Question

The roots of 2x^2-x+k=0 are sinθ and cosθ. Find k.

Answer 1

\[\sin \theta + \cos \theta = \frac{1}{2}\tag{*}\]You need to find \(k\), and you know that \(\frac{k}{2}\) is the product of roots, meaning that \(\frac{k}{2} = \sin\theta \cdot \cos \theta\Rightarrow k = 2\sin \theta \cos \theta\).

So you need to find \(2\sin\theta \cos \theta\). You can do so by squaring \((*)\).

Answer 2

\[\sinθ+\cosθ=\frac{1}{2}\]??? really? im pretty sure its\[\sin^2θ+\cos^2θ=1\]

Answer 3

this is a good resource for the sum and products of the roots of a quadratic polynomial
http://www.mathsisfun.com/algebra/polynomials-sums-products-roots.html

Answer 4

for a quadratic polynomial
ax^2 + bx + c

let r1, r2 be the roots . Then we have

r1 + r2 = -b/a

r1 * r2 = c/a

Answer 5

at some point, do i need to find theta?

Answer 6

None.

Answer 7

If you square both sides in \(\sin \theta + \cos \theta = \frac{1}{2}\), which I got from Vieta's Formulas, what would you get?

Answer 8

ahh

Answer 9

thanks.

Answer 10

Good to see that. :)

What do you get?

Answer 11

-.75