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If there is exactly one solution to the equation \(25x^2\)

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by swerve » Mon Jun 17, 2019 1:23 pm

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E

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If there is exactly one solution to the equation \(25x^2-bx+64=0\), where \(b>0\), what is the value of \(b\)?

A. 26
B. 40
C. 52
D. 80
E. 100

The OA is D

Source: Veritas Prep
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hi

by Scott@TargetTestPrep » Wed Jun 19, 2019 5:34 pm
swerve wrote:If there is exactly one solution to the equation \(25x^2-bx+64=0\), where \(b>0\), what is the value of \(b\)?

A. 26
B. 40
C. 52
D. 80
E. 100

The OA is D

Source: Veritas Prep
Since the quadratic equation has only one solution, it means the left hand side of the equation must be a perfect square. Therefore, we can factor the left hand side as:

(5x + 8)^2 or (5x - 8)^2

Expanding the first perfect square, we have: (5x + 8)^2 = 25x^2 + 80x + 64, so -b = 80 or b = -80. However, b is positive. So b can't be -80. Let's expand the second perfect square:

(5x - 8)^2 = 25x^2 - 80x + 64

-b = -80

b = 80

Answer: D

Scott Woodbury-Stewart
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