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Quadratic Equations — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsQuadratic Equations5 Marks
Question
Solve the following quadratic equations by factorization:
$\frac{\text{x}-4}{\text{x}-5}+\frac{\text{x}-6}{\text{x}-7}=\frac{10}{3},$ $\text{x}\neq5,-7$
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Answer

$\frac{\text{x}-4}{\text{x}-5}+\frac{\text{x}-6}{\text{x}-7}=\frac{10}{3}$$\Rightarrow\frac{(\text{x}-4)(\text{x}-7)+(\text{x}-6)(\text{x}-5)}{(\text{x}-5)(\text{x}-7)}=\frac{10}{3}$
$\Rightarrow\frac{\text{x}^2-11\text{x}+28+\text{x}^2-11\text{x}+30}{\text{x}^2-12\text{x}+35}=\frac{10}{3}$
$\Rightarrow\frac{2\text{x}^2-22\text{x}+58}{\text{x}^2-12\text{x}+35}=\frac{10}{3}$
$\Rightarrow 3(2x^2 - 22x + 58) = 10(x^2 - 12x + 35)$
$\Rightarrow 6x^2 - 66x + 174 = 10x^2 - 120x + 350$
$\Rightarrow 4x^2 - 54x + 176 = 0$
$\Rightarrow 2x^2 - 27x + 88 = 0$
$\Rightarrow 2x^2 - 11x - 16x + 88 = 0$
$\Rightarrow x(2x - 11) - 8(2x - 11) = 0$
$\Rightarrow (x - 8)(2x - 11) = 0$
$\Rightarrow x - 8 = 0 or 2x - 11 = 0$
⇒ x = 8 or $\text{x}=\frac{11}{2}$
Hence, the factors are 8 and $\frac{11}{2}$

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