Question

x^2-8x+16=0
Can someone show me step-by-step how to solve this equation?

Answer 1

e^ln(x^2-8x+16)=e^0
x^2-8x+16=1
x^2-8x+15=0
(x-3)(x-5)=0
x=3,5

Answer 2

Could you just explain what you did there?

Answer 3

Raise both with base e
so you get:

x^2-8x+16 =e^0
x^2 - 8x + 16 = 1
x^2 - 8x + 15 = 0
(x-5) = 0 (x-3) = 0
x = 5 x = 3

Answer 4

You have a perfect square. You are looking for 2 numbers that multiply to +16 and add to -8. Those 2 numbers are -4 and -4 (-4 twice). This fits the form of:

(x - 4)(x - 4) or (x - 4)^2 = 0

NOT 5 and 3.

Answer 5

There really is only one step. It consists of recognizing the completion of a perfect square once one looks at:

x^2 - 8x

At that point, we take half of -8 and square that to get 16. But 16 is what we already have left to complete that square.

You could do:

(x^2 - 8x + 16) + 16 = 0 + 16 and then subtract 16 from both sides, but that step is not necessary.