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What is the sum of all values of that satisfy the equation

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by Ian Stewart » Sat May 04, 2019 11:40 am
We can divide by 4 and get zero on both sides:

4x^2 + 16 = 32x
x^2 + 4 = 8x
x^2 - 8x + 4 = 0

When we factor the left side above, the factorization will look like (x - a)(x - b), where a and b are the two solutions to the quadratic. The numbers a and b will multiply to 4 (since 4 = (-a)(-b) = ab) and will add to 8 (since -a and -b will add to -8), and since a and b are the solutions to the quadratic, a+b = 8 is the answer.
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by Scott@TargetTestPrep » Mon May 13, 2019 5:39 pm
BTGmoderatorLU wrote:Source: Veritas Prep

What is the sum of all values of that satisfy the equation \(4x^2 +16=32x\)?

A. \(−8\)
B. \(−4\sqrt{2}\)
C. \(−4\)
D. \(4\sqrt{2}\)
E. \(8\)

The OA is E
Setting the equation to 0, we have:

4x^2 - 32x + 16 = 0

For a quadratic equation of the form ax^2 + bx + c = 0 (where a ≠ 0), the sum of the roots (or solutions) is always -b/a. Therefore, the sum of the roots in this case is -(-32)/4 = 32/4 = 8.

Answer: E

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