Question

If ${\cos }^4\alpha +a,{\sin }^4\alpha +a$ are roots of the equation $x^2+2kx+k=0$ and ${\cos }^2\alpha +b$, ${\sin }^2\alpha +b$ are roots of the equation $x^2+3x+2=0$ then find the sum of all possible values of $k$.

Answer 1

Answer: 1

Let $x_1={\cos }^4\alpha +a,x_2={\sin }^4\alpha +a,x_3={\cos }^2\alpha +b$ and $x_4={\sin }^2\alpha +b$
Now $x_1-x_2=\cos 2\alpha$ and $x_3-x_4=\cos 2\alpha$
Hence $\left|x_1-x_2\right|=\left|x_3-x_4\right|$
$\Rightarrow {\left|x_1-x_2\right|}^2={\left|x_3-x_4\right|}^2$
$\Rightarrow {\left(x_1+x_2\right)}^2-4x_1{x}_2={\left(x_3+x_4\right)}^2-4x_3{x}_4$
$\Rightarrow 4k^2-4k=9-8=1\Rightarrow 4k^2-4k-1=0$
Hence $k_1+k_2=\frac{4}{4}=1$ Ans.
Note that given that roots of the equation $x^2+2kx+k=0$ are $\left({\cos }^2\alpha +b\right)$ and $\left({\sin }^2\alpha +b\right)$ and difference is $\cos 2\alpha =\sqrt{(5{)}^2-4\cdot 3}=\sqrt{13}$ which is not possible, but if is wrong then is also wrong.