Question

if alpha beta gamma are the zeros of x cube + x square + 6 x minus 1 then find the value of one by Alpha Plus One by beta plus 1 by Gamma​

Answer 1

The value of 1/Alpha + 1/Beta + 1/Gamma = 6

Given that;

alpha, beta, and gamma are the zeros of $x^{3} + x^{2} + 6x^{} - 1$

To find;

The value of 1/Alpha + 1/Beta + 1/Gamma

Solution;

It is given that,

Alpha, Beta, and Gamma are the zeros of $x^{3} + x^{2} + 6x^{} - 1$

P(x) = $x^{3} + x^{2} + 6x^{} - 1$

comparing the above equation with the standard form $ax^{3} + bx^{2} +c x^{} + d$ we get,

a = 1 , b = 1 , c = 6 and d = -1

We know that;

alpha. beta. gamma = $\frac{-d}{a}$....(1) and,

alpha + beta + gamma = $\frac{c}{a}$....(2)

Therefore, 1/alpha + 1/beta + 1/gamma = $\frac{alpha.beta + beta.gamma + gamma.alpha}{alpha.beta.gamma}$

using (1) and (2) we get,

1/alpha + 1/beta + 1/gamma = $\frac{c/a}{-d/a}$

1/alpha + 1/beta + 1/gamma = $\frac{-c}{d}$ = -(-6) = 6

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