# CAT 2019 Set-1 | Question: 99

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The number of solutions to the equation $|x|(6x^{2}+1)=5x^{2}$ is _________

Given that, $\left | x \right | (6x^{2}+1) = 5x^{2}$

We know that, $\left | x \right | = \left\{\begin{matrix} x\; ;& x \geq 0 \\ -x \; ;& x<0 \end{matrix}\right.$

$\textbf{Case 1:} \; x \geq 0$

Now, $x (6x^{2} + 1) = 5x^{2}$

$\Rightarrow 6x^{3} + x = 5x^{2}$

$\Rightarrow 6x^{3} – 5x^{2} + x = 0$

$\Rightarrow x( 6x^{2} -5x +1) = 0$

$\Rightarrow x ( 6x^{2} -3x – 2x +1) = 0$

$\Rightarrow x [3x (2x-1) -1(2x-1)] = 0$

$\Rightarrow x [ (3x-1)(2x-1)] = 0$

$\Rightarrow x=0 , (3x-1)=0 , (2x-1)=0$

$\boxed{x=0,x=\frac{1}{3},x=\frac{1}{2}}$

$\therefore$ The number of solutions $=3$

$\textbf{Case 2:}\; x < 0$

Now, $-x(6x^{2}+1) = 5x^{2}$

$\Rightarrow – 6x^{3} – x – 5x^{2} = 0$

$\Rightarrow 6x^{3} + 5x^{2} + x = 0$

$\Rightarrow x (6x^{2} + 5x + 1) = 0$

$\Rightarrow x (6x^{2} + 3x + 2x + 1) = 0$

$\Rightarrow x [3x(2x+1) +1(2x+1)] = 0$

$\Rightarrow x [(3x+1)(2x+1)] = 0$

$\Rightarrow x=0,(3x+1)=0,(2x+1)=0$

$\boxed {x=0,x=\frac{-1}{3},x=\frac{-1}{2}}$

$\therefore$ The number of solutions $=3$

So, the total number of solutions $= 2 + 3 = 5.$

Correct Answer $:5$

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