If 4x^2 + 8x − 2√(4x^2 + 8x − 3) = 6, what is the product of all possi

Question

Competition Mode Question

If  4x^2 + 8x − 2\sqrt{4x^2 + 8x − 3} = 6 , what is the product of all THE possible values of x?

A. -6
B. -3
C. -1
D. 1
E. 3

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freedom128 · Accepted Answer

IF k=√(4x^2 + 8x − 3), THEN k^2 −3 = (4x^2 + 8x −3) −3 = 4x^2 + 8x −6.

4x^2 + 8x −6 − 2√(4x^2 + 8x − 3) = 0 
k^2 −2k −3=0 --> (k-3)(k+1)=0 
k=3 or k=-1 (N.A)

k=√(4x^2 + 8x − 3) = 3 
k^2 = (4x^2 + 8x − 3) = 9 
4x^2 + 8x − 12 = 0 
4(x^2 + 2x − 3) = 0 --> 4(x+3)(x-1)=0
x=-3 or x=1 (both are valid)

The product of all the possible values of x = -3*1 = -3

FINAL ANSWER IS (B)

nick1816 · Answer

Assume  4x^2+8x = k

k-2\sqrt{k-3} = 6

k-6 = 2\sqrt{k-3}

squaring both sides

k^2-12k+36= 4(k-3)

k^2-12k+36= 4k-12

k^2-16k+48=0

(k-4)(k-12) = 0

k=4 or 12

If,  4x^2+8x=4 
 x^2+2x-1=0

Product of roots = -1

If  4x^2+8x=12 
 x^2+2x-3=0

Product of roots = -3

Product of all possible values of x = (-1)*(-3)=3

yashikaaggarwal · Answer

Putting 4x^2+8x = t and solving equation. We are getting t=4&12. Put values in the equation. We are getting four values. Whose product are equal to -3 so B is the OA

lacktutor · Answer

If  4x^{2}+8x−2\sqrt{4x^{2}+8x−3}=6 , what is the product of all THE possible values of  x ?
Let's say that  \sqrt{4x^{2}+8x-3} =a

-->  a -2√a -3=0 
 (√a+1)(√a-3)=0

(√a+1)  is always greater than zero.
-->  (√a-3)=0 
 4x^{2}+8x-3=9 
 4x^{2}+8x-12=0 
 x^{2}+2x-3 = (x- 1)(x+ 3)=0

Let's check the values of x:
 x=1  -->  4*1+ 8*1- 2( \sqrt{4*1+8*1-3}) = 12-2*3=6  (correct)
 x=-3  -->  4*9+ 8*(-3)-2*(\sqrt{4*9+ 8*(-3)-3}) = 36-24 -2*3 =6  (correct)

The product of all values is  1*(-3)= -3 
 The answer is B.

Siddharthshahani · Answer

Seeing the options and plugging in values of x

only 1 and -3 satisfy

Hence,  B

eakabuah · Answer

Let 4x^2+8x=y
then y-6=2√(y-3)
y^2-12y+36=4y-12
y^2-16y+48=0
(y-4)(y-12)=0
Hence y=4 or y=12
so 4x^2+8x=4
x^2+2x-1=0
(x-1)^2=0, hence x=1
and
4x^2+8x=12
x^2+2x-3=0
(x-1)(x+3)=0, hence x=1, x=-3

Product of all possible values of x = 1*-3 =-3

unraveled · Answer

Thanks  chondro48 
Though i marked B but since one of the roots was negative after assuming  4x^2 + 8x  for some variable, i didn't post a solution.

Thanks for clearing the doubt.
Do you know any similar official question that you can share.??

Shobhit7 · Answer

roots of x are 1, -3, (-1+2^1/2) and (-1-2^1/2)
Product is 3

Ans E

unraveled · Answer

chondro48 
Can you point the mistake where it went wrong, particularly the highlighted part. I had the same approach initially.

ajaygaur319 · Answer

In x^2+2*x−1=0, the 2 roots are  -1 + sqrt(2)  and  -1 - sqrt(2)

As,  sqrt(k-3)  must always be positive,  sqrt(2) - 4  and  -sqrt(2) - 4  are negative and the equation will result imaginary values.

freedom128 · Answer

k=4 is simply NOT a solution since it violates the original equation: 
 k-2\sqrt{k-3} = 6 
 4-2\sqrt{4-3} = 6  (NO!)

Whenever we deal with quadratic or square-root problem, always recheck the solution value against the original equation.

ajargaur319's explanation is also correct, but in real timed test, try to look for simpler original equation.

Posted from my mobile device

nick1816 · Answer

I missed a check, which we should do when dealing with square roots in GMAT. 4x^2+8x-3>0 4x^2+8x+4-7>0 4(x+1)^2>7 (x+1)^2>7/4 x+1>\sqrt{7}/2 x<-2.32 or x+1<-\sqrt{7}/2 x>0.32 x should lie in the above range.

unraveled · Answer

Oh..!! That i missed rather i would say that i tried to be smarter. I thought since product of roots is asked for calculating them is not required. Hence just skipped verifying them individually though had a hunch that something is not right. Also, got a bit lucky to choose B.

aksjain2401 · Answer

k can't be 4 because when you substitute k=4 in the equation given in the question, it gives LHS as 2 and RHS is 6 which is impossible so k = 12 is the only possibility .

If 4x^2 + 8x − 2√(4x^2 + 8x − 3) = 6, what is the product of all possi

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If 4x^2 + 8x − 2√(4x^2 + 8x − 3) = 6, what is the product of all possi

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