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# If (x - 3)^2/(2x + 15) = 3, what is the product of the possible values

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

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10 Aug 2018, 03:55
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55% (hard)

Question Stats:

66% (02:11) correct 34% (02:08) wrong based on 72 sessions

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

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Updated on: 10 Aug 2018, 13:21
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, 04:21.
Last edited by sudarshan22 on 10 Aug 2018, 13:21, 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]

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10 Aug 2018, 04:22
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]

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10 Aug 2018, 04:29
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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Re: If (x - 3)^2/(2x + 15) = 3, what is the product of the possible values  [#permalink]

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10 Aug 2018, 04:33
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]

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10 Aug 2018, 04:42
2
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]

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10 Aug 2018, 06:16
1
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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If (x - 3)^2/(2x + 15) = 3, what is the product of the possible values  [#permalink]

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10 Aug 2018, 06:53
1
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$$= 6(1+ \sqrt{2});6 (1- \sqrt{2})$$

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

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10 Aug 2018, 07:53
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]

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10 Aug 2018, 13:24
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] 10 Aug 2018, 13:24
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