When e = 9, g = 2 When e = 18, g = 1 When e = (-1), g = (-18) When e = (-2), g = (-9) When e = (-3), g = (-6) When e = (-6), g = (-3) When e = (-9), g = (-2) When e = (-18), g = (-1) Our possible factors for this equation are: (x+1) * (x+18) = 0 (x+2) * (x+9) = 0 (x+3) * (x+6) = 0 (x+6) * (x+3) = 0 (x+9) * (x+2) = 0 (x+18) * (x+1) = 0 (x-1) * (x-18) = 0 (x-2) * (x-9) = 0 (x-3) * (x-6) = 0 (x-6) * (x-3) = 0 (x-9) * (x-2) = 0 (x-18) * (x-1) = 0 We know that b = d*g + e*f = -9 We also know that when d and f are both equal to 1 that this reduces to b = e+g We are looking for a set of factors where b = e+g = -9 g+e are the two right terms in (dx+e) * (fx+g) = 0 The first set of factors listed yields b = e+g = 1+18 = 19 The second set of factors listed yields b = e+g = 2+9 = 11 The third set of factors listed yields b = e+g = 3+6 9 the third set yields the right answer but the wrong sign. We immediately look for its negative counterpart. That one is the ninth set of factors listed. The ninth set of factors listed yields b = e+g = (-3)+(-6) = -9 This is the set of factors that we want. The possible factors for this equation are: (x-3) * (x-6) = 0 We multiply (x-3) * (x-6) together to get: x*x = x^2 x*(-6) = -6x (-3)*x = -3x (-3)*(-6) = 18 We add these together and combine like terms to get: x^2 - 9x + 18 = 0 Since this is identical to our equation to be factored, then we solve for x and confirm our results and we are done. The factors are: (x-3)*(x-6) = 0 x-3 = 0 results in x = 3 x-6 = 0 results in x = 6 Our possible roots for this quadratic equation are: x = 3 x = 6 Substitute 3 for x in the original quadratic equation gets: x^2 - 9x + 18 = 3^2 - 9*3 + 18 = 9 - 27 + 18 = 27 - 27 = 0 confirming x = 3 is good. Substitute 6 for x in the original quadratic equation gets: x^2 - 9x + 18 = 6^2 - 9*6 + 18 = 36 - 54 + 18 = 54 - 54 = 0 confirming x = 6 is good. Your answer is that the quadratic equation of x^2 - 9x + 18 = 0 has two roots and they are x = 3 and x = 6 A graph of this equation looks like this: {{{graph(200,200,-7,7,-7,7,x^2 - 9x + 18)}}} The graph confirms that the roots we have calculated are correct. The roots are where the graph of the equation crosses the x-axis. FACTORING EXAMPLE 4 Factor x^2 - 3x - 64 = 10 You need to convert this equation into standard form of ax^2 + bx + c = 0 subtract 10 from both sides of the equation to get: x^2 - 3x - 54 = 0 It is now in standard form. Equation we want to factor is x^2 - 3x - 54 = 0 where: a = 1 b = -3 c = -54 Standard form of the factors for this quadratic equation is: (dx+e) * (fx+g) = 0 where: a = d*f = 1 b = d*g + e*f = -3 c = e*g = -54 d and f are the two left most terms in (dx+e) * (fx+g) = 0 e and g are the two rightmost terms in (dx+e) * (fx+g) = 0 d*g are the two outer terms in (dx+e) * (fx+g) = 0 e*f are the two inner terms in (dx+e) * (fx+g) = 0 We know that a = d*f = 1 Our possible factors to get a = d*f = 1 are: 1*1 Minus times a minus to get a plus is not considered for the a term. When d = 1, f = 1 We know that c = e*g = -54 Our possible factors to get c = e*g = -54 are: 1*(-54) 2*(-27) 3*(-18) 6*(-9) 9*(-6) 18*(-3) 27*(-2) 54*(-1) When e = 1, g = -54 When e = 2, g = -27 When e = 3, g = -18 When e = 6, g = -9 When e = 9, g = -6 When e = 18, g = -3 When e = 27, g = -2 When e = 54, g = -1 Our possible set of factors for this equation are: (x+1)*(x-54) = 0 (x+2)*(x-27) = 0 (x+3)*(x-18) = 0 (x+6)*(x-9) = 0 (x+9)*(x-6) = 0 (x+18)*(x-3) = 0 (x+27)*(x-2) = 0 (x+54)*(x-1) = 0 We know that b = d*g + e*f = -3 We also