Question

$\Large{\underline{\underline{\bf{Question:-}}}}$

If α and β are the zeroes of the polynomial ax² + bx + c, find the value of α² + β²​

Answer 1

Answer:

Answer:

Answer:

Given:

alpha and beta are the zeroes of polynomial ax² + bx + c

To find:

alpha²+ beta²

Pre - requisite Knowledge:

If α and β are the zeros,then,

α + β = -b/a

α * β = c/a

a² + b² = (a+b)² - 2ab

Solving Question:

 We are given the polynomial and are asked to find the value of alpha square + beta square , we could find it by substituting the values in above equations.

Solution:

a² + b² = (a+b)² - 2ab

⇒ α² + β ²=(α + β )² -2αβ

and

α + β = -b/a

α * β = c/a

substitute the values,

⇒ α² + β²= ( -b/a )² -2(c/a)

or,  α² + β²= b²/a² - 2c/a

or,  α² + β²= ( b² - 2ac )/ a²

∴ The value of  α² + β² is ( b² - 2ac )/ a²

Step-by-step explanation:

Abe mujhe unblock kr itna bta answer likhke diya tujhe -_-

Answer 2

$\Large{\underline{\underline{\bf{Required\:Answer:-}}}}$

$\implies$ α+β = $\sf\dfrac{-b}{a}$

$\implies$ α.β = $\sf\dfrac{c}{a}$

we know that:-

$\underline \pink{\boxed{\bf ( \alpha + \beta )^{2} = { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta }}$

hence, $\bf α²+β² = (α+β)²-2αβ$

Substituting the above values in this formula:-

$\implies\sf α²+β² = \bigg(\dfrac{-b}{a}\bigg) ^{2} -2 \bigg( \dfrac{c}{a} \bigg)$

$\implies\sf α²+β² = \dfrac{b^{2}}{a ^{2}} - \dfrac{2c}{a}$

$\implies\sf \green{ α²+β² = \dfrac{b^{2} - 2ac}{a ^{2}} }$

hence, thats our required answer ~ :D