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Quadratic Equations — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsQuadratic Equations4 Marks
Question
Solve the following quadratic equations by factorization:
$\frac{1}{\text{x}-2}+\frac{2}{\text{x}-1}=\frac{6}{\text{x}},$ $\text{x}\neq0$
✓

Answer

We have, $\frac{1}{x-2}+\frac{2}{x-1}=\frac{6}{x}, x \neq 0$
$\Rightarrow \frac{(x-1)+2(x-2)}{(x-2)(x-1)}=\frac{6}{x}$
$\Rightarrow \frac{x-1+2 x-4}{x^2-2 x-x+2}=\frac{6}{x}$
$\Rightarrow \frac{3 x-5}{x^2-3 x+2}=\frac{6}{x}$
$\Rightarrow x(3 x-5)=6\left(x^2-3 x+2\right)$
$\Rightarrow 3 x^2-5 x=6 x^2-18 x+12$
$\Rightarrow 3 x^2-18 x+5 x+12=0$
$\Rightarrow 3 x^2-13 x+12=0$
${[\because 3 \times 18=36 \Rightarrow-9 x-4 \text { and }-13=-9-4]}$
$\Rightarrow 3 x^2-9 x-4 x+12=0$
$\Rightarrow 3 x(x-3)-4(x-3)=0$
$\Rightarrow(x-3)(3 x-4)=0$
$\Rightarrow x=3 \text { or } x=\frac{4}{3}$
$\therefore x=3$ and $x=\frac{4}{3}$ are the two roots of the given equation.

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