know that when d and f are both 1 that this becomes g+e = -3 g+e are the two right terms in (dx+e) * (fx+g) = 0 We are looking for a set of factors that will result in b = e+g = -3 We go down the list until we reach the fourth set of factors listed. That set of factors is: (x+6)*(x-9) = 0 That set of factors yields b = e+g = 6-9 = -3 That is the set of factors that we want. The possible factors for this quadratic equation are: (x+6) * (x-9) = 0 We multiply (x+6) * (x-9) by each other to get: x*x = x^2 x*(-9) = -9x 6*x = 6x 6*(-9)= -54 We add these together and combine like terms to get: x^2 - 3x - 54 = 0 Since the equation we want to factor is identical to this, we solve for x and confirm the results and we are done. x+6 results in x = -6. x-9 results in x = 9. Possible roots for this equation are: x = -6 x = 9 Substituting -6 for x in the original quadratic equation gets: x^2 - 3x - 54 = (-6)^2 - 3(-6) - 54 = 36 + 18 - 54 = 54 - 54 = 0 confirming x = -6 is good. Substituting 9 for x in the original quadratic equation gets: x^2 - 3x - 54 = 9^2 - 3*9 - 54 = 81 - 27 - 54 = 54 - 54 = 0 confirming x = 9 is good. Your answer is: Quadratic Equation of x^2 - 3x - 54 = 0 has two roots and they are x = -6 and x = 9 Graph of this equation is shown below: {{{graph(200,200,-12,12,-200,200,x^2-3x-54)}}} The graph confirms that the roots we calculated are correct. FACTORING EXAMPLE 5 Factor 3x^2 + 18x - 21 Here the a factor looks like it is not 1. The equation, however, has a common factor of 3 that can be used to reduce it into a simpler form. That simpler form becomes: x^2 + 6x - 7 Always try to reduce the equation to its simplest form as much as possible. It makes processing it easier. You could solve this equation without simplifying it but it would be more difficult because then you would be dealing with an equation that has an a factor not equal to 1 which is harder to solve than an equation that has an a factor equal to 1 because you could have a larger number of possible sets of factors to look at, and your b = d*g + e*f would not reduce to g+e as it does when a and f are both equal to 1. We set this equation to standard form of: x^2 + bx + c by making it equal to 0 to get: x^2 + 6x - 7 = 0 Equation we want to factor is: x^2 + 6x - 7 = 0 where: x = 1 b = 6 c = -7 Standard form of the factors for this quadratic equation is: (dx+e) * (fx+g) = 0 where: a = d*f = 1 b = d*g + e*f = 6 c = e*g = -7 d and f are the two left most terms in (dx+e) * (fx+g) = 0 e and g are the two rightmost terms in (dx+e) * (fx+g) = 0 d*g are the two outer terms in (dx+e) * (fx+g) = 0 e*f are the two inner terms in (dx+e) * (fx+g) = 0 We know that c = e*g = -7 Possible factors to get -7 are: 1*(-7) 7*(-1) When e = 1, g = -7 When e = 7, g = -1 We know that a = d*f = 1 Possible factors to get 1 are: 1*1 Minus times a minus to get a plus is not considered for the a term. When d = 1, f = 1 Our possible factors are: (x+1)*(x-7) = 0 (x+7)*(x-1) = 0 We know that b = d*g + e*f = 6 We also know that when d = 1 and f = 1 that this reduces to g+e = 6 We are looking for a set of factors that will make b = g+e = 6 g+e are the two right terms in (dx+e) * (fx+g) = 0 First set of factors yields b = 1-7 = -6. Second set of factors yields b = 7-1 = 6 Second set of factors is the one we want. Our possible factors for this quadratic equation are: (x+7) * (x-1) = 0 We multiply (x+7) * (x-1) by each other to get: x*x = x^2 x*(-1) = -x 7*x = 7x 7*(-1) = -7 We add these together and combine like terms to get: x^2 + 6x - 