Question

if alphaaa and betaa are the zeros ifbpolynomial 3xsquare +2x_6,then find the value of alpga square - Beta square​

Answer 1

Answer:

$\large\boxed{\sf{-\dfrac{4}{9}\sqrt{19}}}$

Step-by-step explanation:

Given a quadratic equation such that,

$3 {x}^{2} + 2x - 6 = 0$

Also, it's given that the zereos are,

• $\alpha$
• $\beta$

Now, we know that,

Sum of zeroes = -b/a

$= > \alpha + \beta = \frac{ - 2}{3}$

And,

Product of zeroes = c/a

$= > \alpha \beta = \frac{ - 6}{3} \\ \\ = > \alpha \beta = - 2$

Now, to find the value of,

• ${ \alpha }^{2} - { \beta }^{2}$

We know that,

$\alpha - \beta = \sqrt{ {( \alpha + \beta )}^{2} - 4 \alpha \beta }$

Substituting the values, we get,

$= > \alpha - \beta = \sqrt{ {( \frac{ - 2}{3}) }^{2} - 4( - 2)} \\ \\ = > \alpha - \beta = \sqrt{ \frac{4}{9} + 8} \\ \\ = > \alpha - \beta = \sqrt{ \frac{76}{9} } \\ \\ = > \alpha - \beta = \sqrt{ \frac{19 \times {2}^{2} }{ {3}^{2} } } \\ \\ = > \alpha - \beta = \frac{2}{3} \sqrt{19}$

Now, we have,

$= > { \alpha }^{2} - { \beta }^{2} = ( \alpha + \beta )( \alpha - \beta ) \\ \\ = > { \alpha }^{2} - { \beta }^{2} = - \frac{2}{3} \times \frac{2}{3} \sqrt{19} \\ \\ = > { \alpha }^{2} - { \beta }^{2} = - \frac{4}{9} \sqrt{19}$

Hence, required value is $\bold{-\dfrac{4}{9}\sqrt{19}}$