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How do you solve #x^2+6x=7 # using completing the square?

Answer 1

Julia Adams

#(x+3)^2 = x^2+6x+9#

So: #(x+3)^2 = 7 + 9 = 16 = 4^2#

Hence #x = -7# or #x = 1#

#(x+3)^2 = x^2+6x+9#

So: #(x+3)^2 = x^2+6x+9 = 7 + 9 = 16 = 4^2#

To account for both scenarios, take the square root of both ends:

#x+3 = +-sqrt(4^2) = +-4#

Subtract #3# from both sides to get:

#x = -3+-4#

That is #x = -7# or #x = 1#

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Answer 2

Savannah Anderson

To solve the equation ( x^2 + 6x = 7 ) using completing the square, follow these steps:

Move the constant term to the other side of the equation: ( x^2 + 6x - 7 = 0 ).
To complete the square, add the square of half the coefficient of ( x ) to both sides of the equation: ( x^2 + 6x + (6/2)^2 = 7 + (6/2)^2 ).
Simplify both sides: ( x^2 + 6x + 9 = 7 + 9 ).
Rewrite the left side as a perfect square trinomial: ( (x + 3)^2 = 16 ).
Take the square root of both sides: ( x + 3 = \pm \sqrt{16} ).
Solve for ( x ): ( x = -3 \pm 4 ).
There are two solutions: ( x = 1 ) and ( x = -7 ).

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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.