Question

Solve:

\frac{x - a^{2} - b^{2}}{c^{2}} + \frac{c^{2}}{x - a^{2} - b^{2}} = 2

Accepted Answer

Given: $\frac{x - a^{2} - b^{2}}{c^{2}} + \frac{c^{2}}{x - a^{2} - b^{2}} = 2$

$\Rightarrow \frac{(x - a^{2} - b^{2}) (x - a^{2} - b^{2}) + c^{2} \times c^{2}}{c^{2} (x - a^{2} - b^{2})} = 2$

$\Rightarrow {(x - a^{2} - b^{2})}^{2} + c^{4} = 2 \times c^{2} (x - a^{2} - b^{2})$

$\Rightarrow {(x - (a^{2} + b^{2}))}^{2} + c^{4} = 2 \times c^{2} (x - a^{2} - b^{2})$

Using the identity: ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$

$\Rightarrow x^{2} + {(a^{2} + b^{2})}^{2} - 2 x (a^{2} + b^{2}) + c^{4} = 2 c^{2} x - 2 a^{2} c^{2} - 2 b^{2} c^{2}$

$\Rightarrow x^{2} + (a^{4} + b^{4} + 2 a^{2} b^{2}) - 2 x a^{2} - 2 x b^{2} + c^{4} - 2 c^{2} x + 2 a^{2} c^{2} + 2 b^{2} c^{2} = 0$

$\Rightarrow x^{2} - 2 x a^{2} - 2 x b^{2} - 2 c^{2} x + a^{4} + b^{4} + 2 a^{2} b^{2} + c^{4} + 2 a^{2} c^{2} + 2 b^{2} c^{2} = 0$

$\Rightarrow x^{2} - 2 x (a^{2} + b^{2} + c^{2}) + (a^{4} + b^{4} + c^{4} + 2 a^{2} b^{2} + 2 a^{2} c^{2} + 2 b^{2} c^{2}) = 0$

Using: $a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a = {(a + b + c)}^{2}$

$\Rightarrow x^{2} - 2 x (a^{2} + b^{2} + c^{2}) + {(a^{2} + b^{2} + c^{2})}^{2} = 0$

Using: $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$

$\Rightarrow {(x - (a^{2} + b^{2} + c^{2}))}^{2} = 0$

$\Rightarrow [x - (a^{2} + b^{2} + c^{2})] = 0$

$∴ x = (a^{2} + b^{2} + c^{2})$

Therefore, the value of $x$ is $(a^{2} + b^{2} + c^{2})$ .

Solve:
$\frac{x - a^{2} - b^{2}}{c^{2}} + \frac{c^{2}}{x - a^{2} - b^{2}} = 2$

Important Questions on Miscellaneous Exercise : Algebra

Learn and Prepare for any exam you want !

Solve: x-a2-b2c2+c2x-a2-b2=2

Important Questions on Miscellaneous Exercise : Algebra

Learn and Prepare for any exam you want !

Solve:
$\frac{x - a^{2} - b^{2}}{c^{2}} + \frac{c^{2}}{x - a^{2} - b^{2}} = 2$