Lesson COMPLETING THE SQUARES
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This lesson covers an overview of the Completing the Squares Method for factoring Quadratic Equations, REFERENCES <a href = "http://www.purplemath.com/modules/sqrquad.htm" target = "_blank">http://www.purplemath.com/modules/sqrquad.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.algebralab.org/lessons/lesson.aspx?file=algebra_completingthesquare.xml" target = "_blank">http://www.algebralab.org/lessons/lesson.aspx?file=algebra_completingthesquare.xml</a> <a href = "http://www.themathpage.com/aPreCalc/complete-the-square.htm" target = "_blank">http://www.themathpage.com/aPreCalc/complete-the-square.htm</a> <a href = "http://en.wikipedia.org/wiki/Completing_the_square" target = "_blank">http://en.wikipedia.org/wiki/Completing_the_square</a> <a href = "http://www.sosmath.com/algebra/quadraticeq/complsquare/complsquare.html" target = "_blank">http://www.sosmath.com/algebra/quadraticeq/complsquare/complsquare.html</a> COMPLETING THE SQUARES METHOD Completing the Squares Method means that we are creating an equation that is the result of the square of its factor. An example would be the equation: {{{x^2 - 2x - 35 = 0}}} (original equation). Once we have squared its factor, this equation becomes: {{{(x-1)^2 = 36}}} That would be the final form of the completing the squares method. You are then ready to take the square root of both sides of this equation to solve for x. The first thing we have to do is to make sure the equation is in the form required by the completing the squares method. The equation of {{{x^2 - 2x - 35 = 0}}} is in standard form. In order to convert it to the form required by the completing the squares method, we need to move the constant term to the right side of the equation and make sure that the x^2 term and the x terms are both on the left side of the equation. For this equation, we add 35 to both sides of it to get: {{{x^2 - 2x = 35}}} The next thing we have to do is to make sure the coefficient of the x^2 term is positive. It is, so we don't have to do anything here. We will see an example later on when it isn't. The next thing we have to do is to make sure the coefficient of the x^2 term is equal to 1. It is, so we don't have to do anything here either. We will see an example later on when it isn't. Our equation is now ready for further processing using the completing the squares method. Our equation is: {{{x^2 - 2x = 35}}} We want to complete the square on the expression that is on the left side of our equation. That expression is {{{x^2 - 2x}}}. We take 1/2 the coefficient of the x term to create a factor of (x-1) which we then square to get: {{{(x-1)^2 = x^2 - 2x + 1}}} We subtract 1 from both sides of this equation to get: {{{(x-1)^2 - 1 = x^2 - 2x}}}. Since {{{(x-1)^2 - 1}}} is equivalent to {{{x^2 - 2x}}}, we can substitute in our equation of: {{{x^2 - 2x = 35}}} to get: {{{(x-1)^2 - 1 = 35}}}. We now add 1 to both sides of this equation to get: {{{(x-1)^2 = 36}}} We have just completed the squares on this equation and all we have to do is solve this equation by taking the square root of both sides of the equation to get: x-1 = +/- {{{sqrt(36)}}} We then add 1 to both sides of this equation to get: x = +/- {{{sqrt(36)}}} + 1. Since {{{sqrt(36) = 6}}}, we get: x = +/- 6 + 1. A graph of our original equation of {{{x^2 - 2x - 35 = 0}}} is shown below: {{{graph(400,200,-10,10,-40,10,x^2 - 2x - 35)}}} The roots of this equation are: x = +/- 6 + 1. This means that: x = 6 + 1 = 7 x = -6 + 1 = -5 The graph confirms that the roots of this equation are x = -5 and x = 7. Those are the points where the graph of the equation crosses the x-axis. MAKING THE EQUATION READY FOR PROCESSING USING THE COMPLETING THE SQUARES METHOD The equation is ready for processing using the completing the squares method when: <pre> > The constant term is on the right side of the equation and the x^2 and x terms