Question

if a and B are the zeroes
the quadratic
polynomial p(x)=2x^2+9x+6 find the value
of a^2B+ ab^2​

Answer 1

Step-by-step explanation:

Given -

• α and β are zeroes of polynomial p(x) = 2x² + 9x + 6

To Find -

• Value of α²β + αβ²

→ αβ(α + β)

Now,

As we know that :-

• α + β = -b/a

→ α + β = -9/2

And

• αβ = c/a

→ αβ = 6/2

Now,

The value of αβ(α + β) is

→ 6/2(-9/2)

→ -54/4

→ -27/2

Hence,

The value of α²β + αβ² is -27/2.

Answer 2

Given:-

• $\alpha$ and $\beta$ are zeroes of polynomial p(x) = 2x² + 9x + 6

To Find:-

• Value of $\sf{ { \alpha}^2 \beta + \alpha { \beta}^2}$

$\implies \; \; \; \; \sf{ \alpha \beta ( \alpha + \beta )}$

Solution:-

$\bullet \; \; \sf{a = 2}$

$\bullet \; \; \sf{b = 9}$

$\bullet \; \; \sf{c = 6}$

★ If $\alpha$ and $\beta$ are zeroes of polynomial then,

$\small \; \; \; \star \; {\underline{\underline{\sf{\purple{Sum\;of\;zeroes\;( \alpha + \beta )\;:}}}}}$

$: \implies \sf{ \alpha + \beta = \dfrac{-b}{a}} \\ \\ : \implies { \alpha + \beta = \dfrac{-9}{2}}$

$\small \; \; \; \star \; {\underline{\underline{\sf{\purple{Product\;of\;zeroes\;( \alpha \beta )\;:}}}}}$

$: \implies \sf{ \alpha \beta = \dfrac{c}{a}} \\ \\ : \implies { \alpha \beta = \dfrac{6}{2}}$

$\rule{200}{3}$

Therefore, we get value of $\sf{( \alpha + \beta ) \; and \; ( \alpha \beta )}$

$\small \; \; \; \star \; {\underline{\underline{\sf{\purple{Then,\;Value\;of\; \alpha \beta ( \alpha + \beta ) \;is:}}}}}$

$: \implies \rm{ \dfrac{6}{4} \bigg( \dfrac{-9}{2} \bigg)}$

$: \implies \rm{ \cancel{\dfrac{-54}{4}}}$

$: \implies \rm{ \cancel{\dfrac{-27}{2}}}$

${\underline{\underline{\boxed{\dag \; { \alpha}^2 \beta + \alpha { \beta}^2}}}}$

$\dag \; {\underline{\underline{\sf{\red{Hence\;Solved!}}}}}$

$\rule{200}{3}$