# If 2+sqrt3 and 2-sqrt3 are solutions of 2x^2+px+q=0, then what is p+q?

Dec 9, 2016

$- 2$

#### Explanation:

We know that

$\left(x + {r}_{1}\right) \left(x + {r}_{2}\right) = {x}^{2} + \left({r}_{1} + {r}_{2}\right) x + {r}_{1} {r}_{2}$ then

$2 \left(x + {r}_{1}\right) \left(x + {r}_{2}\right) = 2 {x}^{2} + 2 \left({r}_{1} + {r}_{2}\right) x + 2 {r}_{1} {r}_{2}$

Calling ${r}_{1} = - \left(2 - \sqrt{3}\right)$ and ${r}_{2} = - \left(2 + \sqrt{3}\right)$ and comparing with

$2 {x}^{2} + p x + q$ we have

$p = 2 \left({r}_{1} + {r}_{2}\right) = - 2 \times 2 = - 4$ and

$q = 2 {r}_{1} {r}_{2} = 2 \left({2}^{2} - 3\right) = 2$ then $p + q = - 2$