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How do you solve using the completing the square method #x^2 + 2x = 7#?

Answer 1

Savannah Anderson

#x = - 1 +- 2sqrt2#

#x^2 + 2x = 7# #x^2 + 2x + 1 = 7 + 1 # #(x + 1)^2 = 8# #x + 1 = +- 2sqrt2# #x = - 1 +- 2sqrt2#

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Answer 2

Abigail Allen

To solve using the completing the square method for the equation (x^2 + 2x = 7), follow these steps:

Move the constant term to the other side of the equation: (x^2 + 2x - 7 = 0)
To complete the square, take half of the coefficient of (x) and square it. Add this value to both sides of the equation: (x^2 + 2x + (2/2)^2 = 7 + (2/2)^2) (x^2 + 2x + 1 = 7 + 1) (x^2 + 2x + 1 = 8)
Rewrite the left side as a perfect square trinomial: ((x + 1)^2 = 8)
Solve for (x): (x + 1 = \pm \sqrt{8}) (x = -1 \pm \sqrt{8})

Thus, the solutions for the equation (x^2 + 2x = 7) are (x = -1 + \sqrt{8}) and (x = -1 - \sqrt{8}).

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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.