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find k

Albert

Well-known member
Jan 25, 2013
1,225
the solutions of :

$x^2+kx+k=0 "

are $ $sin \,\theta \,\,and \,\, cos\, \theta $

please find : $k=?$
 

soroban

Well-known member
Feb 2, 2012
409
Hello, Albert!

[tex]\text{The solutions of: }\: x^2+kx+k\:=\:0[/tex]
[tex]\text{are }\sin\theta\text{ and }\cos\theta[/tex]

[tex]\text{Find }k.[/tex]

Since [tex]k \,=\,\sin\theta\cos\theta[/tex], we see that: .[tex]|k| \,<\,1.[/tex]

Quadratic Formula: .[tex]x \:=\:\frac{-k \pm \sqrt{k^2-4k}}{2}[/tex]

[tex]\text{Let: }\:\begin{Bmatrix}\sin\theta &=& \frac{-k + \sqrt{k^2-4k}}{2} \\ \cos\theta &=& \frac{-k -\sqrt{k^2-4k}}{2} \end{Bmatrix}[/tex]

[tex]\text{Then: }\:\begin{Bmatrix}\sin^2\theta &=& \frac{2k^2 - 4k + 2k\sqrt{k^2-4k}}{4} \\ \cos^2\theta &=& \frac{2k^2 - 4k - 2k\sqrt{k^2-4k}}{4} \end{Bmatrix}[/tex]

[tex]\text{Add: }\:\sin^2\theta + \cos^2\theta \:=\:\frac{4k^2 - 8k}{4} \:=\:1[/tex]

[tex]\text{And we have: }\:k^2 - 2k - 1\:=\:0[/tex]

[tex]\text{Hence: }\:k \:=\:1\pm\sqrt{2}[/tex]


[tex]\text{Since }|k| < 1\!:\;k \:=\:1-\sqrt{2}[/tex]
 
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Pranav

Well-known member
Nov 4, 2013
428

By Vieta's formulas, we have
$$k=\sin\theta \cos\theta$$
$$\sin\theta+\cos\theta=-k$$
Squaring both the sides of second equation,
$$1+2\sin\theta \cos\theta=k^2 \Rightarrow k^2-2k=1 \Rightarrow k^2-2k+1=2 \Rightarrow (k-1)^2=2$$
$$\Rightarrow k=1\pm \sqrt{2}$$
But $|k|<1$, hence, $k=1-\sqrt{2}$.
 

Albert

Well-known member
Jan 25, 2013
1,225
thanks all for your participation:)

your answers are correct !