Question

x' and x'' are the answers of this equation
tan^2(x)-2k*tan(x)+k-1=0 and we know that x'+x''=(3pi/4)
find k.

Answer 1

$$\tan^2(x)-2k\tan(x)+(k-1)=0$$let \(u=\tan(x)\) so this reduces to a quadratic in \(u\): $$u^2-2ku+(k-1)=0$$ now, Vieta's formula relates the solutions \(u_1,u_2\) to this equation with the coefficients of the quadratic, giving us: $$u_1+u_2=2k\\u_1u_2=k-1$$
now consider that \(u_1=\tan(x_1),u_2=\tan(x_2)\) and we're told that \(x_1+x_2=3\pi/4\). consider: $$\tan(x_1+x_2)=\tan(3\pi/4)$$using the sum identity for \(\tan\) we find $$\frac{\tan(x_1)+\tan(x_2)}{1-\tan(x_1)\tan(x_2)}=\tan(3\pi/4)\\\frac{u_1+u_2}{1-u_1u_2}=\tan(3\pi/4)$$substituting in for \(k\) we see $$\frac{2k}{1-(k-1)}=\tan(3\pi/4)\\\frac{2k}{2-k}=\tan(3\pi/4)\\2k=(2-k)\tan(3\pi/4)\\2k+\tan(3\pi/4)k=2\tan(3\pi/4)\\k=\frac{2\tan(3\pi/4)}{2+\tan(3\pi/4)}$$

Answer 2

@ganeshie8 this one is cute