Question

Question 8 of 10

What are the zeros of f(x) = x² − 8x + 16?
A. x= -4 and x = 4
B. x=-2 and x = 8
C. x = 4 only
D. x = -4 only

Answer 1

c ×=4, only

Explanation:

4*-8×+16=0

16-32=-16

-16=-16

Answer 2

D. x=4 only

Explanation:

You can solve using the quadratic equation which is
`x=\frac{-b \pm \sqrt{b^(2)-4ac}}{2a}`

where in this case a=1, b=-8, and c=16. But there is any easier way to solve the equation. You can solve by completing the square which consists of moving the constant to the right side and then adding
`((b)/(2))^2`. To both sides that way you can rewrite the left side as a perfect square binomial. This is because
`(x+b)^2` will expand to
`x^2 + 2xb + b^2`. So when you add
`((b)/(2))^2` to both equations. You're able to rewrite the left side as
`(x+((b)/(2)))^2`. Because once you expand it out, it becomes
`x^2 + 2((b)/(2))x + ((b)/(2))^2`. Which simplifies to
`x^2 + bx + ((b)/(2))^2` which is the original equation. In this case there is no need to move the constant, which in this case is 16, to the other side. Since if you calculate the value of (-8/2)^2 it's equal to 16! So this can already be written as a perfect square binomial.

So the equation becomes

`0 = (x+(b)/(2))^2`

Plug values in

`0 = (x+(-(8)/(2)))^2`

Simplify

`0 = (x-4)^2`

Take the square root of both sides

`0 = x-4`

Add 4 to both sides

`4 = x`