First slide

Theory of expressions

Question

If a ≤ 0, then the root of the equation $x^{2} - 2 a | x - a | - 3 a^{2} = 0$

Moderate

A

$(1 - \sqrt{2}) a$
B

$(- 1 + \sqrt{6}) a$
C

$(1 + \sqrt{2}) a$
D

$- (1 + \sqrt{6}) a$

Solution

If $x - a < 0, | x - a | = - (x - a)$
equation becomes $x^{2} - 2 a (x - a) - 3 a^{2} = 0$
$\begin{array}{lc} \Rightarrow & x^{2} + 2 ax - 5 a^{2} = 0 \\ \Rightarrow & x = - (1 + \sqrt{6}) a, (- 1 + \sqrt{6}) a \\ ∵ x < a \leq 0 \end{array}$
$\begin{array}{lr} ∴ & x = (- 1 + \sqrt{6}) a \\ If & x - a \geq 0, | x - a | = x - a \end{array}$
the equation becomes $x^{2} - 2 a (x - a) - 3 a^{2} = 0$
$\begin{aligned} \Rightarrow x^{2} - 2 ax - a^{2} = 0 \\ \Rightarrow x = (1 + \sqrt{2}) a, (1 - \sqrt{2}) a \\ ∵ x \geq a and a \leq 0 \\ x = (1 - \sqrt{2}) a . \end{aligned}$

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