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Quadratic Equations — Maths STD 10 — Question

Gujarat BoardEnglish MediumSTD 10MathsQuadratic Equations4 Marks
Question
Solve the following quadratic equations by factorization:
$7\text{x}+\frac{3}{\text{x}}=35\frac{3}{5}$
✓

Answer

We have been given,
$7\text{x}+\frac{3}{\text{x}}=35+\frac{3}{5}$
$7\text{x}^2+3=\Big(35+\frac{3}{5}\Big)\text{x}$
$7\text{x}^2-\Big(35+\frac{3}{5}\Big)\text{x}+3=0$
Therefore,
$7\text{x}^2-35\text{x}-\frac{3}{5}\text{x}+3=0$
$7\text{x}(\text{x}-5)-\frac{3}{5}(\text{x}-5)=0$
$\Big(7\text{x}-\frac{3}{5}\Big)(\text{x}-5)=0$
Therefore,
$7\text{x}-\frac{3}{5}=0$
$7\text{x}=\frac{3}{5}$
$\text{x}=\frac{3}{35}$
or, $\text{x}-5=0$
$\text{x}=5$
Hence, $\text{x}=\frac{3}{35}$ or $\text{x}=5$

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