Question

Find a quadratic equation whose roots $\alpha$ and $\beta$ are connected by the relation:
$\alpha +\beta =2$ and $\frac{1-\alpha }{1+\beta }+\frac{1-\beta }{1+\alpha }=2\left(\frac{4{\lambda }^2+15}{4{\lambda }^2-1}\right)$

• A. $x^2-2x-\frac{(4{\lambda }^2+11)}{4}=0$
• B. $x^2+2x-\frac{(4{\lambda }^2-11)}{4}=0$
• C. $x^2-2x+\frac{(-2{\lambda }^2+11)}{4}=0$
• D. None of these

Answer 1

The correct option is A $x^2-2x-\frac{(4{\lambda }^2+11)}{4}=0$

$\alpha +\beta =2$ and let $\alpha \beta =p$
$\therefore$ Equation is $x^2-2x+p=0$ ...(1)
We have to find the value of p.
Now $\frac{1-\alpha }{1+\beta }+\frac{1-\beta }{1+\alpha }=\frac{(1-{\alpha }^2)+(1-{\beta }^2)}{1+(\alpha +\beta )+p}$
or $\frac{2-({\alpha }^2+{\beta }^2)}{1+2+p}=\frac{2-\{{(\alpha +\beta )}^2-2\alpha \beta \}}{3+p}$
or $\frac{2-4+2p}{3+p}or\frac{2(p-1)}{p+3}=2\left(\frac{4{\lambda }^2+15}{4{\lambda }^2-1}\right)$
or $\frac{p-1}{p+3}=\frac{4{\lambda }^2+15}{4{\lambda }^2-1}$
or $p[(4{\lambda }^2-1)-(4{\lambda }^2+15)]=3(4{\lambda }^2+15)+(4{\lambda }^2-1)$
or $-16p=16{\lambda }^2+44=4(4{\lambda }^2+11)$
$\therefore p=-\frac{(4{\lambda }^2+11)}{4}$
Putting for $p$ in (1) we get the required equation as
$x^2-2x-\frac{(4{\lambda }^2+11)}{4}=0$