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Linear Equations in One Variable — Maths STD 8 — Question

Maharashtra BoardEnglish MediumSTD 8MathsLinear Equations in One Variable5 Marks
Question
Solve the following equation and also check your result in case:
$\Big[(2\text{x}+3)+(\text{x+5})\Big]^2+\Big[(2\text{x}+3)-(\text{x}+5)\Big]^2=10\text{x}^2+92$
✓

Answer

$\Big[(2\text{x}+3)+(\text{x+5})\Big]^2+\Big[(2\text{x}+3)-(\text{x}+5)\Big]^2=10\text{x}^2+92$ $(3\text{x}+8)^2+(\text{x}-2)^2=10\text{x}^2+92$ $9\text{x}^2+48\text{x}+64+\text{x}^2-4\text{x}+4=10\text{x}^2+92$ $\Big[(\text{a+b})^2=\text{a}^2+\text{b}^2+2\text{ab and}(\text{a-b)}^2=\text{a}^2+\text{b}^2-2\text{ab}\Big]$ $10\text{x}^2-10\text{x}^2+44\text{x}=92-68$ $\text{x}=\frac{24}{44}$ $\text{x}=\frac{6}{11}$ Thus, $\text{x}=\frac{6}{11}$ is the solution of the given equation. Check:Substituting $\text{x}=\frac{6}{11}$ in the given equation, we get:
$\text{L.H.S.}=\Big[(2\times\frac{6}{11}+3)+(\frac{6}{11}+5)\Big]^2\\\\\\\\\\\\\ +\Big[(2\times\frac{6}{11}+3)-(\frac{6}{11}+5)\Big]^2$ $=\Big[(\frac{45}{11})+(\frac{61}{11})\Big]^2+\Big[(\frac{45}{11})-(\frac{16}{11})\Big]^2$ $=\Big(\frac{106}{11}\Big)^2+\Big(\frac{-16}{11}\Big)^2$ $\text{R.H.S.}=10\times\Big(\frac{6}{11}\Big)+92$ $=\frac{360}{121}+92=\frac{11492}{121}$ $\therefore\text{L.H.S.}=\text{R.H.S.}\text{ for x}=\frac{6}{11}$

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