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} $
$\Rightarrow \frac{b^{2} q^{2}}{a^{2} p^{2}}-\frac{4 b^{2} r}{a^{2} p}=\frac{b^{2} q^{2}}{a^{2} p^{2}}-\frac{4 q^{2} c}{a p^{2}}$
$\Rightarrow \frac{b^{2} r}{a}=\frac{q^{2} c}{p} \Rightarrow \frac{b^{2}}{q^{2}}=\frac{a c}{p r}$
