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Quadratic Equations — Maths STD 10 — Question

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

Answer

$3\Big(\frac{\text{7x}+1}{\text{5x}-3}\Big)-4\Big(\frac{\text{5x}-3}{\text{7x}+1}\Big)=11$
Taking $\frac{\text{7x}+1}{\text{5x}-3}=\text{y},$ we have
$\text{3y}-\frac{4}{\text{y}}=11$
$\Rightarrow\frac{\text{3y}^2-4}{\text{y}}=11$
$\Rightarrow 3y^2 - 4 = 11y$
$\Rightarrow 3y^2 - 11y - 4 = 0$
$\Rightarrow 3y^2 - 12y + y - 4 = 0$
$\Rightarrow 3y(y - 4) + 1(y - 4) = 0$
$\Rightarrow (y - 4)(3y + 1) = 0$
$\Rightarrow y - 4 = 0 or 3y + 1 = 0$
⇒ y = 4 or $\text{y}=\frac{-1}{3}$
$\Rightarrow\frac{\text{7x}+1}{\text{5x}-3}=4$ or $\frac{\text{7x}+1}{\text{5x}-3}=\frac{-1}{3}$
$\Rightarrow 7x + 1 = 20x - 12 or 21x + 3 = -5x + 3$
$\Rightarrow 13x = 13 or 26x = 0$
$\Rightarrow x = 1 or x = 0$

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