Question

If ${\alpha }_1,{\alpha }_2$ are the roots of the equation $x^2-px+1=0$ and ${\beta }_1,{\beta }_2$ are those of the equation $x^2-qx+1=0$ and vector ${\alpha }_1\hat{i}+{\beta }_1\hat{j}$ is parallel to ${\alpha }_2\hat{i}+{\beta }_2\hat{j}$, then

• A. $p=\pm q$
• B. $p=\pm 2q$
• C. $p=2q$
• D. none of these

Answer 1

Answer: (A)

$p=\pm q$

Root of quadratic eq $a{x}^2+bx+c=0$is

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Root of quadratic eq $x^2-px+1=0$ is ${\alpha }_1,{\alpha }_2$

$\alpha =\frac{p\pm \sqrt{p^2-4}}{2}$

${\alpha }_1=\frac{p+\sqrt{p^2-4}}{2}$

${\alpha }_2=\frac{p-\sqrt{p^2-4}}{2}$

Root of quadratic eq $x^2-qx+1=0$ is ${\beta }_1,{\beta }_2$

$\beta =\frac{q\pm \sqrt{q^2-4}}{2}$

${\beta }_1=\frac{q+\sqrt{q^2-4}}{2}$

${\beta }_2=\frac{q-\sqrt{q^2-4}}{2}$

Given is ${\alpha }_1\hat{i}+{\beta }_1\hat{j}$ is parallel to ${\alpha }_2\hat{i}+{\beta }_2\hat{j}$ i.e.

$\left|\right({\alpha }_1\hat{i}+{\beta }_1\hat{j})\times ({\alpha }_2\hat{i}+{\beta }_2\hat{j}\left)\right|=0$

$$\left|\begin{array}{cc}\hat{i}& \hat{j}\\ {\alpha }_1& {\beta }_1\\ {\alpha }_2& {\beta }_2\end{array}\right|$$$=0$

${\alpha }_1{\beta }_2\hat{i}-{\alpha }_2{\beta }_1\hat{j}=0$

${\alpha }_1{\beta }_2=0and{\alpha }_2{\beta }_1=0$

${\alpha }_1{\beta }_2=0$

$\left(\frac{p+\sqrt{p^2-4}}{2}\right)\left(\frac{q-\sqrt{q^2-4}}{2}\right)=0$

$pq-p\sqrt{q^2-4}+q\sqrt{p^2-4}-\sqrt{(p^2-4)(q^2-4)}=0$ -----------------------(1)

${\alpha }_2{\beta }_1=0$

$\left(\frac{p-\sqrt{p^2-4}}{2}\right)\left(\frac{q+\sqrt{q^2-4}}{2}\right)$

$pq+p\sqrt{q^2-4}-q\sqrt{p^2-4}-\sqrt{(p^2-4)(q^2-4)}=0$ -----------------------(2)

comparing eq (1) and (2)

$pq-p\sqrt{q^2-4}+q\sqrt{p^2-4}-\sqrt{(p^2-4)(q^2-4)}=pq+p\sqrt{q^2-4}-q\sqrt{p^2-4}-\sqrt{(p^2-4)(q^2-4)}$

$2p\sqrt{q^2-4}=2q\sqrt{p^2-4}$

$\frac{p}{\sqrt{p^2-4}}=\frac{q}{\sqrt{q^2-4}}$

on squaring both sides

$\frac{p^2}{p^2-4}=\frac{q^2}{q^2-4}$

$\frac{1}{1-\frac{4}{p^2}}=\frac{1}{1-\frac{4}{q^2}}$

$1-\frac{4}{p^2}=1-\frac{4}{q^2}$

$p^2=q^2$

$p=\sqrt{q^2}$

$p=\pm q$