if a and b are the zeros of quadratic polynomial f(×) =a×2+b×+c, then evaluate a2/b+b2/a
Answers
Step-by-step explanation:
Given polynomial, f(x) = ax² + bx + c
If, α and β are the zeros of the quadratic polynomial.
\begin{gathered} \alpha + \beta = \frac{ - b}{a} \\ \\ \alpha \beta = \frac{c}{a} \end{gathered}
α+β=
a
−b
αβ=
a
c
Now, (1)
\begin{gathered}( \alpha - \beta )^{2} = ( \alpha + \beta ) ^{2} - 4 \alpha \beta \\ \\ ( \alpha - \beta ) ^{2} = ( \frac{ - b}{a} ) ^{2} - 4 (\frac{c}{a} ) \\ \\ \alpha - \beta = \frac{\sqrt{ {b}^{2} - 4ac} }{a}\end{gathered}
(α−β)
2
=(α+β)
2
−4αβ
(α−β)
2
=(
a
−b
)
2
−4(
a
c
)
α−β=
a
b
2
−4ac
Now (2)
\begin{gathered} \frac{1}{ \alpha } - \frac{1}{ \beta } = \frac{ \beta - \alpha }{ \alpha \beta } = \frac{ - \sqrt{ {b}^{2} - 4ac} \times a }{c \times a} \\ \\ \frac{1}{ \alpha } - \frac{1}{ \beta } = \frac{ - \sqrt{ {b}^{2} - 4ac}}{c} \end{gathered}
α
1
−
β
1
=
αβ
β−α
=
c×a
−
b
2
−4ac
×a
α
1
−
β
1
=
c
−
b
2
−4ac
Now (3)
\begin{gathered} \frac{1}{ \alpha } + \frac{1}{ \beta } - 2 \alpha \beta \\ \\ = \frac{ \alpha + \beta }{ \alpha \beta } - 2 \alpha \beta \\ \\ = \frac{ - b}{c} - 2( \frac{c}{a} ) \\ \\ = - \frac{ab - {c}^{2} }{ac} \\ \\ = \frac{ {c}^{2} - ab}{ac} \end{gathered}
α
1
+
β
1
−2αβ
=
αβ
α+β
−2αβ
=
c
−b
−2(
a
c
)
=−
ac
ab−c
2
=
ac
c
2
−ab
Now (4)
\begin{gathered} \alpha^{2} \beta + \alpha \beta ^{2} \\ \\ = \alpha \beta ( \alpha + \beta ) \\ \\ = ( \frac{ - b}{a} )( \frac{c}{a} ) \\ \\ = \frac{ - bc}{a ^{2} } \end{gathered}
α
2
β+αβ
2
=αβ(α+β)
=(
a
−b
)(
a
c
)
=
a
2
−bc
Now (5)
\begin{gathered} { \alpha }^{4} + { \beta }^{4} \\ \\ = ( \alpha ^{2} + \beta ^{2} ) ^{2} - 2 { \alpha }^{2} { \beta }^{2} \\ \\ =( ( \alpha + \beta )^{2} - 2 \alpha \beta ) ^{2} - 2( { \alpha \beta })^{2} \\ \\ = {( (\frac{b ^{2} }{ {a}^{2} }) - 2 (\frac{c}{a}) ) }^{2} - 2( \frac{ {c}^{2} }{ {a}^{2} }) \\ \\ = (\frac{ {b}^{2} - 2ac }{ {a}^{2} } ) ^{2} - 2 \frac{ {c}^{2} }{ { a}^{2} } \\ \\ = \frac{ {b}^{4} + 4 {a}^{2} {c}^{2} - 4ac {b}^{2} - 2 {a}^{2} {c}^{2} }{ {a}^{4} } \\ \\ = \frac{ {b}^{2}(b ^{2} - 4ac) + 2 {a}^{2} {c}^{2} }{ {a}^{4} } \end{gathered}
α
4
+β
4
=(α
2
+β
2
)
2
−2α
2
β
2
=((α+β)
2
−2αβ)
2
−2(αβ)
2
=((
a
2
b
2
)−2(
a
c
))
2
−2(
a
2
c
2
)
=(
a
2
b
2
−2ac
)
2
−2
a
2
c
2
=
a
4
b
4
+4a
2
c
2
−4acb
2
−2a
2
c
2
=
a
4
b
2
(b
2
−4ac)+2a
2
c
2
Now (6)
\begin{gathered} \frac{1}{a \alpha + b} + \frac{1}{a \beta + b} \\ \\ = \frac{a( \alpha + \beta ) + 2b}{( a\alpha + b)( a \beta + b)} \\ \\ = \frac{ - b + 2b}{ {a}^{2}( \frac{c}{a} ) + {b}^{2} + ab( \frac{ -b }{a} )} \\ \\ = \frac{b}{ac + {b}^{2} - {b}^{2} } \\ \\ = \frac{b}{ac} \end{gathered}
aα+b
1
+
aβ+b
1
=
(aα+b)(aβ+b)
a(α+β)+2b
=
a
2
(
a
c
)+b
2
+ab(
a
−b
)
−b+2b
=
ac+b
2
−b
2
b
=
ac
b
Now (7)
\begin{gathered} \frac{ \beta }{a \alpha + b} + \frac{ \alpha }{a \beta + b} \\ \\ = \frac{ \beta (a \beta + b) + \alpha (a \alpha + b)}{(a \alpha + b)(a \beta + b)} \\ \\ = \frac{a( { \alpha }^{2} + { \beta }^{2} ) + b( \alpha + \beta )}{ {a}^{2} } \\ \\ = \frac{a( \frac{b ^{2} }{ {a}^{2} } - \frac{2c}{a}) - \frac{ {b}^{2} }{a} }{ {a}^{2} } \\ \\ = \frac{ \frac{ {b}^{2} - 2ac - {b}^{2} }{a} }{ {a}^{2} } \\ \\ = \frac{ - 2c}{ {a}^{2} } \\ \end{gathered}
aα+b
β
+
aβ+b
α
=
(aα+b)(aβ+b)
β(aβ+b)+α(aα+b)
=
a
2
a(α
2
+β
2
)+b(α+β)
=
a
2
a(
a
2
b
2
−
a
2c
)−
a
b
2
=
a
2
a
b
2
−2ac−b
2
=
a
2
−2c
Now (8)
\begin{gathered}a( \frac{ { \alpha }^{2} }{ \beta } + \frac{ { \beta }^{2} }{ \alpha } ) + b( \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } ) \\ \\ = \frac{a( { \alpha }^{3} + { \beta }^{3}) + b( { \alpha }^{2} + { \beta }^{2} ) }{ \alpha \beta } \\ \\ = \frac{a( \frac{ {b}^{3} }{ {a}^{3} } + 3 \frac{bc}{ {a}^{2} } ) + b( \frac{ {b}^{2} }{ {a}^{2} } - 2 \frac{c}{a}) }{ \alpha \beta } \\ \\ = \frac{ {b}^{3} + 3abc + {b}^{3} - 2abc }{ {a}^{2} \times \frac{c}{a} } \\ \\ = \frac{2 {b}^{3} + abc }{ac} \end{gathered}
a(
β
α
2
+
α
β
2
)+b(
β
α
+
α
β
)
=
αβ
a(α
3
+β
3
)+b(α
2
+β
2
)
=
αβ
a(
a
3
b
3
+3
a
2
bc
)+b(
a
2
b
2
−2
a
c
)
=
a
2
×
a
c
b
3
+3abc+b
3
−2abc
=
ac
2b
3
+abc
x=a*2+b*2+c
x=2a+2b+c ans