For $x \in R$ , the number of real roots of the equation $3 x^{2} - 4 | x^{2} - 1 | + x - 1 = 0$ is

Open in App

Solution

Given:
$3 x^{2} - 4 | x^{2} - 1 | + x - 1 = 0$
Case $1 :$ $x^{2} - 1 < 0 \Rightarrow x \in (- 1, 1)$
$∴ 3 x^{2} + 4 (x^{2} - 1) + x - 1 = 0$
$\Rightarrow 7 x^{2} + x - 5 = 0$
$\Rightarrow x = \frac{- 1 \pm \sqrt{141}}{14}$
$\Rightarrow x = \frac{- 1 - \sqrt{141}}{14}, \frac{- 1 + \sqrt{141}}{14}$
Both roots lie in $(- 1, 1)$

Case $2 :$ $x^{2} - 1 \geq 0 \Rightarrow x \in (- \infty, - 1] \cup [1, \infty)$
$∴ 3 x^{2} - 4 (x^{2} - 1) + x - 1 = 0$
$\Rightarrow x^{2} - x - 3 = 0$
$\Rightarrow x = \frac{1 \pm \sqrt{13}}{2}$
$\Rightarrow x = \frac{1 - \sqrt{13}}{2}, \frac{1 + \sqrt{13}}{2}$
Both roots lie in $(- \infty, - 1] \cup [1, \infty)$
Hence, total number of solutions $= 4$

Suggest Corrections

Similar questions

m

17

(x^{2} + x + 1)^{2} - (m - 3) (x^{2} + x + 1) + m = 0, m \in R

4

(x^{2} + x + 1)^{2} - (m - 3) (x^{2} + x + 1) + m = 0, m \in R

m

{(x + \frac{1}{x})}^{2} - 4 = 0

{tan}^{- 1} \sqrt{x (x + 1)} + {sin}^{- 1} \sqrt{x^{2} + x + 1} = \frac{π}{4}

\frac{4 x^{2} + x + 4}{x^{2} + 1} + \frac{x^{2} + 1}{x^{2} + x + 1} = \frac{31}{6}

Join BYJU'S Learning Program