For equation |{x^2}| + |x| - 6 = 0 , roots are

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4-2.Quadratic Equations and Inequations

hard

For the equation $|{x^2}| + |x| - 6 = 0$, the roots are

A

One and only one real number

B

Real with sum one

C

Real with sum zero

D

Real with product zero

Solution

(c) When $x < 0$, $|x| = – x$

Equation is ${x^2} – x – 6 = 0 \Rightarrow x = – 2,\,3$

$\;x < 0,\;\therefore \;x = – 2$is the solution.

When $x \ge 0$,$|x| = x$

$\therefore $ Equation is${x^2} + x – 6 = 0 \Rightarrow x = 2, – 3$

$x \ge 0$, $x = 2$ is the solution.

Hence $x = 2$, $ – 2$ are the solutions and their sum is zero.

Aliter : $|{x^2}| + |x| – 6 = 0$

==> $(|x| + 3)(|x| – 2) = 0$

==> $|x| = – 3$, which is not possible and $|x| = 2$

==> $x = \pm 2$.

Standard 11

Mathematics

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