Question

If alpha and beta are the zeroes of polynomial ax2+bx+c, then evaluate
i) alpha/beta + beta/alpha
Note-genuine answers only.

Answer 1

Step-by-step explanation:

Given -

• α and β are zeroes of the polynomial ax² + bx + c

To Find -

• Value of α/β + β/α

→ α² + β²/αβ

Now,

As we know that :-

• α + β = -b/a

Squaring both sides :

→ (α + β)² = (-b/a)²

→ α² + 2αβ + β² = b²/a²

→ α² + β² = b²/a² - 2c/a

→ α² + β² = b² - 2ac/a²

And

• αβ = c/a

Now,

Value of α² + β²/αβ is

→ b² - 2ac/a² ÷ c/a

→ b²-2ac/a² × a/c

→ b² - 2ac/ac

Hence,

The value of α/β + β/α is b² - 2ac/ac

Answer 2

$\star \; {\underline{\underline{\rm{\pink{GivEn:-}}}}}$

$\bullet \; \sf{ \alpha \; \beta \; are \; the \; zeroes \; of \; polynomial \; ax^2 + bx + c = 0}$

$\star \; {\underline{\underline{\rm{\pink{To \; Find:-}}}}}$

$\bullet \; \sf{Value \; of \; \dfrac{ \alpha}{ \beta} + \dfrac{ \beta}{ \alpha}}$

$: \implies \sf{ \dfrac{ { \alpha}^2 + { \beta}^2}{ \alpha \beta}}$

Now,

$\star \; {\underline{\underline{\sf{\purple{Sum\;of\;zeroes ( \alpha + \beta) = \dfrac{-b}{a}}}}}}$

$: \implies \; \sf{ \alpha + \beta = \dfrac{-b}{a}}$

$\footnotesize {\underline{\sf{\red{\dag \; Squaring\; both\; side:}}}}$

$: \implies \; \sf{ ( \alpha + \beta)^2 = ( \dfrac{-b}{a})^2} \\ \\ : \implies \; \sf{ {\alpha}^2 + 2 \alpha \ beta + { \beta}^2 = \dfrac{b^2}{a^2}}$

$\footnotesize {\underline{\sf{\blue{\dag \;Putting\; \alpha \beta \; as \; \dfrac{c}{a}:}}}}$

$: \implies \; \sf{ { \alpha}^2 + { \beta}^2 = \dfrac{b^2}{a^2} - \dfrac{2c}{a}} \\ \\ \\ : \implies \; { \alpha}^2 + { \beta}^2 = \dfrac{ba^2 - 2a^2c}{a^3}} \\ \\ \\ : \implies \; { \alpha}^2 + { \beta}^2 = \dfrac{a(b^2 - 2ac)}{a^3}} \\ \\ \\ : \implies \; { \alpha}^2 + { \beta}^2 = \dfrac{b^2 - 2ac}{a^2}}}$

And,

$\star \; {\underline{\underline{\sf{\purple{Product\;of\;zeroes\; ( \alpha \beta) = \dfrac{c}{a}}}}}}$

$\footnotesize {\underline{\sf{\red{\dag \;Value\; of \; \dfrac{ { \alpha}^2 + { \beta}^2}{ \alpha \beta} :}}}}$

$: \implies \sf{ \dfrac{ \dfrac{b^2 - 2ac}{a^2}}{ \dfrac{c}{a}}} \\ \\ \\ : \implies \dfrac{b^2 - 2ac}{a^2} \times { \dfrac{c}{a}}} \\ \\ \\ : \implies \dfrac{b^2 - 2ac}{ac}}}}$

${\underline{\underline{\boxed{\sf{\pink{\dag \; Hence, \;The \;value \;of \; \dfrac{ \alpha}{ \beta} + \dfrac{ \beta}{ \alpha} \; is \; \dfrac{b^2 - 2ac.}{ac}}}}}}}$

$\rule{200}{2}$