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How do you solve #x^2 - 8x - 6=0# using completing the square?

Answer 1

Ethan Adkins

First you change the form into #x^2-8x=6#
(add #6# to both sides)

Method: you halve the #x#-coefficient (#-8#) to get #-4#. The square of this is to be added to both sides.

#->x^2-8x+16=6+16#

#(x-4)^2=22->x-4=+-sqrt22#

#x_(1,2)=4+-sqrt22# (two solutions) graph{x^2-8x-6 [-52.02, 52.07, -26, 26]}

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Answer 2

Hazel Anglin

To solve the quadratic equation (x^2 - 8x - 6 = 0) using completing the square:

Move the constant term to the other side: (x^2 - 8x = 6).
Take half of the coefficient of (x), square it, and add it to both sides: (x^2 - 8x + (-8/2)^2 = 6 + (-8/2)^2).
Simplify: (x^2 - 8x + 16 = 6 + 16).
Simplify further: (x^2 - 8x + 16 = 22).
Rewrite the left side as a perfect square: ((x - 4)^2 = 22).
Take the square root of both sides: (x - 4 = \pm \sqrt{22}).
Solve for (x): (x = 4 \pm \sqrt{22}).

So, the solutions to the equation are (x = 4 + \sqrt{22}) and (x = 4 - \sqrt{22}).

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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.