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Quadratic Equations — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsQuadratic Equations4 Marks
Question
Solve for x.
$\frac{\text{x}-1}{\text{x}-2}+\frac{\text{x}-3}{\text{x}-4}=3\frac{1}{3},$ $\text{x}\neq2,4$
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Answer

$\frac{\text{x}-1}{\text{x}-2}+\frac{\text{x}-3}{\text{x}-4}=3\frac{1}{3}$
$\Rightarrow\frac{(\text{x}-1)(\text{x}-4)+(\text{x}-3)(\text{x}-2)}{(\text{x}-2)(\text{x}-4)}=\frac{10}{3}$ [Solving improper fraction]
$\Rightarrow\frac{(\text{x}^2-5\text{x}+4)+(\text{x}^2-5\text{x}+6)}{(\text{x}-2)(\text{x}-4)}=\frac{10}{3}$
This can also be written as,
$\Rightarrow3[2\text{x}^2-10\text{x}+10]=10[(\text{x}-2)(\text{x}-4)]$
$\Rightarrow6\text{x}^2-30\text{x}+30=10\text{x}^2-60\text{x}+80$
By solving them, by taking all to one side, we get
$\Rightarrow(10\text{x}^2-60\text{x}+80)-(6\text{x}^2-30\text{x}+30)=0$
$\Rightarrow4\text{x}^2-30\text{x}+50=0$
Here, $\text{a}=4,\text{b}=-30,\text{c}=50$
Hence we get x by $\text{x} = {-\text{b} \pm \sqrt{\text{b}^2-4\text{ac}} \over 2\text{a}}$
$\Rightarrow\text{x}=-\frac{(-30)+\sqrt{(-30)^2-4\times4\times50}}{2(4)}$
$\Rightarrow\text{x}=\frac{30-\sqrt{900-800}}{8}$
$\Rightarrow\text{x}=\frac{300-\sqrt{100}}{8}=\frac{30-10}{8}$
$\Rightarrow\text{x}=\frac{20}{8}=\frac{5}{2}$
$\Rightarrow\text{x}=-\frac{(-30)+\sqrt{(-30)^2-4\times4\times50}}{2\times4}$
$\Rightarrow\text{x}=\frac{30+\sqrt{900-800}}{8}$
$\Rightarrow\text{x}=\frac{300+\sqrt{100}}{8}=\frac{30+10}{8}$
$\Rightarrow\text{x}=\frac{40}{8}=5$
$\therefore$ The value of c are 5 and $\frac{5}{2}$

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