# Form the quadratic equation whose roots alpha and beta satisfy the relations alpha beta=768 and alpha^2+beta^2=1600?

##### 1 Answer
Jan 6, 2018

${x}^{2} - 56 x + 768 = 0$

#### Explanation:

if $\alpha \text{ and } \beta$ are the roots of a quadratic eqn , the eqn can be written as

${x}^{2} - \left(\alpha + \beta\right) x + \alpha \beta = 0$

we are given

$\textcolor{red}{\alpha \beta = 768}$

$\textcolor{b l u e}{{\alpha}^{2} + {\beta}^{2} = 1600}$

now
${\left(\alpha + \beta\right)}^{2} = \textcolor{b l u e}{{\alpha}^{2} + {\beta}^{2}} + \textcolor{red}{2 \alpha \beta}$

so ${\left(\alpha + \beta\right)}^{2} = \textcolor{b l u e}{1600} + 2 \times \textcolor{red}{768} = 3136$

$\therefore \alpha + \beta = \sqrt{3136} = 56$

the required quadratic is then

${x}^{2} - 56 x + 768 = 0$