7 = 0 Since x^2 + 6x - 7 = 0 is identical to the equation we want to factor, then we solve for x and confirm the results and we are done. The factors are: (x+7) * (x-1) = 0 x+7 = 0 results in x = -7 x-1 = 0 results in x = 1 The possible roots for this quadratic equation are: x = -7 x = 1 Substitute 1 for x in the quadratic equation to get: x^2 + 6x - 7 = 1^2 + 6*1 - 7 = 1 + 6 - 7 = 7 - 7 = 0 confirming x = 1 is good. Substitute (-7) for x in the quadratic equation to get: x^2 + 6x - 7 = (-7)^2 + 6*(-7) - 7 = 49 - 42 - 7 = 49 - 49 = 0 confirming x = -7 is good. Your answer is: The quadratic equation of x^2 + 6x - 7 = 0 has two roots and they are x = 1 and x = -7. a graph of this equation looks like this: {{{graph (200,200,-9,9,-19,19,x^2+6x-7)}}} The graph confirms that the roots we have calculated are correct. FACTORING EXAMPLE 6 Occasionally, you might get a request to factor an equation like: -x^2 -6x + 7 You can factor it this way or you can convert it so the x^2 term is positive. If the x^2 term is positive and both d and f equal to 1, then solving by factoring is easier since you don't need to consider the a term nor its signs and your middle term of b = d*g + e*f reduces to g+e. First you convert this equation to standard form by setting it equal to 0. -x^2 - 6x + 7 = 0 To make the x^2 term positive, you simply multiply both sides of this equation by (-1) to get: x^2 + 6x - 7 = 0 Since we already factored x^2 + 6x - 7 in example 5, we will factor -x^2 -6x + 7 = 0 in this example to show you the difference between this example and example 5. Equation to factor is -x^2 -6x + 7 = 0 where: a = -1 b = -6 c = 7 Standard form of the factors for this quadratic equation is: (dx+e) * (fx+g) = 0 where: a = d*f = -1 b = d*g + e*f = -6 c = e*g = 7 d and f are the two left most terms in (dx+e) * (fx+g) = 0 e and g are the two rightmost terms in (dx+e) * (fx+g) = 0 d*g are the two outer terms in (dx+e) * (fx+g) = 0 e*f are the two inner terms in (dx+e) * (fx+g) = 0 Since we know that a = d*f = -1 then possible factors to get a = d*f = -1 are: 1*(-1) (-1)*1 When d = 1, f = -1 When d = -1, f = 1 Since we know that c = e*g = 7 then possible factors to get c = e*g = 7 are: 1*7 7*1 When e = 1, g = 7 When e = 7, g = 1 Our possible factors for this equation are: (x+1)*(-x+7) = 0 (x+7)*(-x+1) = 0 (-x+1)*(x+7) = 0 (-x+7)*(x+1) = 0 We know that b = d*g + e*f = -6 Since d and f are NOT both equal to 1, we cannot reduce this to e+g = -6 We need to examine the full factors of b = d*g + e*f = -6 d*g are the two outer terms in (dx+e)*(fx+g) = 0 e*f are the two inner terms in (dx+e)*(fx+g) = 0 The first set of factors gives us d*g + e*f equals 1*7 + 1*(-1) = 7 - 1 = 6 The second set of factors give us d*g + e*f equals 1*1 + 7*(-1) = 1 - 7 = -6 The second set of factors is the one that we want. Our possible factors for this quadratic equation are: (x+7)*(-x+1) = 0 We multiply (x+7)*(-x+1) by each other to get: x*(-x) = -x^2 x*1 = x 7*(-x) = -7x 7*1 = 7 We add these together and combine like terms to get: -x^2 -6x + 7 = 0 Since -x^2 -6x + 7 = 0 is identical to the equation we want to factor, we solve for x and confirm the results and we are done. The factors are: (-x+1)*(x+7) = 0 -x+1 = 0 results in -x = -1 multiply both sides of this equation by (-1) to get: x = 1 x+7 = 0 results in x = -7 The possible roots for this quadratic equation are: x = 1 or x = -7 This answer is the same as the answer in example 5. In example 5 we factored x^2 + 6x - 7 = 0 to get x = 1 or x = -7 In this example we