SOLUTION: If the equation {{{sqrt(2)x^2 - sqrt(3)x + k = 0}}} with k a constant has two solutions {{{sin(theta)}}} and {{{cos(theta)}}} (0<=theta<=pi/2),then k = ... Hi, I got {{{3/4sqrt(

Algebra ->  Equations -> SOLUTION: If the equation {{{sqrt(2)x^2 - sqrt(3)x + k = 0}}} with k a constant has two solutions {{{sin(theta)}}} and {{{cos(theta)}}} (0<=theta<=pi/2),then k = ... Hi, I got {{{3/4sqrt(      Log On


   

Question 1065005: If the equation sqrt%282%29x%5E2+-+sqrt%283%29x+%2B+k+=+0 with k a constant has two solutions sin%28theta%29 and cos%28theta%29 (0<=theta<=pi/2),then k = ...
Hi, I got 3%2F4sqrt%282%29+%3E+k. What should I do next? Do I need to pick some lesser number than that?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
I agree that 3%2F4sqrt%282%29+%3E=+k is necessary
to get two real solutions, d and e ,
but we need more than that,
and trying to use the quadratic formula seems awkward in this case.
I would say that it is necessary and sufficient
to have two real solutions, d and e ,
such that system%28d%3E=0%2Ce%3E=0%2Cd%5E2%2Be%5E2=1%29 .
For any quadratic equation, we know that
if d and e are solutions to ax%5E2+%2Bbx+%2B+c+=+0 ,
system%28de=c%2Fa%2C%22and%22%2Cd%2Be=-b%2Fa%29%29 .
In this case,
if d and e are solutions to sqrt%282%29x%5E2+-+sqrt%283%29x+%2B+k+=+0 ,
system%28de=k%2Fsqrt%282%29%2C%22and%22%2Cd%2Be=sqrt%283%29%2Fsqrt%282%29%29 .
With k%3E=0 , we would have de=k%2Fsqrt%282%29%3E0 , meaning d%3E=0 and e%3E=0 .
Let us see if we can find a k%3E=0 out of
system%28d%5E2%2Be%5E2=1%2Cde=k%2Fsqrt%282%29%2Cd%2Be=sqrt%283%29%2Fsqrt%282%29%29 .
system%28d%5E2%2Be%5E2=1%2Cde=k%2Fsqrt%282%29%2Cd%2Be=sqrt%283%29%2Fsqrt%282%29%29 --> system%28d%5E2%2Be%5E2=1%2C2de=2k%2Fsqrt%282%29%2C%28d%2Be%29%5E2=3%2F2%29 --> system%28d%5E2%2Be%5E2=1%2C2de=2k%2Fsqrt%282%29%2Cd%5E2%2Be%5E2%2B2de=3%2F2%29 --> system%28d%5E2%2Be%5E2=1%2C2de=2k%2Fsqrt%282%29%2C1%2B2k%2Fsqrt%282%29=3%2F2%29 --> system%28d%5E2%2Be%5E2=1%2C2de=2k%2Fsqrt%282%29%2C2k%2Fsqrt%282%29=1%2F2%29 --> system%28d%5E2%2Be%5E2=1%2C2de=2k%2Fsqrt%282%29%2Chighlight%28k=sqrt%282%29%2F4%29%29

NOTE:
The solutions to sqrt%282%29x%5E2+-+sqrt%283%29x+%2B+k+=+0 are
x=%28sqrt%283%29+%2B-+1%29%2F%282sqrt%282%29%29 ,
which correspond to
system%28theta=pi%2F12%2C%22or%22%2Ctheta=5pi%2F12%29