Question

If alpha beta are the zeroes of the polynomial 3x^2 + 11x - 4, then the value of alpha^2 + beta ^2 is

Answer 1

Answer:

$\frac{145}{9}$

Step-by-step explanation:

Answer 2

GIVEN :

$\alpha \: and \: \beta \: are \: zeros \: of \: the \: polynomial \\ 3 {x}^{2} + 11x - 4.$

To FIND :

${ \alpha }^{2} + { \beta }^{2}$

SOLUTION :

We have,

$f(x) = 3 {x}^{2} + 11x - 4$

For, a quadratic equation in the form ax²+bx+c, we know

$( \alpha + \beta ) = - \frac{b}{a}$

$( \alpha \beta ) = \frac{c}{a}$

Then, in 3x²+11x - 4; a =3, b =11 & c = - 4,

$\alpha + \beta = - \frac{11}{3}$

$\alpha \beta = - \frac{ 4}{3}$

On solving,

${( \alpha + \beta )}^{2} = { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta$

${ ( - \frac{11}{3} )}^{2} = { \alpha }^{2} + { \beta }^{2} + 2( - \frac{4}{ 3} )$

$\frac{121}{9} = { \alpha }^{2} + { \beta }^{2} - \frac{8}{3}$

$\frac{121}{9} + \frac{8}{3} = { \alpha }^{2} + { \beta }^{2}$

$\frac{363 + 72}{9\times 3} = { \alpha }^{2} + { \beta }^{2}$

$\frac{145}{9} = { \alpha }^{2} + { \beta }^{2}$