Solve the quadratic equation by factorization …
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Sia ? 6 years, 7 months ago
We have,
{tex}\frac{{x + 1}}{{x - 1}} - \frac{{x - 1}}{{x + 1}} = \frac{5}{6}{/tex}
{tex}\Rightarrow{/tex} {tex}\frac{(x+1)^2-(x-1)^2}{(x-1)(x+1)}= \frac56{/tex}
{tex}\Rightarrow{/tex} (x2 + 1 + 2x) - (x2 + 1 - 2x) = {tex}\frac{5}{6}{/tex}(x2 - 12)
{tex}\Rightarrow{/tex} x2 + 1 + 2x - x2 - 1 + 2x = {tex}\frac{5}{6}{/tex}(x2 - 1)
{tex}\Rightarrow{/tex} 4x = {tex}\frac{5}{6}{/tex}(x2 - 1)
{tex}\Rightarrow{/tex} 24x = 5(x2 - 1)
{tex}\Rightarrow{/tex} 24x = 5x2 - 5
{tex}\Rightarrow{/tex} 5x2 - 24x - 5 = 0
{tex}\Rightarrow{/tex} 5x2 - 25x + 1x - 5 = 0
{tex}\Rightarrow{/tex} 5x(x - 5) + 1(x - 5) = 0
{tex}\Rightarrow{/tex} (x - 5)(5x + 1) = 0
{tex}\Rightarrow{/tex} x - 5 = 0 or 5x + 1 = 0
{tex}\Rightarrow{/tex} x = 5 or {tex}x = - \frac { 1 } { 5 }{/tex}
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