MHB Solving Quadratic Equations: Find k

Albert1
Messages
1,221
Reaction score
0
the solutions of :

$x^2+kx+k=0 "

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

please find : $k=?$
 
Mathematics news on Phys.org
Hello, Albert!

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

\text{Find }k.
Since k \,=\,\sin\theta\cos\theta, we see that: .|k| \,<\,1.

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

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

\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}

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

\text{And we have: }\:k^2 - 2k - 1\:=\:0

\text{Hence: }\:k \:=\:1\pm\sqrt{2}\text{Since }|k| < 1\!:\;k \:=\:1-\sqrt{2}
 
Last edited by a moderator:
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, $k=1-\sqrt{2}$.
 
thanks all for your participation:)

your answers are correct !
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

Similar threads

Replies
6
Views
2K
Replies
6
Views
2K
Replies
1
Views
1K
Replies
16
Views
4K
Replies
5
Views
1K
Back
Top