Question

alpha and beta are the zeroes of the polynomial
2x square plus 3x + 4 find the value of 1/alpha square +1/beta square.

Answer 1

iI hope it will help you

Answer 2

Answer:

Value of $\frac{1}{{\alpha}^2}+\frac{1}{{\beta}^2}\:\:is\:\:\frac{-7}{16}$

Step-by-step explanation:

Given Quadratic Polynomial, 2x² + 3x + 4

          α  and β are zeroes

To find: $\frac{1}{{\alpha}^2}+\frac{1}{{\beta}^2}$

we know that Sum of Zeroes = $\frac{-coefficient\:of\:x}{coefficient\:of\:x^2}$

And Product of Zeroes = $\frac{Constant\:term}{coefficient\:of\:x^2}$

⇒ $\alpha+\beta=\frac{-3}{2}\:\:and\:\:\alpha\beta=\frac{4}{2}=2$

we have a identity,

( a + b )² = a² + b² + 2ab

( α + β )² = α² + β² + 2αβ

α² + β² = ( α + β )² - 2αβ

${\alpha}^2+{\beta}^2=(\frac{-3}{2})^2-2\times2$

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

${\alpha}^2+{\beta}^2=\frac{9-16}{4}$

${\alpha}^2+{\beta}^2=\frac{-7}{4}$

Now,

Consider,

$\frac{1}{{\alpha}^2}+\frac{1}{{\beta}^2}$

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

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

$\implies\frac{\frac{-7}{4}}{(2)^2}$

$\implies\frac{\frac{-7}{4}}{4}$

$\implies\frac{-7}{16}$

Therefore, Value of $\frac{1}{{\alpha}^2}+\frac{1}{{\beta}^2}\:\:is\:\:\frac{-7}{16}$