How do you solve by completing the square for #x^2-8x+7#?

Answer 1

You have to take half the #x#-coefficient (=-4) and square it. At the same time you have to balance the 7 that is allready there.

#x^2-8x+16=(x-4)^2#

So we need to add #9# to both sides: #x^2-8x+16=9-># #(x-4)^2=9-># So: #x-4=sqrt9=3->x=7# Or: #x-4=-sqrt9=-3->x=1#

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Answer 2

Hazel Anglin

To solve by completing the square for (x^2 - 8x + 7), follow these steps:

Rewrite the equation as (x^2 - 8x + \left(\frac{-8}{2}\right)^2 - \left(\frac{-8}{2}\right)^2 + 7).
Simplify inside the parentheses: (x^2 - 8x + 16 - 16 + 7).
Combine like terms: (x^2 - 8x + 16 - 9).
Rewrite as a perfect square trinomial: ((x - 4)^2 - 9).
The solution is ((x - 4)^2 - 9 = 0).
Add 9 to both sides: ((x - 4)^2 = 9).
Take the square root of both sides: (x - 4 = \pm 3).
Solve for (x): (x = 4 \pm 3).
The solutions are (x = 4 + 3 = 7) and (x = 4 - 3 = 1).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.