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How do you solve # x^2 = 8x - 16# using the quadratic formula?

Answer 1

Eli Assouline

x = 4

#y = x^2 - 8x + 16 = (x - 4)^2 = 0# There is a double root at x = 4.

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Answer 2

Ethan Adkins

To solve the equation (x^2 = 8x - 16) using the quadratic formula, first, rearrange the equation to have zero on one side: (x^2 - 8x + 16 = 0). Then, identify the coefficients: (a = 1), (b = -8), and (c = 16). Next, substitute these values into the quadratic formula: (x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}). Plugging in the values, you get: (x = \frac{{-(-8) \pm \sqrt{{(-8)^2 - 4(1)(16)}}}}{{2(1)}}). Simplify under the square root: (x = \frac{{8 \pm \sqrt{{64 - 64}}}}{2}). Since (64 - 64 = 0), the expression under the square root becomes zero. Therefore, (x = \frac{{8 \pm 0}}{2}). This simplifies to (x = \frac{8}{2}) or (x = \frac{8}{2}). Hence, (x = 4) is the only solution.

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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.