How to find the quadratic equation using 2 given solutions

Question

Find the quadratic equation $ax^2 + bx + c = 0$, Such that $a=1$ and the solutions are:

$3(\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3})), 2(\cos(\frac{5\pi}{6}) + i\sin(\frac{5\pi}{6}))$

Answer 1

HINT:

How about this quadratic equation: $$\left[x-3\left\{\cos\left(\frac{\pi}{3}\right) + i\sin\left(\frac{\pi}{3}\right)\right\}\right]\cdot \left[x-2\left\{\cos\left(\frac{5\pi}{6}\right) + i\sin\left(\frac{5\pi}{6}\right)\right\}\right]$$

Can you simplify this?

Answer 2

You know that a quadratic polynomial can be represented as the product of its roots $p(x)=(x-a)(x-b)$ where $ a,b $ are the roots. Since you are given the roots; simply go ahead and reconstruct the quadratic. In this case however $b$ and $c$ may turn out to be complex numbers.

Answer 3

If $x = x_0$ and $x= x_1$ are the two solutions to $ax^2 + bx + c = 0$ and $a = 1$ then

$ax^2 + bx + c = (x-x_0)(x-x_1) = x^2 + (-x_0 + x_1)x + x_0x_1 = 0$

And $a = 1$ $b = (-x_0 + x_1)$ and c= $(x_0x_1)$.

So if $x_0 = 3(\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3}))$ and $x_1 = 2(\cos(\frac{5\pi}{6}) + i\sin(\frac{5\pi}{6}))$

Then $b = -3(\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3}))- 2(\cos(\frac{5\pi}{6}) + i\sin(\frac{5\pi}{6}))$ and $c = 3(\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3}))* 2(\cos(\frac{5\pi}{6}) + i\sin(\frac{5\pi}{6}))$