Find the value of sum of
roots
of given Polynomial 2x²-4x+6
Answers
If α and β are Zeroes of a Quadratic Polynomial, ax² + bx + c then :
● α and β are roots of the Quadratic Equation, ax² + bx + c = 0
● \mathsf{Sum\;of\;the\;Roots\;(\alpha + \beta) = \dfrac{-b}{a}}SumoftheRoots(α+β)=
a
−b
● \mathsf{Product\;of\;the\;Roots\;(\alpha.\beta) = \dfrac{c}{a}}ProductoftheRoots(α.β)=
a
c
Given : α and β are zeros of Quadratic Polynomial 2x² - 4x + 5
Here : a = 2 and b = -4 and c = 5
● α and β are roots of the Quadratic Equation, 2x² - 4x + 5 = 0
● \mathsf{Sum\;of\;the\;Roots\;(\alpha + \beta) = \dfrac{4}{2} = 2}SumoftheRoots(α+β)=
2
4
=2
● \mathsf{Product\;of\;the\;Roots\;(\alpha.\beta) = \dfrac{5}{2}}ProductoftheRoots(α.β)=
2
5
Consider : (α + β)³
:\implies:⟹ (α + β)³ = α³ + 3α²β + 3αβ² + β³
:\implies:⟹ (α + β)³ = α³ + 3αβ(α + β) + β³
:\implies:⟹ α³ + β³ = (α + β)³ - 3αβ(α + β)
\mathsf{\implies \alpha^3 + \beta^3 = 2^3 - 3\bigg(\dfrac{5}{2}\bigg)(2)}⟹α
3
+β
3
=2
3
−3(
2
5
)(2)
\mathsf{\implies \alpha^3 + \beta^3 = 8 - (3\times 5)}⟹α
3
+β
3
=8−(3×5)
\mathsf{\implies \alpha^3 + \beta^3 = 8 - 15}⟹α
3
+β
3
=8−15
\mathsf{\implies \alpha^3 + \beta^3 = -7}⟹α
3
+β
3
=−7