Question

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

• A. $\frac{s^2-sq+s}{r+(s-q)p}$
• B. $\frac{s^2+sq+s}{r+(s+p)q}$
• C. $\frac{s^2-sq+s}{r-(s-p)q}$
• D. $\frac{s^2-sq-s}{r-(s-q)p}$

Answer 1

Answer: (A)

$\frac{s^2-sq+s}{r+(s-q)p}$

Let fourth root be $k$

Then sum of roots is $\tan A+\tan B+\tan C+k=p$

also product of all the roots is $k.\tan A.\tan B.\tan C=s$

Now we know In any $\Delta ABC$, $\tan A+\tan B+\tan C=\tan A.\tan B.\tan C$

$\Rightarrow p-k=\frac{s}{k}\Rightarrow k^2=pk-s$ (i)

$k$ will also satisfy $k^4-p{k}^3+q{k}^2-rk+s=0$

$\Rightarrow k^2(pk-s)-p{k}^3+q{k}^2-rk+s=0$ using (i)

$\Rightarrow (q-s)k^2-rk+s=0\Rightarrow (q-s)(pk-s)-rk+s=0$

$\Rightarrow s^2-sq+s-k(r+ps-pq)=0\Rightarrow k=\frac{s^2-sq+s}{r+(s-q)p}$