M02-13

Question

If  y = \sqrt{64}  and  x^2 - 10x = (-4y^3 + 64y)*\frac{1}{96} , what is the minimum possible value of  x ?

A. 2
B. 4
C. 8
D. 12
E. 16

Bunuel · Accepted Answer

No. When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is: \sqrt{9} = 3 , NOT +3 or -3; \sqrt {16} = 2 , NOT +2 or -2; Notice that in contrast, the equation x^2 = 9 has TWO solutions, +3 and -3. Because x^2 = 9 means that x =-\sqrt{9}=-3 or x=\sqrt{9}=3 . Theory on Number Properties: math-number-theory-88376.html Tips on Numper Properties: number-properties-tips-and-hints-174996.html All DS Number Properties Problems to practice: search.php?search_id=tag&tag_id=38 All PS Number Properties Problems to practice: search.php?search_id=tag&tag_id=59 Hope it helps.

Bunuel · Answer

Official Solution:

If  y = \sqrt{64}  and  x^2 - 10x = (-4y^3 + 64y)*\frac{1}{96} , what is the minimum possible value of  x ?

A. 2
B. 4
C. 8
D. 12
E. 16

y = \sqrt{64}=8  (recall that  √  symbol always means non-negative root). Plug this into the second equation:
 
 x^2-10x = \frac{-4*8^3 + 64*8}{96}  
 
 x^2-10x = \frac{-4*8^3 + 8^3}{96}  
 
 x^2-10x = \frac{-3*8^3}{8*4*3} 
 
 x^2-10x = -16 
 
 x^2-10x +16 = 0 
 
 (x-8)(x-2)=0 
 
The possible values of  x  are 8 and 2. Therefore the minimum value of  x  is 2.

Answer: A

healthjunkie · Answer

Couldn't Y also equal -8? I know that when you plug in y=-8 you end up getting x^2 - 10x -16 = 0 which you can't factor out, but I didnt realize that until after I plugged in -8 (I plugged in -8 since we were trying to find the smallest value of X, I figured a negative Y would give us smaller values for X)

adityadon · Answer

Hi Bunuel ,

Please correct me if i amwrong

Does this mean when gmat gives sqare root sign itself in question stem , then it means we should consider only positive root 
and on the other hand gmat gave me a quadratic eqaution in question stem and when i solve it i should consider both +ve and - ve root ?

right ?

Bunuel · Answer

\sqrt{}  sign always means non-negative root.

In contrast x^2 = 4 gives two solutions: 2 and -2.

damduy · Answer

I think this is a  high-quality  question and I agree with explanation.

Prostar · Answer

How did you get The numerator to -3*8^3? ->>>  x^2-10x = \frac{-4*8^3 + 8^3}{8*4*3} = \frac{-3*8^3}{8*4*3} = -16

Bunuel · Answer

It's basic manipulation:  -4*8^3 + 8^3 = 8^3(-4+1) = 3*8^3 . The same way -4x + x = -3x.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

Ash9910 · Answer

why are we taking the roots when the question's asked about minimum value of the quadratic equation?

Bunuel · Answer

Please review the question:

x  and  y  are positive integers. If  y = \sqrt{64}  and  x^2 - 10x = (-4y^3 + 64y)*\frac{1}{96} ,  what is the minimum possible value of  x  ?

OneAttemptFzLLC · Answer

I like the solution - it’s helpful.

prachz · Answer

hello, why couldn't I factor out the x on LHS? I reached till x^2 - 10x = -16, but instead of making it a quadratic equation, I did x(x-10) = -16 and solved for 2 values, -16 and -6. why is this approach wrong?

Bunuel · Answer

How did you go from x(x - 10) = -16 to x = -16 or x = -6? You cannot set x and (x - 10) equal to -16 separately, because -16 is not 0. If you plug in x = -16 or x = -6, neither value satisfies the equation.

M02-13

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