find sum of zeroes for polynomial (2x2-5x-8)
Answers
Answer:
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Step-by-step explanation:
Given:
x^2-5x+8x2−5x+8
\textbf{To find:}To find:
\text{Sum of the squares of the zeros}Sum of the squares of the zeros
\textbf{Solution:}Solution:
\text{Let the zeros of $x^2-5x+8$ be $\alpha$ and $\beta$}Let the zeros of x2−5x+8 be α and β
\text{Then,}Then,
\alpha+\beta=\dfrac{-b}{a}α+β=a−b
\alpha+\beta=\dfrac{5}{1}α+β=15
\implies\boxed{\alpha+\beta=5}⟹α+β=5
\alpha\,\beta=\dfrac{c}{a}αβ=ac
\alpha\,\beta=\dfrac{8}{1}αβ=18
\implies\boxed{\alpha\,\beta=8}⟹αβ=8
\text{Now,}Now,
{\alpha}^2+{\beta}^2α2+β2
=(\alpha+\beta)^2-2\,\alpha\beta=(α+β)2−2αβ
=(5)^2-2(8)=(5)2−2(8)
=25-16=25−16
=9=9
\implies\boxed{{\alpha}^2+{\beta}^2=9}⟹α2+β2=9
\therefore\textbf{Sum of the squares of zeros of $x^2-5x+8$ is 9}∴Sum of the squares of zeros of x2−5x+8 is 9
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d) 2x^2 + x + 3 = 0
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2x² - 5x - 8 = 0
a = 2 , b = -5 , c = -8
Sum of the zeroes