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Quadratic Equations — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsQuadratic Equations5 Marks
Question
In the following, determine whether the given values are solution of the given equation or not:
$\text{a}^2\text{x}^2-3\text{abx}+2\text{b}^2=0,$ $\text{x}=\frac{\text{a}}{\text{b}},\text{x}=\frac{\text{b}}{\text{a}}$
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Answer

$\text{a}^2\text{x}^2-3\text{abx}+2\text{b}^2=0,$ $\text{x}=\frac{\text{a}}{\text{b}}$ and $\text{x}=\frac{\text{b}}{\text{a}}$Here, $L.H.S. a^2x^2 - 3abx + 2b^2 = 0$ and R.H.S.
Substitute $\text{x}=\frac{\text{a}}{\text{b}}$ and $\text{x}=\frac{\text{b}}{\text{a}}$ in L.H.S.
$\Rightarrow\text{a}^2\Big(\frac{\text{a}}{\text{b}}\Big)^2-3\text{ab}\Big(\frac{\text{a}}{\text{b}}\Big)+2\text{b}^2$ and $\text{a}^2\Big(\frac{\text{b}}{\text{a}}\Big)^2-3\text{ab}\Big(\frac{\text{b}}{\text{a}}\Big)+2\text{b}^2$
$\Rightarrow\text{a}^2\Big(\frac{\text{a}^2}{\text{b}^2}\Big)-3\text{a}\times\text{a}+2\text{b}^2$ and $\text{a}^2\times\frac{\text{b}^2}{\text{a}^2}-3\text{b}\times\text{b}+2\text{b}^2$
$\Rightarrow\frac{\text{a}^4}{\text{b}^2}-3\text{a}^2+2\text{b}^2$ and $\text{b}^2-3\text{b}^2+2\text{b}^2$
$\Rightarrow\frac{\text{a}^4}{\text{b}^2}-3\text{a}^2+2\text{b}^2$ and $3\text{b}^2-3\text{b}^2=0$
$\Rightarrow\neq$ R.H.S. = R.H.S
$\therefore\text{x}=\frac{\text{b}}{\text{a}}$ is a solution and $\text{x}=\frac{\text{a}}{\text{b}}$ is not a solution for the given equation.

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