factored -x^2 - 6x + 7 = 0 to get x = 1 or x = -7 We got the same answer whether we factored x^2 + 6x - 7 = 0 or whether we factored -x^2 - 6x + 7 = 0 The bottom line is: If you get an equation such as -x^2 + bx + c = 0, you can multiply both sides of the equation by -1 to get an equivalent equation of x^2 -bx - c = 0 and solve for that, or you can leave the equation as is and solve it with the x^2 term being negative. Either way will get you the same answer. You have a choice. This is because -ax^2 - bx - c = 0 is equivalent to ax^2 + bx + c = 0. If you multiply both sides of either equation by (-1), you will get the other. FACTORING EXAMPLE 7 Find the roots of the equation 2x^2 + 7x - 4 by using the factoring method. Set the equation to standard form by setting it equal to 0. Equation becomes: 2x^2 + 7x - 4 = 0 Equation we want to factor is: 2x^2 + 7x - 4 = 0 where: a = 2 b = 7 c = -4 Standard form of the factors for this quadratic equation is: (dx+e) * (fx+g) = 0 where: a = d*f = 2 b = d*g + e*f = 7 c = e*g = -4 d and f are the two left most terms in (dx+e) * (fx+g) = 0 e and g are the two rightmost terms in (dx+e) * (fx+g) = 0 d*g are the two outer terms in (dx+e) * (fx+g) = 0 e*f are the two inner terms in (dx+e) * (fx+g) = 0 Since we know that c = e*g = -4, then possible combination of factors to get c = e*g = -4 are: 1 * (-4) 2 * (-2) 4 * (-1) When e = 1, g = -4 When e = 2, g = -2 When e = 4, g = -1 Since we know that a = d*f = 2, then possible combination of factors to get a = d*f = 2 are: 1 * 2 2 * 1 When d = 1, f = 2 When d = 2, f = 1 Our possible factors for this equation are: (x+1)*(2x-4) (x+2)*(2x-2) (x+4)*(2x-1) (2x+1)*(x-4) (2x+2)*(x-2) (2x+4)*(x-1) We know that b = d*g + e*f = 7 d and f are not both equal to 1, so we can't reduce this equation any further. We are looking for a set of factors that will make b = d*g + e*f = 7 d*g are the outer terms in (dx+e) * (fx+g) = 0 e*f are the inner terms in (dx+e) * (fx+g) = 0 The first set of factors yields b = d*g+e*f = 1*(-4) + 1*2 = -4 + 2 = -2 The second set of factors yields b = d*g+e*f = 1*(-2) + 2*2 = -2 + 4 = 2 The third set of factors yields b = d*g+e*f = 1*(-1) + 4*2 = -1 + 8 = 7 The third set of factors is the one that we want. Our possible factors to this quadratic equation are: (x+4) * (2x-1) = 0 We multiply (x+4) * (2x-1) by each other to get: x*2x = 2x^2 x*(-1) = -x 4*2x = 8x 4*(-1) = -4 Add these together and combine like terms to get: 2x^2 + 7x - 4 = 0 2x^2 + 7x - 4 = 0 is identical to the equation we want to factor so we solve for x and confirm the results and we are done. 2x-1 = 0 results in x = 1/2 x+4 = 0 results in x = -4 Possible roots for this quadratic equation are: x = 1/2 x = -4 Plug (1/2) for x in the original equation and you get: 2x^2 + 7x - 4 = 2*(1/2)^2 + 7*(1/2) - 4 = (1/2) + (7/2) - 4 = (8/2) - 4 = 4-4 = 0 confirming x = 1/2 is good. Plug -4 for x in the original equation and you get: 2x^2 + 7x - 4 = 2*(-4)^2 + 7*(-4) -4 = 32 + (-28) - 4 = 4-4 = 0 confirming x = -4 is good. FACTORING EXAMPLE 8 Find the roots of 10x^2 - 12x - 64 by factoring. Place into standard form by setting it equal to 0 to get: 10x^2 - 12x - 64 = 0 Divide by 2 to get: 5x^2 - 6x - 32 = 0 There is no other common factor so this is equation to be factored. Equation to be factored is 5x^2 - 6x - 32 = 0 where: a = 5 b = -6 c = -32 Standard form of the factors for this quadratic equation is: (dx+e) * (fx+g) = 0 where: a = d*f = 5 b = d*g + e*f = -6 c = e*g = -32 d and f are the two left most terms in (dx+e) * (fx+g) = 