Just a quadratic equation problem

Algebra Level pending

If $\sin \theta$ and $\cos \theta$ are the roots of the equation $4x^{2}-kx-1$ ,(k>0)then the value of k is:

3\sqrt{2}

4\sqrt{2}

2\sqrt{2}

1 solution

Akshay Mujumdar
Nov 27, 2014

There are many methods to solve this, the most common one being product of roots = sin t * cos t = -1/4. But there is one more method which can be used which is helpful in other particularly difficult problems. We know that sin^2 t + cos^2 t = 1. So if a and b are the roots of this equation, we will find equation with roots a^2 and b^2. Let a^2 = p. Hence, a = sqroot(p). Since a is a root of given equaton, we should get zero on putting x = a = sqroot(p). Hence, 4p - ksqroot(p) -1 = 0. Rearranging and squaring both sides, we get 16p^2 - (8+k^2)p+1 = 0. This is the equation having roots sin^2 t and cos^2 t. Sum of roots = 1. Hence, 8+k^2 = 16. Hence, k = 2sqroot(2).

Just a quadratic equation problem

1 solution

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