Question

Three roots of the equation $x^4-p{x}^3+q{x}^2-rx+s=0$ are tan A,tan B and tan C where A,B,C are the angles of a triangle. The fourth roots 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, the bi-quadratic equation $x^4-p{x}^3+q{x}^2-rx+s=0⟶\left(1\right)$ has three roots $\tan A,\tan B,\tan C$ and let the fourth root is $\alpha$.

$\because A,B,C$ are the angles of a triangle,

$\therefore A+B+C=\pi \Rightarrow A+B=\pi -C\Rightarrow \tan (A+B)=\tan (\pi -C)\Rightarrow \frac{\tan A+\tan B}{1-\tan A\tan B}=-\tan C\Rightarrow \tan A+\tan B+\tan C=\tan A\tan B\tan C\Rightarrow a+b+c=abc[taking\tan A=a,\tan B=b,\tan C=c]$

Now, sum of the roots of the equation,

$\alpha +a+b+c=p⟶\left(2\right)$

roots taken two at a time,

$\alpha [a+b+c]+ab+bc+ca=q⟶\left(3\right)$

roots taken three at a time,

$\alpha [ab+bc+ca]+abc=r⟶\left(4\right)$

and, product of the roots,

$\alpha \cdot a\cdot b\cdot c=s⟶\left(5\right)$

Again let, $a+b+c=abc=k$

Now, from equation $\left(2\right)$,

$\alpha +k=p$

from equation $\left(3\right)$,

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

from equation $\left(5\right)$,

$k\alpha =s$

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

Thus, $\alpha [ab+bc+ca]+abc=r\Rightarrow \alpha [ab+bc+ca]+k=r\Rightarrow \alpha [q-s]+k=r\Rightarrow \alpha [q-s]+p-\alpha =r[\because \alpha +k=p]\Rightarrow \alpha [q-s-1]=r-p\Rightarrow \alpha =\frac{r-p}{q-s-1}\Rightarrow \alpha =\frac{p-r}{1-q+s}[multiplyingthenumeratorandthedenominatorby-1]$

Hence, the fourth root of the bi-quadratic equation is $\frac{p-r}{1-q+s}.$