Matrices and Determinants 4 Question 27
28. Let $\alpha _1, \alpha _2, \beta _1, \beta _2$ be the roots of $a x^{2}+b x+c=0$ and $p x^{2}+q x+r=0$, respectively. If the system of equations $\alpha _1 y+\alpha _2 z=0$ and $\beta _1 y+\beta _2 z=0$ has a non-trivial solution, then prove that $\frac{b^{2}}{q^{2}}=\frac{a c}{p r}$.
(1987, 3M)
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Answer:
Correct Answer: 28. $x=\frac{4-5 k}{7}, y=\frac{13 k-9}{7}, z=k$
Solution:
- Since, $\alpha _1, \alpha _2$ are the roots of $a x^{2}+b x+c=0$.
$\Rightarrow \quad \alpha _1+\alpha _2=-\frac{b}{a}$ and $\alpha _1 \alpha _2=\frac{c}{a}$
Also, $\beta _1, \beta _2$ are the roots of $p x^{2}+q x+r=0$.
$\Rightarrow \quad \beta _1+\beta _2=-\frac{q}{p}$ and $\beta _1 \beta _2=\frac{r}{p}$
Given system of equations
$ \alpha _1 y+\alpha _2 z=0 $
and $\beta _1 y+\beta _2 z=0$, has non-trivial solution.
$\therefore\left|\begin{array}{ll}\alpha _1 & \alpha _2 \\ \beta _1 & \beta _2\end{array}\right|=0 \Rightarrow \frac{\alpha _1}{\alpha _2}=\frac{\beta _1}{\beta _2}$
Applying componendo-dividendo, $\frac{\alpha _1+\alpha _2}{\alpha _1-\alpha _2}=\frac{\beta _1+\beta _2}{\beta _1-\beta _2}$
$\Rightarrow \alpha _1+\alpha _2 \beta _1-\beta _2 =\alpha _1-\alpha _2\beta _1+\beta _2$
$\Rightarrow(\alpha _1+\alpha _2)^{2}{(\beta _1+\beta _2)^{2}-4 \beta _2 \beta _2}$
$ =(\beta _1+\beta _2)^{2}{(\alpha _1+\alpha _2)^{2}-4 \alpha _1 \alpha _2} $
From Eqs. (i) and (ii), we get
$ \frac{b^{2}}{a^{2}} \frac{q^{2}}{p^{2}}-\frac{4 r}{p}=\frac{q^{2}}{p^{2}} \quad \frac{b^{2}}{a^{2}}-\frac{4 c}{a} $
$\Rightarrow \frac{b^{2} q^{2}}{a^{2} p^{2}}-\frac{4 b^{2} r}{a^{2} p}=\frac{b^{2} q^{2}}{a^{2} p^{2}}-\frac{4 q^{2} c}{a p^{2}}$
$\Rightarrow \frac{b^{2} r}{a}=\frac{q^{2} c}{p} \Rightarrow \frac{b^{2}}{q^{2}}=\frac{a c}{p r}$
