Quadratic equation?

Question

Say I have an equation, $\sqrt 2 x^2 - \sqrt 3 x + k = 0$, $k$ is a constant and there are 2 solutions $\sin \theta$ and $\cos\theta$ in the interval $0\le \theta \le \pi/2$.

What is the value of $k$?

How should I view this question? Equaling two equations where one is $f(x) = \sin\theta$ and another $f(x) = \cos\theta$? :p

It'd be great if someone could be of guidance.

score 3 · Accepted Answer · answered May 14 '17 at 14:52

3

$$\sqrt2x^2-\sqrt3x+k=0$$

We have $$\sin\theta+\cos\theta=\dfrac{\sqrt3}{\sqrt2}\ \ \ \ (1)$$

$$\sin\theta\cos\theta=\dfrac k{\sqrt2}\ \ \ \ (2)$$

Take the square of $(1)$

Can you take it from here?

answered May 14 '17 at 14:52

lab bhattacharjee

274,582

Ah..found the answer. So for (1) it's sum of solutions equals -b/a and (2) product of solutions c/a am I correct? – cartmanbrahhhhh May 14 '17 at 15:11
@cartmanbrahhhhh, See https://brilliant.org/wiki/vietas-formula/ – lab bhattacharjee May 14 '17 at 15:13
Call it sum and product of roots. Sum of roots is $\frac{-b}{a}$ and product of roots is $\frac{c}{a}$ – MonK May 15 '17 at 13:17

Quadratic equation?

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