The product of all real roots of the equation | vedclass

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Question

(A) Given equation: $x^2 - |x| - 6 = 0$Case $1$: If $x \ge 0$,then $|x| = x$. The equation becomes $x^2 - x - 6 = 0$.$(x - 3)(x + 2) = 0$.This gives $

Vedclass · Accepted Answer

$-9$

Vedclass · Answer

(A) Given equation: $x^2 - |x| - 6 = 0$Case $1$: If $x \ge 0$,then $|x| = x$. The equation becomes $x^2 - x - 6 = 0$.$(x - 3)(x + 2) = 0$.This gives $x = 3$ or $x = -2$. Since $x \ge 0$,we have $x = 3$.Case $2$: If $x $(x + 3)(x - 2) = 0$.This gives $x = -3$ or $x = 2$. Since $x The real roots are $3$ and $-3$.The product of all real roots is $3 	imes (-3) = -9$.

The product of all real roots of the equation $x^2 - |x| - 6 = 0$ is

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