How many value of x satisfy |x|(6x^2 + 1 ) = 5x^2 ?

Question

How many value of x satisfy  |x|(6x^2 + 1 ) = 5x^2  ?

A. 2
B. 3
C. 4
D. 5
E. 6

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GMATinsight · Accepted Answer

|x|(6x^2 + 1 ) = 5x^2

Clearly 0 satisfies the equation so keeping it aside and now

Dividing by |x| both sides we get

i.e.  (6x^2 + 1 ) = 5|x|

let, |x| = y

(6y^2 + 1 ) = 5y

(6y^2 -5y + 1 ) = 0 
i.e.  (6y^2 - 3y - 2y + 1 ) = 0

i.e.  (3y-1)(2y-1)=0

i.e. y = 1/2 and 1/3

i.e. x =  + 1/3 and  + 1/2 and 0

i.e. 5 values

Answer: Option D

GMATWhizTeam · Answer

To solve this question, we'll have to consider two cases:

|x| * (6x^2 + 1) = 5x^2

Case 1 : x ≥ 0
 x*(6x^2 +1) – 5x^2 = 0 
⟹  x * (6x^2 -5x + 1) = 0 
⟹  x* (3x -1 ) (2x – 1) = 0 
This will give us three values: x =  0, \frac{1}{3}, \frac{1}{2}

Case 2 : x < 0
 -x * (6x^2 + 1) – 5x^2 = 0 
⟹  -x ( 6x^2 +5x + 1) = 0 
⟹  -x (3x + 1) (2x + 1) = 0 
This will also give us three values: x =  0 (discard this, since x < 0),  - \frac{1}{3}, - \frac{1}{2} 
Thus, total possible cases are:  -\frac{1}{2}, -\frac{1}{3}, 0, \frac{1}{3}, \frac{1}{2}

Thus, the correct answer is   Option D

preetamsaha · Answer

case I : let, x ≥ 0 , |x|=x
|x|(6x^2+1)=5x^2
or, x(6x^2+1)=5x^2
or, x(6x^2+1-5x)=0
or,x(3x-1)(2x-1)=0
or,x= 0,1/3,1/2  (all accepted)

case II : let, x<0 , |x|=-x
|x|(6x^2+1)=5x^2
or, -x(6x^2+1)=5x^2
or, x(6x^2+1+5x)=0
or,x(3x+1)(2x+1)=0
or,x=0, -1/3,-1/2  (-1/3,-1/2 accepted)

no of solutions of x= 5 
 correct answer D

Kritisood · Answer

Hi I did as below:

case I : let, x ≥ 0 , |x|=x
|x|(6x^2+1)=5x^2 
or, x(6x^2+1)=5x (cancelling LHS x with RHS x)
or, 6x^2-5x+1=0
or, (3x-1)(2x-1)=0
or,x=1/3,1/2 (all accepted)

case II : let, x<0 , |x|=-x
|x|(6x^2+1)=5x^2
or, -x(6x^2+1)=5x^2 
or, (6x^2+1)=-5x (cancelled LHS x with RHS x)
or, 6x^2+5x+1=0
or, (3x+1)(2x+1)=0
or,x-1/3,-1/2

I am getting only 4 values?

preetamsaha · Answer

Kritisood  in x ≥ 0 case, u have rejected x=0, so considering x=0, you will get 5 values of x.

Kritisood · Answer

Understood. Thanks! 
Could you help if otherwise, the method followed is correct? I can cancel the x from LHS and RHS as I did?

preetamsaha · Answer

Kritisood  since x=0 , u cannot cancel x from both sides, u have to take x out as common .

GMATinsight · Answer

Kritisood We can cancel a variable in an Equation only if the variable is NON-ZERO. If we cancel the variable with less care without anticipating that the variable may assume a value zero then we lose a solution In case of inequation the cancellation of variable gets even more complicated - If variable is > 0, then you can cancel it - If variable is < 0, then you can cancel it but the inequality sign needs to be reversed - If variable can be Zero too then it's preferable that you don't cancel the variable. I hope this helps!!!

kpstudies · Answer

How do you go from 5x^2 on the RHS and then reduce it to 5x when you move it to the LHS?

eakabuah · Answer

Alternate solution

|x|(6x^2+1)=5x^2 
 |x|=\frac{5x^2}{(6x^2+1)} 
square both side to take care of the absolute value
 x^2 = (\frac{5x^2}{(6x^2+1)})^2 
Hence  (x+\frac{5x^2}{(6x^2+1)})(x-\frac{5x^2}{(6x^2+1)})=0 
Implying  (x+\frac{5x^2}{(6x^2+1)})=0  or  (x-\frac{5x^2}{(6x^2+1)})=0 
When  (x+\frac{5x^2}{(6x^2+1)})=0 
Then  x(6x^2+5x+1)=0 
 x(6x^2+3x+2x+1)=0 
 x(3x+1)(2x+1)=0 
Roots are  x=-\frac{1}{2}, x=-\frac{1}{3}, x=0

When  (x-\frac{5x^2}{(6x^2+1)})=0 
then  x(6x^2-5x+1)=0 
 x(3x-1)(2x-1)=0 
Roots are  x=0, \frac{1}{3}, \frac{1}{2}

Combined roots are  -\frac{1}{2}, -\frac{1}{3}, 0, \frac{1}{3}, \frac{1}{2} 
There are 5 Roots.

D is the answer.

CrackverbalGMAT · Answer

Hello kpstudies

That’s a pretty simple step. 
|x| can be substituted with either x or -x. As such, if you say |x| (6 x^2 -1), you are essentially talking about (6 x^2 -1) being multiplied with x or -x.

When 5 x^2  is taken to the LHS, x can be taken as the common factor. This is how it is done:
x(6 x^2  – 1) – 5 x^2  = 0. Taking x common, we have, 
x  x^2 -1) – 5x] = 0.

A simple hack to understand this better is this. Let’s keep (6 x^2  – 1) as is, let’s replace x (outside the brackets) with a number. Say, x = 7. How does the original equation look like? It looks like 7(6 x^2 -1) = 5* 7^2 .

Taking 5* 7^2  on to the LHS, what’s common? 7 is common. So the expression becomes, 
7 x^2 -1) – 5*7] = 0.

A similar thing happens when you replace x with -x on the LHS side. Hope that answers your query.
Thanks!

Crytiocanalyst · Answer

Straight out of the bat we get x=0

let us assume x>0 then through simplification we get 
x=1/2 , x=1/3

Then let us assume the second possibility for x<0 we get 
x=-1/2 , x=-1/3

Therefore the total number of possibilities = 5

Therefore IMO D

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How many value of x satisfy |x|(6x^2 + 1 ) = 5x^2 ?

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How many value of x satisfy |x|(6x^2 + 1 ) = 5x^2 ?

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