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If \(\dfrac{(x+3)^2}{x+15}=6,\) then the difference between the two possible values of \(x\) is:

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Re: If \(\dfrac{(x+3)^2}{x+15}=6,\) then the difference between the two possible values of \(x\) is:

by Jay@ManhattanReview » Mon Apr 27, 2020 12:38 am
M7MBA wrote: ↑
Sun Apr 26, 2020 9:33 am
 If \(\dfrac{(x+3)^2}{x+15}=6,\) then the difference between the two possible values of \(x\) is:

A. 0
B. 6
C. 12
D. 18
E. 24

[spoiler]OA=D[/spoiler]

Source: Veritas Prep
Let's expand \(\dfrac{(x+3)^2}{x+15}=6,\). Upon expansion, we get x = ±9

Difference between the two possible values of \(x=9-(-9)=18\)

The correct answer: D

Hope this helps!

-Jay
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Re: If \(\dfrac{(x+3)^2}{x+15}=6,\) then the difference between the two possible values of \(x\) is:

by Scott@TargetTestPrep » Fri May 01, 2020 3:18 pm
M7MBA wrote: ↑
Sun Apr 26, 2020 9:33 am
 If \(\dfrac{(x+3)^2}{x+15}=6,\) then the difference between the two possible values of \(x\) is:

A. 0
B. 6
C. 12
D. 18
E. 24

[spoiler]OA=D[/spoiler]

Source: Veritas Prep
Simplifying, we have:

(x + 3)^2 = 6(x + 15)

x^2 + 6x + 9 = 6x + 90

x^2 = 81

x = 9 or - 9

Therefore, the difference between the two possible values of x is 9 - (-9) = 18.

Answer: D

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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