www.VirtualNerd.com

How Do You Solve a Quadratic Equation by Completing the Square?

Solve x2 + 12x + 32 = 0 by completing the square.

Summary

  1. The goal is to get the left side to look like (x+?)2
  2. Note that 6+6=12 and 6•6=36, so we can add 36 to the left hand side to give us a perfect square trinomial
  3. A perfect square trinomial can be factored into a binomial squared
  4. Adding 36 to both sides gives us 4 on the right
  5. Taking the square root will get rid of the squared term on the left
  6. The symbol over the (x+6)2 and the 4 is the symbol for square root
  7. ± means 'plus or minus'
  8. Since both 22 and (-2)2 equal 4, we will have two possible solutions

Notes

    1. We want to turn the left hand side into a perfect square trinomial
    2. Then we will be able to factor it into something squared
    3. Put the x-terms by themselves on one side to make it easier to find a perfect square
    4. Then we'll be able to pick a number to add to both sides that makes the trinomial on the left a perfect square
    1. The equation we're working with is:
    2. x2+12x=-32
    3. Remember, we're trying to turn the left hand side into a perfect square
    4. So we need to figure out what number to add to x2+12x to make a perfect square trinomial
    1. Remember, 'c' comes from the general form of a quadratic equation: ax2+bx+c
    2. We want something we can add to x2+12x so that we can factor it into (x+)2
    3. When we FOIL two binomials together, the middle terms add up to give us 'b'
    4. 'b' is 12 in our given equation
    5. Those two middle terms are the same when we square a binomial, so we need a number we can add to itself to get 12
    1. This works out great -- our 'c' value will be 36!
    2. Remember, 'c' comes from the general form of a quadratic equation:
    3. ax2+bx+c
    4. (x+6)•(x+6) = x2+12x+36
    1. We figured out that 36 is our value for 'c' on the left side of the equation
    2. If we add 36 to one side of the equation, we need to add 36 to the other!
    3. In Step 1, we subtracted 32 from both sides, which gave us a -32 on the right side
    4. So if we add 36 to -32 we get 4 on the right
    1. We have x2+12x+36 on the left side of our equation
    2. This factors into (x+6)•(x+6)
    3. Notice (x+6)•(x+6) is the same as (x+6)2, which is something squared, like we wanted!
    1. The left side of our equation is squared
    2. Taking the square root will get rid of the exponent '2' and let us solve for 'x'
    3. '±' means 'plus or minus'
    4. Note that both 22 and (-2)2 equal 4, so the square root of 4 is ±2
    5. Since we don't know what x is, we need to consider both possibilities
    1. Since the square-root of 4 is ±2, we have two possible solutions
    2. '±' means 'plus or minus'
    3. When x+6 = 2, subtract 6 from both sides to get x = -4
    4. When x+6 = -2, subtract 6 from both sides to get x = -8
    1. When x+6 = 2, subtract 6 from both sides to get x = -4
    2. When x+6 = -2, subtract 6 from both sides to get x = -8
    1. We found that 'x' can be '-4' and '-8' in this problem
    2. We can plug these values back into our original equation to check them
    3. (-4)2+12(-4)+32 = 16-48+32 = 0
    4. (-8)2+12(-8)+32 = 64-96+32 = 0
    5. So our solutions of 'x=-4' and 'x=-8' are correct!