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Quadratic Equations — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsQuadratic Equations5 Marks
Question
Solve the following quadratic equations by factorization:
$3\Big(\frac{7\text{x}+1}{5\text{x}-3}\Big)-4\Big(\frac{5\text{x}-3}{7\text{x}+1}\Big)=11,$ $\text{x}\neq\frac{3}{5},-\frac{1}{7}$
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Answer

$\Rightarrow3\Big(\frac{7\text{x}+1}{5\text{x}-3}\Big)-4\Big(\frac{5\text{x}-3}{7\text{x}+1}\Big)=11$
$\Rightarrow\frac{3(7\text{x}+1)^2-4(5\text{x}-3)^2}{(5\text{x}-3)(7\text{x}+1)}=11$
$\Rightarrow\frac{3(49\text{x}^2+1+14\text{x})-4(25\text{x}^2+9-30\text{x})}{35\text{x}^2+5\text{x}-21\text{x}-3}=11$
$\Rightarrow\frac{147\text{x}^2+3+42\text{x}-100\text{x}^2-36+120\text{x}}{35\text{x}^2-16\text{x}-3}=11$
$\Rightarrow47\text{x}^2+162\text{x}-33\\=11(35\text{x}^2-16\text{x}-3)$
$\Rightarrow47\text{x}^2+162\text{x}-33\\=385\text{x}^2-176\text{x}-33$
$\Rightarrow385\text{x}^2-47\text{x}^2-176\text{x}\\-162\text{x}-33+33=0$
$\Rightarrow338\text{x}^2-338\text{x}=0$
$\Rightarrow\text{x}^2-\text{x}=0$
$\Rightarrow\text{x}(\text{x}-1)=0$
$\Rightarrow\text{x}=0$ or $\text{x}-1=0$
$\Rightarrow\text{x}=0$ or $\text{x}=1$
Hence, the factors are 0 and 1.

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