Question

If alpha, beta and gamma are the zeroes of the polynomial 6x^3 + 3x^2 -5x + 1, find the value of alpha^-1 + beta^-1 + gamma^-1​

Answer 1

Answer:

Value of $\bold{\alpha^{-1}+\beta^{-1}+\gamma^{-1}=5}$

Step-by-step explanation:

p(x) = 6x³ + 3x² - 5x + 1

We know,

If ɑ, β and ɣ are the zeroes of the polynomial p(x) = ax³ + bx² + cx + d, then

• ɑ + β + ɣ = -b/a

• ɑβ + βɣ + ɣɑ = c/a

• ɑβɣ = -d/a

∴ ɑβ + βɣ + ɣɑ = c/a = -5/6

ɑβɣ = -d/a = -1/6

$\hrulefill$

Now, we have to find the value of $\bold{\alpha^{-1}+\beta^{-1}+\gamma^{-1}}$

This can be rewritten as,

☛ $\bold{\dfrac{1}{\alpha}+\dfrac{1}{\beta}+\dfrac{1}{\gamma}=\dfrac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma}}$

Applying the values we found, we get,

☛ $\bold{\dfrac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma}=\dfrac{-5}{\cancel{6}}\times{}\dfrac{\cancel{6}}{-1}=5}$

i.e., $\bold{\alpha^{-1}+\beta^{-1}+\gamma^{-1}=5}$