0 e and g are the two rightmost terms in (dx+e) * (fx+g) = 0 d*g are the two outer terms in (dx+e) * (fx+g) = 0 e*f are the two inner terms in (dx+e) * (fx+g) = 0 c = e*g Combination of Factors to get the c term of 32 can be: 1 * 32 2 * 16 4 * 8 8 * 4 16 * 2 32 * 1 When e = 1, g = 32 When e = 2, g = 16 When e = 4, g = 8 When e = 8, g = 4 When e = 16, g = 2 When e = 32, g = 1 a = d*f Combination of factors to get the a term of 5 can be: 1*5 5*1 When d = 1, f = 5 When d = 5, f = 1 The possible factors for this quadratic equation are: (x-1) * (5x+32) (x-2) * (5x+16) (x-4) * (5x+8) (x-8) * (5x+4) (x-16) * (5x+2) (x-32) * (5x+1) (5x-1) * (x+32) (5x-2) * (x+16) (5x-4) * (x+8) (5x-8) * (x+4) (5x-16) * (x+2) (5x-32) * (x+1) We know that b = d*g+e*f = -6 d*g are the two outer terms in (dx+e)*(fx+g) = 0 e*f are the two inner terms in (dx+e)*(fx+g) = 0 We look for a combination of d*g + e*f that will yield -6 The second set of factors of (x-2) * (5x+16) yields b = d*g + e*f = 1*16 + (-2)*5 = 16 - 10 = 6 This is the right number but the wrong sign. We immediately look for the combination of factors of (x+2) * (5x-16) because this would provide us with b = d*g + e*f = 1*(-16) + 2*5 = -16 + 10 = -6 The eleventh set of factors of (5x-16) * (x+2) is the same as (x+2) * (5x-16) so this is the one that we want. Our possible factors for this quadratic equation are: (5x-16) * (x+2) = 0 We multiply (5x-16) * (x+2) by each other to get: 5x*x = 5x^2 5x*2 = 10x (-16)*x = -16x (-16)*2 = -32 We add these together and combine like terms to get: 5x^2 - 6x - 32 = 0 Since 5x^2 - 6x - 32 = 0 is identical to the quadratic equation we are trying to factor, we solve for x and confirm the results and we are done. Our factors are: (5x-16) * (x+2) = 0 5x-16 = 0 becomes 5x = 16 which results in x = 16/5. x+2 = 0 results in x = -2 Possible roots for this quadratic equation are: x = 16/5 x = -2 Substitute 16/5 for x in the original quadratic equation to get: 5x^2 - 6x - 32 = 5*(16/5)^2 - 6*(16/5) - 32 = 51.2 - 19.2 - 32 = 32 - 32 = 0 confirming x = 16/2 is good. Substitute -2 for x in the original quadratic equation to get: 5x^2 - 6x - 32 = 5*(-2)^2 - 6*(-2) - 32 = 20 + 12 - 32 = 32 - 32 = 0 confirming x = -2 is good. Your answer is: The quadratic equation of 5x^2 - 6x - 32 = 0 has two roots and they are x = 16/5 and x = -2. FACTORING EXAMPLE 9 Factor 15x^2 + 11x - 12 = 0 Since this is already in standard form of ax^2 + bx + c = 0, we don't have to do anything. This equation cannot be simplified any further because there are no common factors that we can divide into each term to get a simplified expression. Our equation to be factored is 15x^2 + 11x - 12 = 0 where: a = 15 b = 11 c = -12 Standard form of the factors for this quadratic equation is: (dx+e) * (fx+g) = 0 where: a = d*f = 15 b = d*g + e*f = 11 c = e*g = -12 d and f are the two left most terms in (dx+e) * (fx+g) = 0 e and g are the two rightmost terms in (dx+e) * (fx+g) = 0 d*g are the two outer terms in (dx+e) * (fx+g) = 0 e*f are the two inner terms in (dx+e) * (fx+g) = 0 Possible factors to get a = d*f = 15 are: 1*15 3*5 5*3 15*1 When d = 1, f = 15 When d = 3, f = 5 When d = 5, f = 3 When d = 15, f = 1 Possible factors to get c = e*g = -12 are: 1*(-12) 2*(-6) 3*(-4) 4*(-3) 6*(-2) 12*(-1) When c = 1, g = -12 When c = 2, g = -6 When c = 3, g = -4 When c = 4, g = -3 When c = 6, g = -2 When c = 12, g = -1 Possible factors for this equation are: (x+1)*(15x-12) (x+2)*(15x-6) (x+3)*(15x-4) (x+4)*(15x-3) (x+6)*(15x-2) (x+12)*(15x-1) (3x+1)*(5x-12) (3x+2)*(5x-6) (3x+3)*(5x-4) (3x+4)*(5x-3) (3x+6)*(5x-2) (3x+12)*(5x-1) (5x+1)*(3x-12) (5x+2)*(3x-6) (5x+3)*(3x-4) (5x+4)*(3x-3) (5x+6)*(3x-2) (5x+12)*(3x-1) (15x+1)*(x-12) (15x+2)*(x-6) (15x+3)*(x-4) (15x+4)*(x-3) (15x+6)*(x-2) (15x+12)*(x-1) We know that b = d*g+e*f = 11 d*g are the two outer terms in (dx+e)*(fx+g) = 0 e*f are the two inner terms in (dx+e)*(fx+g) = 0 We look for a combination of d*g+e*f that equals 11. We know that 11 is positive. We know that the combination will be a plus and a minus We know that we need a larger plus and a smaller minus in order to get the result to be plus. We know that d*g will be minus. This is because all the possible d terms are positive and all the possible g terms are negative. The factors were purposely listed this way to keep positive and negative separate as much as possible which allows this type of analysis to occur. All the possible d terms are in the left hand side of the first factor. All the possible g terms are in the right hand side of the second factor. We know that e*f will be positive. This is because all the possible e terms are positive and all the possible f terms are positive. All the possible e terms are in the right hand side of the first factor. All the possible f terms are in the left hand side of the second factor. When d = 15 (left term in the first factor), we can see that the answer will be negative so we reject this. When f = 15 (left term in the second factor), we can see that the answer will be too positive so we reject this. We started checking each set of factors when d = 3 We are looking for a set of factors that will provide b = d*g + e*f = 11 (3x+1)*(5x-12) yields b = d*g+e*f = 3*(-12) + 1*5 = -24 + 5 = -19 (3x+2)*(5x-6) yields b = d*g+e*f = 3*(-6) + 2*5 = -18 + 10 = -8 (3x+3)*(5x-4) yields b = d*g+e*f = 3*(-4) + 3*5 = -12 + 15 = 3 (3x+4)*(5x-3) yields b = d*g+e*f = 3*(-3) + 4*5 = -9 + 20 = 11 (3x+4)*(5x-3) is the set of factors we are looking for. Possible factors for this equation are: (3x+4) * (5x-3) = 0 We multiply (3x+4) * (5x-3) by each other to get: 3x*5x = 15x^2 3x*(-3) = -9x 4*5x = 20x 4*(-3) = -12 We add these together and combine like terms to get: 15x^2 + 11x - 12 = 0 Since 15x^2 + 11x - 12 = 0 is identical to the equation we want to factor, we solve for x and confirm the results and we are done. Our factors for this equation are: (3x+4) * (5x-3) = 0 3x+4 = 0 results in x = -4/3 5x-3 = 0 results in x = 3/5 Our possible roots for this equation are: x = -4/3 x = 3/5 We substitute x = -4/3 in the original quadratic equation to get: 15x^2 + 11x - 12 = (15 * (-4/3)^2) + (11 * (-4/3)) - 12 = (26.666... - 14.666...) - 12 = 12 - 12 = 0 confirming x = (-4/3) is good. We substitute x = 3/5 in the the original quadratic equation to get: 15x^2 + 11x - 12 = (15 * (3/5)^2) + (11 * (3/5)) - 12 = (5.4 + 6.6) - 12 = 12 - 12 = 0 confirming x = (3/5) is good. This equation was the most complicated because the possible number of sets of factors is greater. This is because the a factor had multiple combinations possible and the c factor had multiple combinations possible. It takes a while but you will get the result if the quadratic equation has integer factors that can be determined through the factor method. If you have to find the factors using the factor method, then go through the analysis and you will eventually succeed as long as the problem was originally set up so the factors could be found. If you don't have to find the factors using the factor method and you try to find them using