Question

(a) If each pair of the the three equations $x^2+p_{1}x+q_1=0,x^2+p_{2}x+q_2=0$ and
$x^2+p_{3}x+q_3=0$ has a common root, then prove
that $p_1^2{p}_2^2+p_3^2+4(q_1+q_2+q_3)$
$=2(p_1{p}_2+p_2{p}_3+p_3{p}_1)$
(b) If every pair of equations $x^2+ax+bc=0,x^2+bx+ca=0,x^2+cx+ab=0$ has a common root, then find the sum and product of these common roots.
(c) If the three equations $x^2+ax+12=0,x^2+bx+15=0$ and $x^2+(a+b)x+36=0$ have a common possible root, then find a and b and the roots.

Answer 1

(a) Since each pair has a common root, the root of the three equations be taken as $\alpha ,\beta ;\beta ,\gamma$ and $\gamma ,\alpha$ respectively.
$\therefore$ $\alpha +\beta =-p_1,\beta +\gamma =-p_2,\gamma +\alpha =-p_3$
$\alpha \beta =q_1,\beta \gamma =q_2,\gamma \alpha =q_3$.
The given result can be written as on adding $z\sum p_1{p}_2$ in both sides.
${(p_1+p_2+p_3)}^2=4[\sum p_1{p}_2-\sum q_1]$
L.H.S.$={\left[2\right(\alpha +\beta +\gamma \left)\right]}^2=4{(\alpha +\beta +\gamma )}^2$
R.H.S.$=4\left[\right(\alpha +\beta \left)\right(\beta +\gamma )+(\beta +\gamma \left)\right(\gamma +\alpha )+(\gamma +\alpha \left)\right(\alpha +\beta )-(\alpha \beta +\beta \gamma +\gamma \alpha \left)\right]$
$=4[{\alpha }^2+{\beta }^2+{\gamma }^2+2\alpha \beta +2\beta \gamma +2\gamma \alpha ]$
$=4{[\alpha +\beta +\gamma ]}^2=$ L.H.S.
(b) Since each pair has common root, let the roots be
$\alpha ,\beta$ for I; $\beta ,\gamma$ for II; and $\gamma ,\alpha$ for III.
$\therefore$ $\alpha +\beta =-a$ $\alpha \beta =bc$
$\beta +\gamma =-b$, $\beta \gamma =ca$
$\gamma +\alpha =-c$ $\gamma \alpha =ab$
Adding and multiplying,we get
$2(\alpha +\beta +\gamma )=-(a+b+c)$
$\therefore$ $\sum \alpha =-\frac{1}{2}\sum a=sum$
${\alpha }^2{\beta }^2{\gamma }^2=a^2{b}^2{c}^2$
$\therefore$ $\alpha \alpha \gamma =abc$= Product
(c) Let $\alpha$ be the common root of the three equations and their other roots be $\beta ,\gamma ,\delta$ respectively
$\therefore$ $\alpha +\beta =-a,\alpha \beta =12$
$\alpha +\gamma =-b,\alpha \gamma =15$
$\alpha +\delta =-(a+b),a\delta =36$
$\therefore$ $(\alpha +\beta )+(\alpha +\gamma )=-(a+b)=\alpha +\delta$
$\therefore$ $\alpha +\beta +\gamma =\delta$...........(1)
Again $\alpha (\beta +\gamma +\delta )=12+15+36=63$
or $\alpha (2\delta -\alpha )=63$ by (1)
or $2\alpha \delta -{\alpha }^2-63$ or $72-{\alpha }^2=63\because \alpha \delta =36$
$\therefore$ ${\alpha }^2=9\therefore \alpha =3,-3$
$\alpha =3\Rightarrow \beta =4,\gamma =5,\delta =12$
$\alpha =-3\Rightarrow \beta =-4,\gamma =-5,\delta =-12$
$\therefore$ $a=-(\alpha +\beta )=7,-7;b=-(\alpha +\gamma )=8,-8$