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

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

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?

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

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

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

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!!!

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

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.

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 [(6\(x^2\)-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[(6\(x^2\)-1) – 5*7] = 0.

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

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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