Theory of Equations 1 Question 50
51. For $a \leq 0$, determine all real roots of the equation
$ x^{2}-2 a|x-a|-3 a^{2}=0 $
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Answer:
Correct Answer: 51. $(x={a(1-\sqrt{2}), a(\sqrt{6}-1)})$
Solution:
- Here, $a \leq 0$
Given, $\quad x^{2}-2 a|x-a|-3 a^{2}=0$
Case I When $x \geq a$
$ \begin{aligned} \Rightarrow & x^{2}-2 a(x-a)-3 a^{2} =0 \\ \Rightarrow & x^{2}-2 a x-a^{2} =0 \\ \Rightarrow & x =a \pm \sqrt{2} a \end{aligned} $
[as $a(1+\sqrt{2})<a$ and $a(1-\sqrt{2})>a$ ]
$\therefore$ Neglecting $x=a(1+\sqrt{2})$ as $x \geq a$
$ \Rightarrow \quad x=a(1-\sqrt{2}) $
Case II When $x<a \Rightarrow x^{2}+2 a(x-a)-3 a^{2}=0$
$ \Rightarrow \quad x^{2}+2 a x-5 a^{2}=0 \Rightarrow x=-a \pm \sqrt{6} a $
$ \text { [as } a(\sqrt{6}-1)<a \text { and } a(-1-\sqrt{6})>a \text { ] } $
$\therefore$ Neglecting $x=a(-1-\sqrt{6}) \Rightarrow x=a(\sqrt{6}-1)$
From Eqs. (i) and (ii), we get
$ x={a(1-\sqrt{2}), a(\sqrt{6}-1)} $
