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anonymous
 one year ago
x' and x'' are the answers of this equation
tan^2(x)2k*tan(x)+k1=0 and we know that x'+x''=(3pi/4)
find k.
anonymous
 one year ago
x' and x'' are the answers of this equation tan^2(x)2k*tan(x)+k1=0 and we know that x'+x''=(3pi/4) find k.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$\tan^2(x)2k\tan(x)+(k1)=0$$let \(u=\tan(x)\) so this reduces to a quadratic in \(u\): $$u^22ku+(k1)=0$$ now, Vieta's formula relates the solutions \(u_1,u_2\) to this equation with the coefficients of the quadratic, giving us: $$u_1+u_2=2k\\u_1u_2=k1$$ 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 $$\frac{\tan(x_1)+\tan(x_2)}{1\tan(x_1)\tan(x_2)}=\tan(3\pi/4)\\\frac{u_1+u_2}{1u_1u_2}=\tan(3\pi/4)$$substituting in for \(k\) we see $$\frac{2k}{1(k1)}=\tan(3\pi/4)\\\frac{2k}{2k}=\tan(3\pi/4)\\2k=(2k)\tan(3\pi/4)\\2k+\tan(3\pi/4)k=2\tan(3\pi/4)\\k=\frac{2\tan(3\pi/4)}{2+\tan(3\pi/4)}$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 this one is cute
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