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Quadratic Equations — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsQuadratic Equations5 Marks
Question
Solve the following quadratic equations by factorization:
$\frac{5+\text{x}}{5-\text{x}}-\frac{5-\text{x}}{5+\text{x}}=3\frac{3}{4},$ $\text{x}\neq5,-5$
✓

Answer

$\frac{5+\text{x}}{5-\text{x}}-\frac{5-\text{x}}{5+\text{x}}=3\frac{3}{4}$
$\frac{(5+\text{x})^2-(5-\text{x})^2}{(5-\text{x})(5+\text{x})}=\frac{15}{4}$
$\frac{25+\text{x}^2+10\text{x}-25-\text{x}^2+10\text{x}}{25-\text{x}^2}=\frac{15}{4}$
$\frac{20\text{x}}{25-\text{x}^2}=\frac{15}{4}$
$\Rightarrow 80x = 375 - 15x^2$
$\Rightarrow 15x^2 + 80x - 375 = 0$
$\Rightarrow 3x^2 + 16x - 75 = 0$
$\Rightarrow 3x^2 + 25x - 9x - 75 = 0$
$\Rightarrow x(3x + 25) - 3(3x + 25) = 0$
$\Rightarrow (x - 3)(3x + 25) = 0$
Either x - 3 = 0
$\therefore$ x = 3
or 3x + 25 = 0
⇒ 3x = -25
$\text{x}=-\frac{25}{3}$

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