If alpha and beta are the zeroes of the polynomial 2x^2+7x+5, then value of alpha+beta+alpha*beta
Answers
Step-by-step explanation:
give th root of equation
2x
2
−7x+5=0 are \alpha ,\betaα,β
Compare it with:
ax^2+bx+c=0ax
2
+bx+c=0
So:
a=2
,b=-7
and c=5
We know that:
\alpha+\beta=-b/aα+β=−b/a
=7/2...............(1)
\alpha*\beta=c/aα∗β=c/a
=5/2.............(2)
Given
\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}
β
α
2
+
α
β
2
\implies \frac{\alpha^3+\beta^3}{\alpha*\beta}⟹
α∗β
α
3
+β
3
\implies \frac{(\alpha+\beta)(\alpha^2-\alpha*\beta+\beta^2)}{\alpha*\beta}⟹
α∗β
(α+β)(α
2
−α∗β+β
2
)
\implies \frac{(\alpha+\beta)([\alpha+\beta]^2-3\alpha*\beta)}{\alpha*\beta}⟹
α∗β
(α+β)([α+β]
2
−3α∗β)
From 1 and 2
\implies \frac{(7/2)(49/4-3*5/2)}{5/2}⟹
5/2
(7/2)(49/4−3∗5/2)
\implies \frac{(7/2)(19/4)}{5/2}⟹
5/2
(7/2)(19/4)
\implies \frac{133/8}{5/2}⟹
5/2
133/8
\implies 133/8*2/5⟹133/8∗2/5
\implies 133/20⟹133/20
The answer is 133/20
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