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Quadratic Equations — Maths STD 10 — Question

Gujarat BoardEnglish MediumSTD 10MathsQuadratic Equations4 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 80 x=375-15 x^2 $
$ \Rightarrow 15 x^2+80 x-375=0 $
$ \Rightarrow 3 x^2+16 x-75=0 $
$ \Rightarrow 3 x^2+25 x-9 x-75=0 $
$ \Rightarrow x(3 x+25)-3(3 x+25)=0 $
$ \Rightarrow(x-3)(3 x+25)=0$
$\text { Either } x-3=0$
$\therefore x = 3$
or $3x + 25 = 0$
$⇒ 3x = -25$
$\text{x}=-\frac{25}{3}$

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