Question

If p(x) = ax2 + bx + c has zeroes as α and β, find the value of 1/a + 1/b​

Answer 1

(3abc-b³)/c³

Step-by-step explanation:

Given,

α and β are the zeroes of the polynomial p(x) = ax² + bx + c.

To Find :-

• 1/α³ + 1/β³

Solution :-

We know that :-

Sum of the roots = -(Coefficient of 'x' term)/Constant term

Product of the roots = coefficient of 'x²' term/Constant term.

According to Question :-

α+β = -b/a [ Let it be equation 1]

• $α\sf\times β = c/a$[Let it be equation 2]

As we need to find :-

$\tt\dfrac{1} { \alpha ^{3} } + \dfrac{1}{ { \beta }^{3} }$

Taking L.C.M :-

$= \tt\dfrac{\beta^{3}+\alpha^{3}}{\alpha^{3}\times \beta^{3}}$

We know that :-

• ★a³+b³ = (a+b)³ - 3ab(a + b)

• ★ a³(b³) = (ab)³

(α×β) 3 (α+β) 3 −3αβ(α+β)

$\sf= \dfrac{ \bigg( \dfrac{ - b}{a} \bigg) ^{3} - 3 \times \dfrac{c}{a} \bigg( \dfrac{ - b}{a} \bigg) }{ \bigg(\dfrac{c}{a} \bigg) ^{3} }$

$\sf= \dfrac{ \dfrac{ { - b}^{3} }{ {a}^{3} } + \dfrac{3bc}{ {a}^{2} } }{ \dfrac{ {c}^{3} }{ {a}^{3} } }$

$\tt= \dfrac{ \dfrac{ - {b}^{3} + 3abc }{ {a}^{3}} }{ \dfrac{ {c}^{3} }{ {a}^{3} } }$

$\sf★\dfrac{3abc - {b}^{3} }{ {c}^{3} } ★$

Answer 2

Step-by-step explanation:

Hope it helps you ❣️✔️❤️✔️✔️✔️