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Quadratic Equations — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsQuadratic Equations4 Marks
Question
Solve the following quadratic equations by factorization:
$\frac{3}{\text{x}+1}+\frac{4}{\text{x}-1}=\frac{29}{4\text{x}-1},$ $\text{x}\neq-1,-1,\frac{1}{4}$
✓

Answer

$\frac{3}{\text{x}+1}+\frac{4}{\text{x}-1}=\frac{29}{4\text{x}-1}$
$\frac{3\text{x}-3+4\text{x}+4}{(\text{x}^2-1)}=\frac{29}{4\text{x}-1}$
$\Rightarrow\frac{7\text{x}+1}{\text{x}^2-1}=\frac{29}{4\text{x}-1}$
$\Rightarrow (7x + 1)(4x - 1) = 29(x^2 - 1)$
$\Rightarrow 28x^2 - 7x + 4x - 1 = 29x^2 - 29$
$\Rightarrow 29x^2- 29 - 28x^2 + 7x - 4x + 1 = 0$
$\Rightarrow x^2 + 3x - 28 = 0$
$\begin{Bmatrix}\because-28=7\times(-4) \\3=7-4\end{Bmatrix}$
$\Rightarrow x^2 + 7x - 4x - 28 = 0$
$\Rightarrow x(x + 7) - 4(x + 7) = 0$
$\Rightarrow (x + 7)(x - 4) = 0$
Either$ x + 7 = 0$, then $x = -7$
or $x - 4 = 0$, then $x = 4$
$\therefore$ $x = 4, -7$

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