SOLUTION: what 3 techniques can be used tp solve a quadratic equation? demonstrate these techniques on the eqution "x^2 -10x-39 =0"

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Question 203984: what 3 techniques can be used tp solve a quadratic equation? demonstrate these techniques on the eqution "x^2 -10x-39 =0"
Found 2 solutions by nerdybill, jsmallt9:
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
Three techniques could be:
1. factoring
2. completing the square
3. quadratic equation
.
1. factoring:
x^2 -10x-39 =0
(x-13)(x+3) = 0
x = {-3, 13}
.
2. completing the square
x^2 -10x-39 =0
x^2 -10x = 39
x^2 -10x+25 = 39+25
(x-5)^2 = 64
x-5 = +-8
x = 5+-8
x = {-3, 13}
.
3. quadratic equation
x^2 -10x-39 =0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-10x%2B-39+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-10%29%5E2-4%2A1%2A-39=256.

Discriminant d=256 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--10%2B-sqrt%28+256+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-10%29%2Bsqrt%28+256+%29%29%2F2%5C1+=+13
x%5B2%5D+=+%28-%28-10%29-sqrt%28+256+%29%29%2F2%5C1+=+-3

Quadratic expression 1x%5E2%2B-10x%2B-39 can be factored:
1x%5E2%2B-10x%2B-39+=+1%28x-13%29%2A%28x--3%29
Again, the answer is: 13, -3. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-10%2Ax%2B-39+%29

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
What 3 techniques can be used tp solve a quadratic equation? Demonstrate these techniques on the eqution x%5E2+-10x-39+=0
  1. Solve by factoring.
    1. Simplify and make one side of the equation zero.
      Your equation is already in this form.
    2. Factor the non-zero side. There are many techniques which can be used in factoring. These include:
      • Greatest common Factor (GCF). If the GCF is not 1, this should always be done first.
      • Factoring by patterns. Some of the more commonly taught patterns are:
        • Difference of squares: a%5E2+-+b%5E2+=+%28a%2Bb%29%28a-b%29
        • Difference of Cubes: a%5E3+-+b%5E3+=+%28a-b%29%28a%5E2+%2Bab+%2Bb%5E2%29
        • Sum of Cubes: a%5E3+%2B+b%5E3+=+%28a%2Bb%29%28a%5E2+-ab+%2B+b%5E2%29
        • Perfect Square Trinomials:
          • a%5E2+%2B+2ab+%2B+b%5E2+=+%28a%2Bb%29%28a%2Bb%29+=+%28a%2Bb%29%5E2
          • a%5E2+-+2ab+%2B+b%5E2+=+%28a-b%29%28a-b%29+=+%28a-b%29%5E2
      • Trinomial Factoring
      • Factoring by Grouping
      • Factoring by trial and error of the possible rational roots
      All techniques should be used as many times as possible to factor an expression until it cannot be factored any further. (And sometimes factoring out a -1 is helpful.)
      The GCF of the left side of your equation is 1 and we don't usually factor out 1. The left side does not fit any of the patterns but it is a trinomial we can factor:
      x%5E2+-10x+-39+=+0
      becomes
      %28x-13%29%28x%2B3%29+=+0
    3. Use the Zero Product property (which states that a product can equal zero only if one of the factors is zero) to write equations where each factor which has a variable in it is equal to zero.
      Your factors are (x-13) and (x+3) so one of these must be zero:
      x+-+13+=+0 or x+%2B+3+=+0
    4. Solve each of these equations.
      x+=+13 or x+=+-3
  2. Solve by using the Quadratic Formula
    1. Simplify and make one side of the equation zero.

