(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 $

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(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 $3 \times (-3) = -9$.

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

Medium

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

A
$-9$
B
$6$
C
$9$
D
$36$

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

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