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If (x - 3)^2/(2x + 15) = 3, what is the product of the possible values
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Bunuel
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If (x - 3)^2/(2x + 15) = 3, what is the product of the possible values [#permalink] New post  10 Aug 2018, 02:55
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If \(\frac{(x - 3)^2}{(2x + 15)} = 3\), what is the product of the possible values of x?


A. −36
B. −6
C. 4
D. 6
E. 36
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sudarshan22
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If (x - 3)^2/(2x + 15) = 3, what is the product of the possible values [#permalink] New post  Updated on: 10 Aug 2018, 12:21
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Bunuel wrote:
If \(\frac{(x - 3)^2}{(2x + 15)} = 3\), what is the product of the possible values of x?

(x - 3)^2 / (2x + 15) = 3
x^2 - 6x + 9 = 6x + 45
x^2 - 12x - 36 = 0
Product of roots of equation of form ax^2+bx+c = 0 is c/a
Therefore,
Product = -36/1 = -36

Hence, A.
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Originally posted by sudarshan22 on 10 Aug 2018, 03:21.
Last edited by sudarshan22 on 10 Aug 2018, 12:21, edited 2 times in total.
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Re: If (x - 3)^2/(2x + 15) = 3, what is the product of the possible values [#permalink] New post  10 Aug 2018, 03:22
Bunuel wrote:
If \(\frac{(x - 3)^2}{(2x + 15)} = 3\), what is the product of the possible values of x?


A. −36
B. −6
C. 4
D. 6
E. 36


\(\frac{(x - 3)^2}{(2x + 15)} = 3\)
Or,\(\frac{\left(x-3\right)^2}{2x+15}\left(2x+15\right)=3\left(2x+15\right)\)
Or, \(\left(x-3\right)^2=3\left(2x+15\right)\)
Or, \(x=6\left(1+\sqrt{2}\right),\:x=6\left(1-\sqrt{2}\right)\)

\(Product=36*(1+\sqrt{2})(1+\sqrt{2})=36*(1^2-(\sqrt{2})^2)=36(1-2)=36*(-1)=-36\)

Ans. (A)
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Re: If (x - 3)^2/(2x + 15) = 3, what is the product of the possible values [#permalink] New post  10 Aug 2018, 03:29
sudarshan22 wrote:
Bunuel wrote:
If \(\frac{(x - 3)^2}{(2x + 15)} = 3\), what is the product of the possible values of x?

(x - 3)^2 / (2x + 15) = 3
x^2 - 6x + 9 = 6x + 45
x^2 - 12x + 36 = 0
On solving x = 6, -6
Product = -36

Hence, A.



Bro, how did u get + 36.
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sudarshan22
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Re: If (x - 3)^2/(2x + 15) = 3, what is the product of the possible values [#permalink] New post  10 Aug 2018, 03:33
selim wrote:
sudarshan22 wrote:
Bunuel wrote:
If \(\frac{(x - 3)^2}{(2x + 15)} = 3\), what is the product of the possible values of x?

(x - 3)^2 / (2x + 15) = 3
x^2 - 6x + 9 = 6x + 45
x^2 - 12x + 36 = 0
On solving x = 6, -6
Product = -36

Hence, A.



Bro, how did u get + 36.


Typo it was, answer is still the same. Thanks again.
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If (x - 3)^2/(2x + 15) = 3, what is the product of the possible values [#permalink] New post  10 Aug 2018, 03:42
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PKN wrote:
Bunuel wrote:
If \(\frac{(x - 3)^2}{(2x + 15)} = 3\), what is the product of the possible values of x?


A. −36
B. −6
C. 4
D. 6
E. 36


\(\frac{(x - 3)^2}{(2x + 15)} = 3\)
Or,\(\frac{\left(x-3\right)^2}{2x+15}\left(2x+15\right)=3\left(2x+15\right)\)
Or, \(\left(x-3\right)^2=3\left(2x+15\right)\)
Or, \(x=6\left(1+\sqrt{2}\right),\:x=6\left(1-\sqrt{2}\right)\)

\(Product=36*(1+\sqrt{2})(1+\sqrt{2})=36*(1^2-(\sqrt{2})^2)=36(1-2)=36*(-1)=-36\)

Ans. (A)

