Question

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Answer 1

Question:-

• If $\alpha \: \: \: \beta$are roots of $2 {x}^{2} - 3x - 1 = 0$
• Then, the value of ${ \alpha }^{2} + \beta ^{2}$ is

Solution:-

Given,

$\alpha \: \: \beta$ are roots of $2 {x}^{2} - 3x - 1 = 0$

Here,

a = 2 , b = -3 , c = -1

We have,

Sum of roots $\alpha + \beta = \frac{ - b}{a} = \frac{ - ( - 3)}{2} = \frac{3}{2}$

Product of roots $\alpha \beta = \frac{c}{a}$$= \frac{ - 1}{2}$

Square on both sides

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

${ \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta = \frac{9}{4}$

We have,

$\alpha \beta = \frac{ - 1}{2}$

Substitute it

${ \alpha }^{2} + { \beta }^{2} + 2( \frac{ - 1}{2} ) = \frac{9}{4}$

${ \alpha }^{2} + { \beta }^{2} - 1 = \frac{9}{4}$

${ \alpha }^{2} + { \beta }^{2} = \frac{9}{4} - 1$

${ \alpha }^{2} + { \beta }^{2} = \frac{5}{4}$