How do you solve #x^2-8x=9# by completing the square?

Answer 1

You take half the #x#-coefficient (#-8->-4#) and square it (#16#)

This means you have to add #16# to both sides of the equation: #x^2-8x+16=9+16-># #(x-4)^2=25-># So: #x-4=sqrt25=5->x=9# Or: #x-4=-sqrt25=-5->x=-1#

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Answer 2

Ethan Adkins

Solve y = x^2 - 8x - 9 = 0

In this case, you don't need to solve by completing the squares. Use the shortcut. When (a - b + c = 0) --> 2 real roots: (-1) and (-c/a = 9).

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Answer 3

Eva Abramson

To solve the equation (x^2 - 8x = 9) by completing the square, follow these steps:

Move the constant term to the other side of the equation: (x^2 - 8x - 9 = 0)
To complete the square, take half of the coefficient of (x) (which is (-8/2 = -4)) and square it (((-4)^2 = 16)). Add this value to both sides of the equation: (x^2 - 8x + 16 = 9 + 16)
Factor the left side as a perfect square: ((x - 4)^2 = 25)
Take the square root of both sides: (x - 4 = \pm 5)
Solve for (x): (x = 4 \pm 5)
Therefore, the solutions are (x = 9) and (x = -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.