Instead of finding roots and multiplying them, we can get product directly..

x^2-6x+9 = 6x+45
x^2-12x-36 = 0
Product of roots for equation ax^2+bx+c=0 is c/a
So, product of roots = -36
Hence option A
This can save time in exam

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If (x - 3)^2/(2x + 15) = 3, what is the product of the possible values [#permalink] New post  10 Aug 2018, 05:16
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I am confused sudarshan22. Thought the roots should be 6(1+√2) and 6(1-√2) with product of -36 (A). Not 6 and -6 even though their product is -36.

sudarshan22 wrote:
Bunuel wrote:
If \(\frac{(x - 3)^2}{(2x + 15)} = 3\), what is the product of the possible values of x?

(x - 3)^2 / (2x + 15) = 3
x^2 - 6x + 9 = 6x + 45
x^2 - 12x - 36 = 0
On solving x = 6, -6
Product = -36

Hence, A.
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Bismarck
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If (x - 3)^2/(2x + 15) = 3, what is the product of the possible values [#permalink] New post  10 Aug 2018, 05:53
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funsogu wrote:
I am confused sudarshan22. Thought the roots should be 6(1+√2) and 6(1-√2) with product of -36 (A). Not 6 and -6 even though their product is -36.

sudarshan22 wrote:
Bunuel wrote:
If \(\frac{(x - 3)^2}{(2x + 15)} = 3\), what is the product of the possible values of x?

(x - 3)^2 / (2x + 15) = 3
x^2 - 6x + 9 = 6x + 45
x^2 - 12x - 36 = 0
On solving x = 6, -6
Product = -36

Hence, A.

funsogu
you are right
\(\frac{(x - 3)^2}{(2x + 15)} = 3\)
\((x - 3)^2=3(2x + 15)\)
\(x^2 + 9 -6x =6x +45\)
\(x^2 + 9-45 -6x-6x =0\)
\(x^2 -12x -36 =0\)

Roots \(=\frac {-b±\sqrt{b^2-4ac}}{2a}\) where \(ax^2+bx+c=0\)
\(a=1,b=-12,c=-36\)
Roots\(= \frac{12+ \sqrt{(-12)^2 - 4*1*(-36)}}{2*1}; \frac{12- \sqrt{(-12)^2 - 4*1*(-36)}}{2*1}\)
Roots\(= \frac{12+ \sqrt{2*12^2}}{2}; \frac{12- \sqrt{2*12^2}}{2}\)
Roots\(= 6(1+ \sqrt{2});6 (1- \sqrt{2})\)

It is better to follow the method as followed by pradeep15793


Quote:
Instead of finding roots and multiplying them, we can get product directly.
\(x^2-6x+9 = 6x+45\)
\(x^2-12x-36 = 0\)
Product of roots for equation \(ax^2+bx+c=0\) is \(\frac{c}{a}\)
So, product of roots\(= -36\)
Hence option A
This can save time in exam
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Abhishek009
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Re: If (x - 3)^2/(2x + 15) = 3, what is the product of the possible values [#permalink] New post  10 Aug 2018, 06:53
Bunuel wrote:
If \(\frac{(x - 3)^2}{(2x + 15)} = 3\), what is the product of the possible values of x?


A. −36
B. −6
C. 4
D. 6
E. 36

\(\frac{(x - 3)^2}{(2x + 15)} = 3\).

Or, \(x^2 - 6x + 9 = 6x + 45\)

Or, \(x^2 - 12x -36 =\)

x must be +6 & -6 , Thus the product of x will be - 36 , Answer must be (A) -36
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Re: If (x - 3)^2/(2x + 15) = 3, what is the product of the possible values [#permalink] New post  10 Aug 2018, 12:24
funsogu Bismarck
Thanks for rectifying the blunder, it was a mistake indeed.
As mentioned Pradeep's approach seems to to be more legit, I have updated my post accordingly.
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Re: If (x - 3)^2/(2x + 15) = 3, what is the product of the possible values [#permalink] New post  13 Mar 2021, 13:48
Solving, final equation reduces to: x^2-12x-36=0
Apply c/a which is the product for any quadratic equation, gives -36. So A

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Re: If (x - 3)^2/(2x + 15) = 3, what is the product of the possible values [#permalink]
13 Mar 2021, 13:48
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