      2. Your equation is already in this form.
    2. Use the quadratic formula: x+=+%28-b+%2B-+sqrt%28b%5E2+-+4ac%29%29%2F%282a%29 getting the a, b and c from the starndard form: ax%5E2+%2B+bx+%2B+c+=+0
      In your equation a = 1, b = -10 and c = -39. Substituting these into the Quadratic formula we get:
      x+=+%28-%28-10%29+%2B-+sqrt%28%28-10%29%5E2+-+4%281%29%28-39%29%29%29%2F%282%281%29%29
      Simplifying the expression in the square root:
      x+=+%28-%28-10%29+%2B-+sqrt%28100+-+4%281%29%28-39%29%29%29%2F%282%281%29%29
      x+=+%28-%28-10%29+%2B-+sqrt%28100+%2B+156%29%29%2F%282%281%29%29
      x+=+%28-%28-10%29+%2B-+sqrt%28256%29%29%2F%282%281%29%29
      x+=+%28-%28-10%29+%2B-+16%29%2F%282%281%29%29
      Simplifying the rest:
      x+=+%2810+%2B-+16%29%2F2
      Now we'll split the +- into separate equations:
      x+=+%2810+%2B+16%29%2F2 or x+=+%2810+-+16%29%2F2
      Simplifying:
      x+=+26%2F2 or x+=+%28-6%29%2F2
      x+=+13 or x+=+-3
  3. Solve by "Completing the Square"
    1. Simplify the equation.
      Your equation is already simplified.
    2. Gather the variable terms on one side of the equation and the constant term on the other side.
      Your x terms are together on the left side. So all we need to do is move the constant term to the other side by adding 39 to both sides:
      x%5E2+-+10x+=+39
    3. Complete the square. We are trying to match one of the Perfect Square Trinomial patterns. (See factoring patterns above.)
      1. Ensure there is a 1 in front of the x^2 term. If there is a different number which I will call "q" for now, then multiply both sides of the equation by the reciprocal of "q"
        Your equation has a 1 so you're all set to proceed.
      2. Determine what 1/2 of the coefficient of the x term (not the x^2 term) is. We'll call this number "h" (for half).
        Your equation has a 10 in front of x. 1/2 of 10 is 5.
      3. Add the square of "h" to both sides.
        In your equation we will add %285%29%5E2+=+25 to both sides:
        x%5E2+-10x+%2B+25+=+64
      4. The variable side should now fit one of the Perfect Square Trinomial patterns where the "a" from the pattern is your variable and the "b" from the pattern is your "h". Rewrite the variable side of the equation according to the appropriate pattern. If your middle term has a "+" in front use the first pattern. Otherwise use the second pattern.
        Since your equation has a "-" in front of the middle term, we'll use the second pattern: a%5E2+-+2ab+%2B+b%5E2+=+%28a-b%29%5E2:
        x%5E2+-10x+%2B+25+=+64
        %28x%29%5E2+-2%28x%29%285%29+%2B+%285%29%5E2+=+64
        %28x+-+5%29%5E2+=+64
      5. Find the square root of each side, putting absolute value around the variable expression.
        sqrt%28%28x+-+5%29%5E2%29+=+sqrt%2864%29
        abs%28x+-+5%29+=+8
      6. Solve the absolute value equation.
        x+-+5+=+8 or x+-+5+=+-8
      7. Solve each equation.
        x+=+13 or x+=+-3

An example of "completing the square" with a coefficient other than 1 in front of the squared term:
2x%5E2+%2B7x+%2B+3+=+0
Simplfy: done
Variable terms on one side, constant term on the other:
2x%5E2+%2B7x+=+-3
If the coefficient of the squared term is not a 1 (which we called "q"), multiply both sides of the equation by the reciprocal of "q".
%281%2F2%29%282x%5E2+%2B7x%29+=+%281%2F2%29%28-3%29
x%5E2+%2B+%287%2F2%29x+=+-3%2F2
Determine 1/2 of the coefficient of the x term (which we called "h" above). We want 1/2 of 7/2. So "h" is 7/4.
"Add the square of "h" to both sides.
x%5E2+%2B+%287%2F2%29x+%2B+%287%2F4%29%5E2+=+-3%2F2+%2B+%287%2F4%29%5E2
Simplifying the right side:
x%5E2+%2B+%287%2F2%29x+%2B+%287%2F4%29%5E2+=+-3%2F2+%2B+49%2F16
x%5E2+%2B+%287%2F2%29x+%2B+%287%2F4%29%5E2+=+-3%2F2+%2B+49%2F16
x%5E2+%2B+%287%2F2%29x+%2B+%287%2F4%29%5E2+=+25%2F16%29
Rewrite the variable side as a perfect square:
%28x%29%5E2+%2B+2%28x%29%287%2F4%29+%2B+%287%2F4%29%5E2+=+25%2F16%29
%28x+%2B+7%2F4%29%5E2+=+25%2F16%29
Find the square root of each side:
sqrt%28%28x+%2B+7%2F4%29%5E2%29+=+sqrt%2825%2F16%29%29
abs%28x+%2B+7%2F4%29+=+5%2F4%29
Solve the absolute value equation
x+%2B+7%2F4+=+5%2F4%29 or x+%2B+7%2F4+=+-5%2F4
Solve each:
x+=+-2%2F4+=+-1%2F2 or x+=+-12%2F4+=+-3