Question

These rotts of the equation, $x^4-p{x}^3+q{x}^2-rx+s=0$ are $\tan A,\tan B,\&\tan A,$ where A, B, C are the angles of a triangle. The fourth root of the biquadratic is -

• A. $\frac{p-r}{1-q+s}$
• B. $\frac{p-r}{1+q-s}$
• C. $\frac{p+r}{1-q+s}$
• D. $\frac{p+r}{1+q-s}$

Answer 1

The correct option is A $\frac{p-r}{1-q+s}$

Given a biquadratic equation

$x^4-p{x}^3+q{x}^2-rx+s=0$

Let $\tan A,\tan B,\tan C$ and $\alpha$ be its roots

$\Rightarrow A+B+C=\pi$

$\Rightarrow A+B=\pi -C$

$\Rightarrow \tan (A+B)=\tan (\pi -C)=-\tan C$

$\Rightarrow \frac{\tan A+\tan B}{1-\tan A\tan B}=-\tan C$

$\Rightarrow \tan A+\tan B=-\tan C(1-\tan A\tan B)$

$\Rightarrow \tan A+\tan B=-\tan C+\tan A\tan B\tan C$

$\Rightarrow \tan A+\tan B+\tan C=\tan A\tan B\tan C$

Let $\tan A=a,\tan B=b,\tan C=c$

Sum of the roots$=\alpha +a+b+c=p$

Sum of the roots taken two at a time$=\alpha (a+b+c)+ab+bc+ca=q$

Sum of the roots taken two at a time$=\alpha (ab+bc+ca)+abc=r$

Sum of the roots taken all at a time$=\alpha abc=s$

Now,$\tan A+\tan B+\tan C=\tan A\tan B\tan C$

$\Rightarrow a+b+c=abc=k\left(say\right)$

Consider $\alpha +a+b+c=p$

$\Rightarrow \alpha +k=p$

Consider $\alpha (a+b+c)+ab+bc+ca=q$

$\Rightarrow \alpha k+ab+bc+ca=q$

$\Rightarrow ab+bc+ca=q-\alpha k$

Consider $\alpha abc=s$

$\Rightarrow \alpha k=s$

$\Rightarrow ab+bc+ca=q-s$

Consider the Sum of the roots taken two at a time$=\alpha (ab+bc+ca)+abc=r$

$\Rightarrow \alpha (q-s)+k=r$ where $ab+bc+ca=q-s$

$\Rightarrow \alpha =\frac{r-k}{q-s}$

Again,consider the Sum of the roots$=\alpha +a+b+c=p$

$\Rightarrow \alpha +k=p$

$\Rightarrow k=p-\alpha$

Now,$\alpha (ab+bc+ca)+k=r$

$\Rightarrow \alpha (q-s)+k=r$

$\Rightarrow \alpha (q-s)+p-\alpha =r$ where $k=p-\alpha$

$\Rightarrow \alpha (q-s-1)=r-p$

$\therefore \alpha =\frac{p-r}{1-q+s}$