the factor method but are having trouble, then revert to your fall back position. The easiest fall back position to use is the quadratic formula because you don't have to do any analysis except remember what the quadratic formula is and plug in the a factors and the b factors and the c factors and solve. If you are more comfortable with the completing the squares method, by all means use that. PARABOLAS If you need to learn about how to solve problems involving parabolas where you are asked to find the minimum or maximum point in the graph of the quadratic equation, plus other items relating to parabolas such as axis of symmetry, etc., then see the lesson on PARABOLAS. EQUATIONS THAT ARE NOT IN QUADRATIC FORM THAT CAN BE CONVERTED TO QUADRATIC FORM The basic form of the quadratic equation is ax^2 + bx + c = 0 If you have an equation that looks like something like ax^3 + ax^2 + cx = 0, that equation can be translated into the quadratic equation form by doing the following: Factor an x out of the equation to get x*(ax^2 + ax + c) = 0. You will find up with something like x*(dx+e)*(fx+g) = 0 Each one of these factors can be set equal to 0 to solve the equation once it has been factored. Another form that can be converted would be something like ax^4 + bx^2 + c = 0. Since x^4 is the same as (x^2)^2 and x^2 is the same as (x^2)^1, you can convert this equation to quadratic form by setting a new variable, such as m equal to x^2 to get am^2 +bm + c = 0 and solve for m. Once you find m, you can solve for m = x^2. Another example would be an equation in the form of a(x-d)^2 + b(x-d) + c = 0 You set (x-1) = z, for example, to convert this equation to z^2 + z + c = 0 and solve for z. Once you have solved for z, then you solve for x. Tutorials with examples of equations that can be converted to quadratic equation form can be found in the following references: <a href = "http://www.purplemath.com/modules/factquad4.htm" target = "_blank">http://www.purplemath.com/modules/factquad4.htm</a> <a href = "http://tutorial.math.lamar.edu/Classes/Alg/ReducibleToQuadratic.aspx" target = "_blank">http://tutorial.math.lamar.edu/Classes/Alg/ReducibleToQuadratic.aspx</a> <a href = "http://www.kwiznet.com/p/takeQuiz.php?ChapterID=11297&CurriculumID=48&Num=6.11" target = "_blank">http://www.kwiznet.com/p/takeQuiz.php?ChapterID=11297&CurriculumID=48&Num=6.11</a> <a href = "http://www.mhhe.com/math/precalc/barnettpc5/graphics/barnett05pcfg/ch01/others/bpc5_ch01-07.pdf" target = "_blank">http://www.mhhe.com/math/precalc/barnettpc5/graphics/barnett05pcfg/ch01/others/bpc5_ch01-07.pdf</a> <a href = "http://infinity.cos.edu/algebra/Blakely%20Text/Chapter%2010/10.4.pdf" target = "_blank">http://infinity.cos.edu/algebra/Blakely%20Text/Chapter%2010/10.4.pdf/Blakely</a> Some of these are in pdf format. If you do not have a pdf viewer, you should download one. They are free. Foxit viewer is about 5 megs. Sumatra viewer is about 4 megs. Adobe is the king of the heap but is a large download, somewhere around 35 megs. To find these viewers, go to <a href = "http://www.google.com" target = "_blank">www.google.com</a> or <a href = "http://www.yahoo.com" target = "_blank">www.yahoo.com</a> and do a search on "pdf viewer". Before you do, though, just try and link to the web addresses above that contain pdf at the end and see if you can view them. If you can, then you already have a pdf viewer installed on your machine. Most computers come with the Adobe Viewer already installed. You may not have to do anything.
