Question

If α,β,γ are the zeroes of polynomial x3-3x2+5x+7then1/α+1/β+1/γ?

Answer 1

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${\huge{\purple{\mathbb{\bold{ANSWER}}}}}$

${\large{\blue{\bold{\underline{Given:-}}}}}$

▪︎ Polynomial:- x³-3x²+5x+7

▪︎ α, β, γ are zeroes os above polynomial.

${\large{\blue{\bold{\underline{To\:Find:-}}}}}$

▪︎ The value of 1/α+1/β+1/γ.

${\large{\blue{\bold{\underline{Concepts\:Used:-}}}}}$

▪︎ In a cubic polynomial,

${\small{\orange{Sum\:of\:zeroes\:=\:-b/a}}}$

Where b is coefficient of x² and a is coefficient of x³.

=》 α+β+γ = -b/a.

=》 α+β+γ = -(3)/1

=》 α+β+γ = -3 ---(1).

${\small{\orange{Product\:of\:zeroes\:taken\:2\:at\:a\:time}}}$

${\small{\orange{=\:c/a}}}$

Where c is the coefficient of x.

=》 αβ+βγ+αγ = c/a.

=》 αβ+βγ+αγ = 5/1.

=》 αβ+βγ+αγ = 5 ---(2).

${\small{\orange{Product\:of\:zeroes\:=\:-d/a}}}$

Where d is the constant term.

=》 αβγ = -d/a

=》 αβγ = -(7)/1

=》 αβγ = -7 ---(3).

${\large{\blue{\bold{\underline{Therfore,}}}}}$

=》 1/α+1/β+1/γ

= αβ+ βγ+ αγ/ αβγ

{By taking LCM}

= 5/(-7)

{From (2) and (3)}

${\large{\red{\boxed{\bold{=\:-5/7}}}}}$

Therefore 1/α+1/β+1/γ = -5/7.

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