The roots of the equation {x^4} - 8{x^2} - 9 | vedclass

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Question

(B) Given equation: ${x^4} - 8{x^2} - 9 = 0$Let ${x^2} = y$. Then the equation becomes ${y^2} - 8y - 9 = 0$.Factoring the quadratic equation: ${y^2} -

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$\pm 3, \pm i$

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(B) Given equation: ${x^4} - 8{x^2} - 9 = 0$Let ${x^2} = y$. 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 back ${x^2} = y$:Case $1$: ${x^2} = 9 \Rightarrow x = \pm 3$.Case $2$: ${x^2} = -1 \Rightarrow x = \pm \sqrt{-1} = \pm i$.Thus,the roots are $\pm 3, \pm i$.

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

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