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
What is the sum of all values of that satisfy the equation
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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.
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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Setting the equation to 0, we have: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
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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