Question

(a) If $a\ne 1$ and $a\ne -2$, show that the roots of the equation $(a^2+a-2)x^2+(2a^2+a+3)x+a^2-1=0$ are rational. Hence solve the equation.
(b) Let A,B,C be three angles such that $A=\frac{\pi }{4}$ and $tanBtanC=p$. Find all possible values of p such that A,B,C are the angles of a triangle.
(c) If $tanA$ and $tanB$ are the roots of the quadratic equation $x^2-px+q=0$, then the value of ${sin}^2(A+B)$ is

Answer 1

(a) Here $\Delta =B^2-4AC$

$={(2a^2+a+3)}^2-4(a^2+a-2)(a^2-1)$

$=25a^2+10a+1$ (after simplifications)

$={(5a+1)}^2$

Since $\Delta$ is a perfect square, roots are rational.

The roots are given by

$x=\frac{-(2a^2+a+3)\pm (5a+1)}{2(a^2+a-2)}=-\frac{2{(a-1)}^2}{2(a-1)(a+2)}$

or $\frac{-2(a+2)(a+1)}{2(a-1)(a+2)}$

Since $a\ne 1$ and $a\ne -2$, the roots are

$x_1=-\frac{a-1}{a+2},x_2=-\frac{a+1}{a-1}$

(b) $A+B+C=\pi$,

$\therefore$ $B+C=\pi -\frac{\pi }{4}=\frac{3\pi }{4}$.......(1)

$tan(B+C)=-1$

or $tanB+tanC=-1(1-tanBtanC)$

or $tanB+tanC=-1+p$

or $tanB+tan\{\frac{3\pi }{4}-B\}=-1+p$

or $t+\frac{-1-t}{1-t}=-1+p$.

$\because$ $tan\frac{3\pi }{4}=-1,t=tanB$

or $t^2-t(p-1)+p=0$

Since $t-tanB$ is real $\therefore$ ${(p-1)}^2-4p\ge 0$

$p^2-6p+1\ge 0$ $p=a,b$ $a<b$;

Where $a=3-2\sqrt{2,}$ $b=3-2\sqrt{2}$

$\therefore$ $p\le a$ or $p\ge b$

$\therefore$ $pϵ]-\infty ,3-2\sqrt{2}]\cup [3+2\sqrt{2,}\infty [$

(c) Ans. (d).

$tanA+tanB=p,tanAtanB=q$

$\therefore$ $tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}=\frac{p}{1-q}=T$........(1)

Now ${sin}^2(A+B)=\frac{{tan}^2(A+B)}{{sec}^2(A+B)}=\frac{T^2}{1+T^2}$

$=\frac{p^2}{{(1-q)}^2+p^2}$ by (1)$\Rightarrow$(d)