Answer: Appropriate Question :- If α and β are the zeroes of the quadratic polynomial f(x) = x² – px + q, then find the value of α² + β². Given :- α and β are the zeroes of the quadratic polynomial f(x) = x² – px + q. To Find :- What is the value of α² + β². Solution :- Given equation : \dashrightarrow \sf\bold{x^2 – px + q}⇢x 2 −px+q where, a = 1 b = – p c = q Now, we have to find the sum and product of the zeroes : \clubsuit♣ Sum of Zeroes : \begin{gathered}\longmapsto \sf\boxed{\bold{\pink{Sum\: of\: Zeroes\: (\alpha + \beta) =\: \dfrac{- b}{a}}}}\\\end{gathered} ⟼ SumofZeroes(α+β)= a −b Then, \implies \sf \alpha + \beta =\: \dfrac{- (- p)}{1}⟹α+β= 1 −(−p) \begin{gathered}\implies \sf \alpha + \beta =\: \dfrac{p}{1}\: \bigg\lgroup \bold{\purple{- \times – =\: +}} \bigg\rgroup\\\end{gathered} ⟹α+β= 1 p ⎩ ⎪ ⎪ ⎪ ⎧ −×−=+ ⎭ ⎪ ⎪ ⎪ ⎫ \implies \sf\bold{\green{\alpha + \beta =\: p}}⟹α+β=p Again, \clubsuit♣ Product of Zeroes : \begin{gathered}\longmapsto \sf\boxed{\bold{\pink{Product\: of\: Zeroes\: (\alpha\beta) =\: \dfrac{c}{a}}}}\\\end{gathered} ⟼ ProductofZeroes(αβ)= a c Then, \implies \sf \alpha\beta =\: \dfrac{q}{1}⟹αβ= 1 q \implies \sf\bold{\green{\alpha\beta =\: q}}⟹αβ=q Now, we have to find the value of α² + β² : As we know that : \longmapsto \sf\boxed{\bold{\pink{a^2 + b^2 =\: (a + b)^2 – 2ab}}}⟼ a 2 +b 2 =(a+b) 2 −2ab We have : α + β = p αβ = q According to the question by using the formula we get, \leadsto \sf \alpha^2 + \beta^2 =\: (\alpha + \beta)^2 – 2\alpha\beta⇝α 2 +β 2 =(α+β) 2 −2αβ \leadsto \sf \alpha^2 + \beta^2 =\: (p)^2 – 2q⇝α 2 +β 2 =(p) 2 −2q \leadsto \sf\bold{\red{\alpha^2 + \beta^2 =\: p^2 – 2q}}⇝α 2 +β 2 =p 2 −2q \therefore∴ The value of α² + β² is p² – 2q. Reply
Answer: Step by Step Solution STEP1: Equation at the end of step 1 ((((x4) – 22×3) – x2) + 16x) – 12 STEP2: Polynomial Roots Calculator : 2.1 Find roots (zeroes) of : F(x) = x4-4×3-x2+16x-12 Polynomial Roots Calculator is a set of methods aimed at finding values of x for which F(x)=0 hσpє ít’ѕ hєlpful tσ чσu Reply
Answer:
Appropriate Question :-
If α and β are the zeroes of the quadratic polynomial f(x) = x² – px + q, then find the value of α² + β².
Given :-
α and β are the zeroes of the quadratic polynomial f(x) = x² – px + q.
To Find :-
What is the value of α² + β².
Solution :-
Given equation :
\dashrightarrow \sf\bold{x^2 – px + q}⇢x
2
−px+q
where,
a = 1
b = – p
c = q
Now, we have to find the sum and product of the zeroes :
\clubsuit♣ Sum of Zeroes :
\begin{gathered}\longmapsto \sf\boxed{\bold{\pink{Sum\: of\: Zeroes\: (\alpha + \beta) =\: \dfrac{- b}{a}}}}\\\end{gathered}
⟼
SumofZeroes(α+β)=
a
−b
Then,
\implies \sf \alpha + \beta =\: \dfrac{- (- p)}{1}⟹α+β=
1
−(−p)
\begin{gathered}\implies \sf \alpha + \beta =\: \dfrac{p}{1}\: \bigg\lgroup \bold{\purple{- \times – =\: +}} \bigg\rgroup\\\end{gathered}
⟹α+β=
1
p
⎩
⎪
⎪
⎪
⎧
−×−=+
⎭
⎪
⎪
⎪
⎫
\implies \sf\bold{\green{\alpha + \beta =\: p}}⟹α+β=p
Again,
\clubsuit♣ Product of Zeroes :
\begin{gathered}\longmapsto \sf\boxed{\bold{\pink{Product\: of\: Zeroes\: (\alpha\beta) =\: \dfrac{c}{a}}}}\\\end{gathered}
⟼
ProductofZeroes(αβ)=
a
c
Then,
\implies \sf \alpha\beta =\: \dfrac{q}{1}⟹αβ=
1
q
\implies \sf\bold{\green{\alpha\beta =\: q}}⟹αβ=q
Now, we have to find the value of α² + β² :
As we know that :
\longmapsto \sf\boxed{\bold{\pink{a^2 + b^2 =\: (a + b)^2 – 2ab}}}⟼
a
2
+b
2
=(a+b)
2
−2ab
We have :
α + β = p
αβ = q
According to the question by using the formula we get,
\leadsto \sf \alpha^2 + \beta^2 =\: (\alpha + \beta)^2 – 2\alpha\beta⇝α
2
+β
2
=(α+β)
2
−2αβ
\leadsto \sf \alpha^2 + \beta^2 =\: (p)^2 – 2q⇝α
2
+β
2
=(p)
2
−2q
\leadsto \sf\bold{\red{\alpha^2 + \beta^2 =\: p^2 – 2q}}⇝α
2
+β
2
=p
2
−2q
\therefore∴ The value of α² + β² is p² – 2q.
Answer:
Step by Step Solution
STEP1:
Equation at the end of step 1
((((x4) – 22×3) – x2) + 16x) – 12
STEP2:
Polynomial Roots Calculator :
2.1 Find roots (zeroes) of : F(x) = x4-4×3-x2+16x-12
Polynomial Roots Calculator is a set of methods aimed at finding values of x for which F(x)=0
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