# The quadratic equation in x is x2 + 2x.cos(A) + K=0. &also given summation and difference of solutions of above equation are -1 & -3 respectively. Hence find K & A?

May 5, 2018

$A = {60}^{\circ}$
$K = - 2$

#### Explanation:

${x}^{2} + 2 x \cos \left(A\right) + K = 0$

Let the solutions of the quadratic equation be $\alpha$ and $\beta$.

$\alpha + \beta = - 1$

$\alpha - \beta = - 3$

We also know that $\alpha + \beta = - \frac{b}{a}$ of the quadratic equation.

$- 1 = - \frac{2 \cos \left(A\right)}{1}$

Simplify and solve,

$2 \cos \left(A\right) = 1$

$\cos \left(A\right) = \frac{1}{2}$

$A = {60}^{\circ}$

Substitute $2 \cos \left(A\right) = 1$ into the equation, and we get an updated quadratic equation,

${x}^{2} + x + K = 0$

Using the difference and sum of roots,

$\left(\alpha + \beta\right) - \left(\alpha - \beta\right) = \left(- 1\right) - \left(- 3\right)$

$2 \beta = 2$

$\beta = 1$

When $\beta = 1$,

$\alpha = - 2$

When the roots are $1$ and $- 2$, we can get a quadratic equation as follows,

$\left(x - 1\right) \left(x + 2\right)$

$= {x}^{2} + x - 2$

By comparison,

$K = - 2$