Quadratic equation, only one real root.

Algebra Level 4

$\dfrac{2a^2+x^2}{a^3-x^3}-\dfrac{2x}{ax+a^2+x^2}+\dfrac{1}{x-a}=0$

Find the real values of $a$ for which there exists a unique real solution to the above equation.

(-1,\infty)

(-\infty,1)

{0}

\mathbb{R} -\{0\}

(-1,1)

1 solution

U Z
Feb 9, 2015

$\dfrac{2a^2 + x^2}{a^3 - x^3} - \dfrac{2x(a - x)}{(a - x)(a^2 + ax + x^2)} + \dfrac{(a^2 + ax + x^2)}{(x - a)(a^2 + ax + x^2)} = 0$

$\dfrac{2a^2 + x^2}{a^3 - x^3} - \dfrac{2x(a - x)}{a^3 - x^3} + \dfrac{(a^2 + ax + x^2)}{x^3 - a^3} = 0$

$\dfrac{ 2a^2 + x^2 - 2ax + 2x^2 - ( a^2 + ax + x^2)}{a^3 - x^3} = 0$

$\dfrac{ a^2 - 2ax + x^2 + (x^2 - ax)}{a^3 - x^3} = 0$

$\dfrac{ (a - x)^2 - x(a - x)}{a^3 - x^3} =0$

$\implies a^3 - x^3 \neq 0$

$\implies (a - x)^2 - x(a - x) = 0$

$(a - x)( a - 2x) = 0$

$\implies a = 2x (a=2x \neq 0)$

$Thus~ Domain ~of~a~=~ \mathbb R - \{0\}$

brilliant solution

Ayush Verma - 6 years, 4 months ago

Nice. Specially the last four lines.

Niranjan Khanderia - 6 years, 3 months ago

Good solution...

B.s. Ashwin - 6 years, 3 months ago