Solve the following equations.
$x^{4} - 3 x^{2} + 2 = 0$ , roots are

x = \pm \sqrt{3}, \pm 1

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x = \pm \sqrt{2}, \pm 1

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x = \pm \sqrt{2}, \pm 3

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none of the above

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Solution

The correct option is A $x = \pm \sqrt{2}, \pm 1$
Given equation is $x^{4} - 3 x^{2} + 2 = 0$
$x^{4} - 2 x^{2} - x^{2} + 2 = 0$
$(x^{2} - 1) (x^{2} - 2) = 0$
$∴ x^{2} = 2 \Rightarrow x = \pm \sqrt{2}$

$∴ x^{2} = 1 \Rightarrow x = \pm 1$

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Similar questions

x^{2} - (\sqrt{3} + 1) x + \sqrt{3} = 0

Solve the following pair of equations:
$\frac{5}{x - 1} + \frac{1}{y - 2} = 2$
$\frac{6}{x - 1} - \frac{3}{y - 2} = 1$
(where $x \neq 1, y \neq 2)$

\frac{5}{x - 1} + \frac{1}{y - 2} = 2

\frac{6}{x - 1} - \frac{3}{y - 2} = 1

x \neq 1, y \neq 2)

3, 1

\frac{3}{x + 1} - \frac{1}{2} = \frac{2}{3 x - 1}, x \neq - 1, \frac{1}{3}

- 2, 1

\frac{4}{x} - 3 = \frac{5}{2 x + 3}, x \neq 0, - \frac{3}{2}

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