1.

For \( a \leq 0 \), the roots of the equation \( x^{2}-2 a|x-a|-3 a^{2}=0 \)

Answer»

\(x^2 - 2a |x - a| - 3a^2 = 0\)

Case I: \(x < a \)

then \(|x - a| = -(x - a)\) 

\(\therefore x^2 - 2a|x - a| -3a^2 = 0\)

⇒ \(x^2 + 2a(x - a) - 3a^2 = 0\) 

⇒ \(x^2 + 2ax - 5a^2 = 0\)

⇒ \(x = \frac {-2a \pm \sqrt{4a^2 + 20a^2}}{2}\)

\(= \frac {-2a \pm 2\sqrt 6 a}2\)

\(= ( - 1 \pm \sqrt 6)a\)  \((\because a \le 0)\)

(\(\therefore\) only possibility \(( - 1 \pm \sqrt 6)a\))

Case II: \(x > a\)

then \(|x - a| = x - a\)

\(\therefore x^2 - 2a|x -a| - 3a^2 = 0\)

⇒ \( x^2 - 2a(x -a)- 3a^2 = 0\) 

⇒ \(x^2 - 2ax + 2a^2 - 3a^2 = 0\)

⇒ \(x^2 - 2ax - a^2 = 0\) 

⇒ \(x = \frac {2a \pm \sqrt{4a^2 + 4a^2}}{2} \)

\(= \frac {2a \pm 2\sqrt 2a}2\)

\(= a\pm \sqrt 2 a\)

\(= (1\pm \sqrt 2)a\)

only possibility is \(( 1- \sqrt2)a\) as \(x > a\) & \(a \le0\).

Hence, possible roots are \(( - 1 \pm \sqrt 6)a\) & \(( 1- \sqrt2)a\).


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