How do you solve the quadratic equation by completing the square: #y^2 + 16y = 2#?

Question

Julia Adams · Accepted Answer

#y= -8+-sqrt(66)#

You apply the following formula to finish the square:

#ax^2+bx+c#

a has to equal 1.

#c=(b/2)^2#

The square when finished is:

#(x+b/2)^2#

Now, the x in the general formula is represented by the y in your function:

#y^2 + 16y = 2#

#y^2 + 16y +underbrace(c = 2+c)# we add c to both sides so we don't alter the equation

now resolve c:

#c=(b/2)^2 = (16/2)^2=64#

#y^2 + 16y +64 = 2+64#

now finish the square:

#(y+8)^2 = 66#

Now resolve:

#sqrt((y+8)^2) = +-sqrt(66)#

#y+8 = +-sqrt(66)#

#y= -8+-sqrt(66)#

Eva Adamson · Answer

undefined To solve the quadratic equation by completing the square, follow these steps:

1. Rewrite the equation in the form $ y^2 + 16y + \text{(some number)} = 2 + \text{(same number)} $ by adding and subtracting the square of half of the coefficient of $ y $.
2. Factor the trinomial as a perfect square.
3. Solve for $ y $ by taking the square root of both sides.
4. Simplify the solution.

Applying these steps to the equation $ y^2 + 16y = 2 $, we get:

1. Add and subtract $ (16/2)^2 = 64 $:
   $ y^2 + 16y + 64 - 64 = 2 $
2. Factor the trinomial as a perfect square:
   $ (y + 8)^2 - 64 = 2 $
3. Solve for $ y $ by taking the square root of both sides:
   $ y + 8 = \pm \sqrt{66} $
4. Simplify the solution:
   $ y = -8 \pm \sqrt{66} $