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4-2.Quadratic Equations and Inequations
medium
The roots of $|x - 2{|^2} + |x - 2| - 6 = 0$are
A
$0, 4$
B
$-1, 3$
C
$4, 2$
D
$5, 1$
Solution
(a) When $x < 2,$ ${(x – 2)^2} – (x – 2) – 6 = 0$
$ \Rightarrow {x^2} – 4x + 4 – x + 2 – 6 = 0$$ \Rightarrow $ ${x^2} – 5x = 0$
$ \Rightarrow $ $x(x – 5) = 0$ $ \Rightarrow $ $x = 0$
When $x \ge 2$; ${(x – 2)^2} + (x – 2)\, – 6 = 0$
$ \Rightarrow {x^2} – 4x + 4 + x – 2 – 6 = 0$
$ \Rightarrow $ ${x^2} – 3x – 4 = 0$
$ \Rightarrow $ ${x^2} – 4x + x – 4 = 0$
$ \Rightarrow $ $(x – 4)(x + 1) = 0$
$ \Rightarrow $ $x = 4$.
Standard 11
Mathematics
