Question

Find the condition that in the equations
$a{x}^2+bx+c=0$ and $d{x}^2+b^{\prime }x+c^{\prime }=0$
(a) One root be common.
(b) A root of first be reciprocal of a root of the second.
(c) Both have the same pair of roots.

Answer 1

(a) Let $\alpha$ be a common root which will satisfy both the equations

$\therefore$ $a{\alpha }^2+b\alpha +c=0$

and $d{\alpha }^2+b^{\prime }\alpha +c^{\prime }=0$.

Solving by the method of cross multiplication we get

$=\frac{{\alpha }^2}{b{c}^{\prime }-b^{\prime }c}=\frac{\alpha }{cd-c^{\prime }a}=\frac{1}{a{b}^{\prime }-db}$

or $\frac{{\alpha }^2}{P}=\frac{\alpha }{Q}=\frac{1}{R}$, say

$\therefore$ $\frac{{\alpha }^2}{\alpha }=\frac{b{c}^{\prime }-b^{\prime }c}{c{d}^{\prime }-c^{\prime }a}=\frac{P}{Q}$

$\alpha =\frac{\alpha }{1}=\frac{c{d}^{\prime }-c^{\prime }a}{a{b}^{\prime }-db}=\frac{Q}{R}$..........(1)

Since $\alpha =\alpha$,

$\therefore$ $=\frac{b{c}^{\prime }-b^{\prime }c}{cd-c^{\prime }a}=\frac{cd-c^{\prime }a}{a{b}^{\prime }-db}$

Required condition is

$(b{c}^{\prime }-b^{\prime }c)(a{b}^{\prime }-db)={(cd-c^{\prime }a)}^2$

or $PR=Q^2$

The value of the common root is given by (1).

Note : The value of common root.

Make the coefficient of $x^2$ in the two equations same or unity and then subtract the two.

${\alpha }^2+\frac{b}{a}\alpha +\frac{c}{a}=0$

${\alpha }^2+\frac{b^{\prime }}{d}\alpha +\frac{c^{\prime }}{d}=0$

Now subtract

$\therefore$ $(\frac{b}{a}-\frac{b^{\prime }}{d})\alpha +(\frac{c}{a}-\frac{c^{\prime }}{d})=0$

or $\alpha =\frac{cd-a{c}^{\prime }}{a{b}^{\prime }-db}=\frac{Q}{R}=\frac{P}{Q}$ as $PR=Q^2$

(b) If $\alpha$ be a root of first equation then $\alpha$ will be a root of the second.

$\therefore$ $a{\alpha }^2+b\alpha +c=0$

and $d.{\left(1/\alpha \right)}^2+b^{\prime }.\left(1/\alpha \right)+c^{\prime }=0$

or $c^{\prime }{\alpha }^2+b^{\prime }\alpha +d=0$

$\therefore$ $\frac{{\alpha }^2}{bd-c{b}^{\prime }}=\frac{\alpha }{c{c}^{\prime }-ad}=\frac{1}{a{b}^{\prime }-b{c}^{\prime }}$

$\therefore$ ${(c{c}^{\prime }-ad)}^2=(bd-c{b}^{\prime })(a{b}^{\prime }-b{c}^{\prime })$

is the required condition.

(c) If both the roots are common then their sum and product will be same.

$\therefore$ $S=-b/a=-b^{\prime }/d$ $\therefore$ $a/d=b/b^{\prime }$

$P=c/a=c^{\prime }/d$ or $a/d=c/c^{\prime }$.

Hence the condition is $a/d=b/b^{\prime }=c/c^{\prime }$.