(B) Given equation: $x^4 - 8x^2 - 9 = 0$ Let $y = x^2$,then the equation becomes $y^2 - 8y - 9 = 0$. Factoring the quadratic equation: $y^2 - 9y + y -

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(B) Given equation: $x^4 - 8x^2 - 9 = 0$ Let $y = x^2$,then the equation becomes $y^2 - 8y - 9 = 0$. Factoring the quadratic equation: $y^2 - 9y + y - 9 = 0$ $y(y - 9) + 1(y - 9) = 0$ $(y + 1)(y - 9) = 0$ So,$y = -1$ or $y = 9$. Substituting $y = x^2$ back: Case $1$: $x^2 = -1 \Rightarrow x = \pm i$ Case $2$: $x^2 = 9 \Rightarrow x = \pm 3$ Thus,the roots are $x = \pm 3, \pm i$.

Class 11 Mathematics 4-2.Quadratic Equations and Inequations Solution of quadratic equations and Nature of roots

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The roots of the equation $x^4 - 8x^2 - 9 = 0$ are

A
$ \pm 3, \pm 1 $
B
$ \pm 3, \pm i $
C
$ \pm 2, \pm i $
D
None of these

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4-2.Quadratic Equations and Inequations

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Solution of quadratic equations and Nature of roots

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The roots of the equation $x^4 - 8x^2 - 9 = 0$ are

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