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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