are on the left side of the equation. > The sign of the coefficient of the x^2 term is positive. > The coefficient of the x^2 term is equal to 1. </pre> REMOVING A NEGATIVE COEFFICIENT IN THE X^2 TERM Your equation might be something like {{{-x^2 + 2x = -35}}} You cannot complete the squares while the equation is like this because the coefficient of the x^2 term is negative. What you have to do in this case is to take your equation and multiply both sides of it by -1 to get: {{{x^2 - 2x = 35}}} Now you can complete the squares because the coefficient of the x^2 term on the left side of your equation is positive as required. EXAMPLES OF USING THE COMPLETING THE SQUARES METHOD The following examples show you how the process is used to solve the different types of problems you will encounter. EXAMPLE 1 Original equation is: {{{x^2 - 2x - 5 = 0}}} Add 5 to both sides of this equation to get: {{{x^2 - 2x = 5}}} Now the constant term is on the right side of the equation as required by the completing the squares method. Since the coefficient of the x^2 term is positive, we do not need to multiply both sides of this equation by -1. Since the coefficient of the x^2 term equals 1, we do not need to factor the expression on the left side of the equation by the coefficient of the x^2 term. The equation is now ready for further processing using the completing the squares method. We take 1/2 the coefficient of the x term to get -1. We create our factor using 1/2 the coefficient of the x term to get {{{(x-1)^2}}}. Since {{{(x-1)^2 = x^2 - 2x + 1}}}, we subtract 1 from both sides of this equation to get: {{{(x-1)^2 - 1 = x^2 - 2x}}}. Since they are now equivalent, we replace {{{x^2 - 2x}}} with {{{(x-1)^2 - 1}}} in our original equation of {{{x^2 - 2x = 5}}} to get: {{{(x-1)^2 - 1 = 5}}} We add 1 to both sides of this equation to get: {{{(x-1)^2 = 6}}} We take the square root of both sides of this equation to get: x-1 = +/- {{{sqrt(6)}}} We add 1 to both sides of this equation to get: x = +/- {{{sqrt(6)}}} + 1. Graph of our original equation of y = {{{x^2 - 2x - 5}}} looks like this: {{{graph(400,200,-5,5,-10,5,x^2 - 2x - 5)}}} The roots of this equation are x = +/- {{{sqrt(6)}}} + 1. {{{sqrt(6)}}} is roughly equal to 2.4. This means that: x is roughly equal to 2.4 + 1 which equals 3.4. x is roughly equal to -2.4 + 1 which equals -1.4. This is confirmed by the graph of the equation crossing the x-axis at x = 3.4 and x = -1.4. EXAMPLE 2 Original equation is: {{{x^2 + 4x - 10 = 0}}} Add 10 to both sides of this equation to get: {{{x^2 + 4x = 10}}} Now the constant term is on the right side of the equation as required by the completing the squares method. Since the coefficient of the x^2 term is positive, we do not need to multiply both sides of this equation by -1. Since the coefficient of the x^2 term equals 1, we do not need to factor the expression on the left side of the equation by the coefficient of the x^2 term. We take 1/2 the coefficient of the x term to get 2. We create our factor using 1/2 the coefficient of the x term to get {{{(x+2)^2}}}. Since {{{(x+2)^2 = x^2 + 4x + 4}}}, we subtract 4 from both sides of this equation to get: {{{(x+2)^2 - 4 = x^2 + 4x}}}. Since they are now equivalent, we replace {{{x^2 + 4x}}} with {{{(x+2)^2 - 4}}} in our original equation of {{{x^2 + 4x = 10}}} to get: {{{(x+2)^2 - 4 = 10}}} We add 4 to both sides of this equation to get: {{{(x+2)^2 = 14}}} We take the square root of both sides of this equation to get: x+2 = +/- {{{sqrt(14)}}} We subtract 2 from both sides of this equation to get: x = +/- {{{sqrt(14)}}} - 2. Graph of our original equation of y = {{{x^2 + 4x - 10}}} looks like this: {{{graph(400,200,-10,5,-15,5,x^2 + 4x - 10)}}} The roots of this equation are x = +/- {{{sqrt(14)}}} - 2. {{{sqrt(14)}}} is roughly equal to 3.7. This means that: x is roughly equal to 3.7 - 2 = 1.7 x is roughly equal to -3.7 - 2 = -5.7 This is confirmed by the graph of the equation