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

Answer 1

The solutions are:
#color(green)(x = 2# ,
# color(green)(x = -4#

#x^2 + 2x- 8 =0 #
#x^2 + 2x = 8#

We add one to both sides in order to write the left-hand side as a perfect square:

#x^2 + 2x + 1 = 8 + 1#
#x^2 + 2* x * 1 + 1 = 9#
Using the Identity #color(blue)((a+b)^2 = a^2 + 2ab + b^2#, we get
#(x+1)^2 = 9#
#x + 1 = sqrt9# or #x + 1 = -sqrt9#
#color(green)(x = 3 -1 = 2# or # color(green)(x = -3 - 1 = -4#
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Answer 2

To solve (x^2 + 2x - 8 = 0) using the completing the square method, follow these steps:

  1. Move the constant term to the other side of the equation: (x^2 + 2x = 8)

  2. Take half of the coefficient of (x), square it, and add it to both sides of the equation: (x^2 + 2x + (2/2)^2 = 8 + (2/2)^2) (x^2 + 2x + 1 = 8 + 1)

  3. Rewrite the left side as a perfect square: ((x + 1)^2 = 9)

  4. Take the square root of both sides: (x + 1 = ±\sqrt{9})

  5. Solve for (x): (x = -1 ± 3)

So, the solutions are (x = -1 + 3) and (x = -1 - 3), which simplifies to (x = 2) and (x = -4).

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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.

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.

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.

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.

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