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Quadratic Equations — Maths STD 10 — Question

Gujarat BoardEnglish MediumSTD 10MathsQuadratic Equations4 Marks
Question
Solve the following quadratic equations by factorization:
$\frac{\text{m}}{\text{n}}\text{x}^2+\frac{\text{n}}{\text{m}}=1-2\text{x}$
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Answer

We have been given
$\Rightarrow\frac{\text{m}}{\text{n}}\text{x}^2+\frac{\text{n}}{\text{m}}=1-2\text{x}$
$\Rightarrow\frac{\text{m}^2\text{x}^2+\text{n}^2}{\text{mn}}=1-2\text{x}$
$\Rightarrow\text{m}^2\text{x}^2+2\text{mnx}+(\text{n}^2-\text{mn})=0$
$\Rightarrow\text{m}^2\text{x}^2+\text{mnx}+\text{mnx}+\big[\text{n}^2-(\sqrt{\text{mn}})^2\big]=0$
$\Rightarrow\text{m}^2\text{x}^2+\text{mnx}+\text{mnx}+(\text{n}+\sqrt{\text{mn}})(\text{n}-\sqrt{\text{mn}})\\+(\text{m}\sqrt{\text{mnx}}-\text{m}\sqrt{\text{mnx}})=0$
$\Rightarrow\big[\text{m}^2\text{x}^2+\text{mnx}+\text{m}\sqrt{\text{mnx}}\big]+\big[\text{mnx}-\text{m}\sqrt{\text{mnx}}\\+(\text{n}+\sqrt{\text{mn}})(\text{n}-\sqrt{\text{mn}})\big]=0$
$\Rightarrow\big[\text{m}^2\text{x}^2+\text{mnx}+\text{m}\sqrt{\text{mnx}}\big]+\big[(\text{mx}(\text{n}-\sqrt{\text{mn}})\\+(\text{n}+\sqrt{\text{mn}})(\text{n}-\sqrt{\text{mn}})\big]=0$
$\Rightarrow(\text{mx})\big(\text{mx}+\text{n}+\sqrt{\text{mn}}\big)+\big(\text{n}-\sqrt{\text{mn}}\big)\\\big(\text{mx}+\text{n}+\sqrt{\text{mn}}\big)=0$
$\Rightarrow\big(\text{mx}+\text{n}+\sqrt{\text{mn}})\big(\text{mx}+\text{n}-\sqrt{\text{mn}}\big)=0$
Therefore,
$\Rightarrow\text{mx}+\text{n}+\sqrt{\text{mn}}=0$
$\Rightarrow\text{mx}=-\text{n}-\sqrt{\text{mn}}$
$\Rightarrow\text{x}=\frac{-\text{n}-\sqrt{\text{mn}}}{\text{m}}$
or, $\text{mx}+\text{n}-\sqrt{\text{mn}}=0$
$\Rightarrow\text{mx}=-\text{n}+\sqrt{\text{mn}}$
$\Rightarrow\text{x}=\frac{-\text{n}+\sqrt{\text{mn}}}{\text{m}}$
Hence, $\text{x}=\frac{-\text{n}-\sqrt{\text{mn}}}{\text{m}}$ or $\text{x}=\frac{-\text{n}+\sqrt{\text{mn}}}{\text{m}}$

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