crossing the x-axis at x = 1.7 and x = -5.7. EXAMPLE 3 Original equation is: {{{-x^2 + 10x - 6 = 0}}} Add 6 to both sides of this equation to get: {{{-x^2 + 10x = 6}}} The constant term is now on the right side of the equation as required. Since the coefficient of the x^2 term is negative, we have to multiply both sides of the equation by -1 in order to make it positive. We get: {{{x^2 - 10x = -6}}} Since the coefficient of the x^2 term equals 1, we do not need to factor the expression on the left side of the equation by the coefficient of the x^2 term. The equation is now ready for further processing using the completing the squares method. We take 1/2 the coefficient of the x term to get -5. We create our factor using 1/2 the coefficient of the x term to get {{{(x-5)^2}}}. Since {{{(x-5)^2 = x^2 - 10x + 25}}}, we subtract 25 from both sides of this equation to get: {{{(x-5)^2 - 25 = x^2 - 10x}}}. Since they are now equivalent, we replace {{{x^2 - 10x}}} with {{{(x-5)^2 - 25}}} in our original equation of {{{x^2 - 10x = -6}}} to get: {{{(x-5)^2 - 25 = -6}}}. We add 25 to both sides of this equation to get: {{{(x-5)^2 = 19}}} We take the square root of both sides of this equation to get: x-5 = +/- {{{sqrt(19)}}} We add 5 to both sides of this equation to get: x = +/- {{{sqrt(19)}}} + 5. Graph of our original equation of y = {{{-x^2 + 10x - 6}}} looks like this: {{{graph(400,200,-5,15,-10,30,-x^2 + 10x - 6)}}} The roots of this equation are x = +/- {{{sqrt(19)}}} + 5. {{{sqrt(19)}}} is roughly equal to 4.4. This means that: x is roughly equal to 4.4 + 5 = 9.4 x is roughly equal to -4.4 + 5 = .6 This is confirmed by the graph of the equation crossing the x-axis at x = 9.4 and x = .6. EXAMPLE 4 Original equation is: {{{-3x^2 + 30x - 18 = 0}}} Add 18 to both sides of this equation to get: {{{-3x^2 + 30x = 18}}} the constant term is now on the right side of this equation as required. Since the coefficient of the x^2 term is negative, we have to multiply both sides of the equation by -1 in order to make it positive. We get: {{{3x^2 - 30x = -18}}} Since the coefficient of the x^2 term is not equal to 1, we have to factor the expression on the left side of the equation by the coefficient of the x^2 term to get: {{{3 * (x^2 - 10x) = -18}}} We take 1/2 the coefficient of the x term to get -5. We create our factor using 1/2 the coefficient of the x term to get {{{(x-5)^2}}}. Since {{{(x-5)^2}}} = {{{x^2 - 10x + 25}}}, we subtract 25 from both sides of this equation to get {{{(x-5)^2 - 25 = x^2 - 10x}}}. Since they are now equivalent, we can replace {{{x^2 - 10x}}} with {{{(x-5)^2 - 25}}} to get: {{{3 * ((x-5)^2 - 25) = -18}}} We simplify by removing parentheses to get: {{{3*(x-5)^2 - 75 = -18}}} We add 75 to both sides of this equation to get: {{{3*(x-5)^2 = 57}}} We divide both sides of this equation by 3 to get: {{{(x-5)^2 = 19}}} We take the square root of both sides of this equation to get: x-5 = +/- {{{sqrt(19)}}} We add 5 to both sides of this equation to get: x = +/- {{{sqrt(19)}}} + 5. Graph of our original equation of y = {{{-3x^2 + 30x - 18}}} looks like this: {{{graph(400,200,-5,15,-10,60,-3x^2 + 30x - 18)}}} The roots of this equation are x = +/- {{{sqrt(19) + 5}}}. {{{sqrt(19)}}} is roughly equal to 4.4. This means that: {{{x = 5 + sqrt(19)}}} is roughly equal to 4.4 + 5 = 9.4 {{{x = 5 - sqrt(19)}}} is roughly equal to -4.4 + 5 = .6 This is confirmed by the graph of the equation crossing the x-axis at x = 9.4 and x = .6. Note that the roots of the equation in example 4 are exactly the same as the roots of the equation in example 3. This is because the equation in example 4 is an exact multiple of the equation in example 3. The equation in example 3 is {{{-x^2 + 10x - 6 = 0}}}. The equation in example 4 is {{{-3x^2 + 30x - 18 = 0}}} If you divide both sides of the equation in example 4 by 3, you will get the equation in example 3. This is why the roots are the same in those two equations.
