If ava+bvb=108 and avb+b√a= 36 then find the value of a² + 2? (a, b are positive real numbers]
Answers
Answer:
We know that, (a + b)2 = a2 + b2 + 2ab
= 234 + 2 x 108 = 450
(a - b)2 = a2 + b2 - 2ab
= 234 - 2 x 108 = 18
∴ (a + b)2/(a - b)2 = 450/18 = 25
⇒ [(a + b)/(a - b)]2 = 25
∴ (a + b)/(a - b) = √25 = 5
Answer:
Correct Question:
Verify that -1, 1/2, 1/3 are the zeroes of the polynomial 6x³ + x² – 4x + 1 and verify the relationship between the zeroes and the coefficients.
SOLUTION:
Given :
Polynomial P(x) = 6x³ + x² – 4x + 1
Three number : (-1), (1/2) and (1/3)
To prove,
Verify that -1, 1/2, 1/3 are the zeroes of the P(x) = 6x³ + x² - 4x + 1.
Verify the relationship between the zeroes and the coefficients.
So,
P(-1) = 0 or not
P(-1) = 6(-1)³ + (-1)² - 4(-1) + 1.
= 6 × (-1) + 1 + 4 + 1.
= -6 + 1 + 4 + 1
= -6 + 6
= 0
P(-1) = 0, so, (-1) is the zero of polyniomial P(x).
\rule{200}2
\bf P(\frac{1}{2}) = 0\ or\ notP(
2
1
)=0 or not
\bf P(\frac{1}{2}) =6 (\frac{1}{2})^3+(\frac{1}{2} )^2 -4 (\frac{1}{2} )+1P(
2
1
)=6(
2
1
)
3
+(
2
1
)
2
−4(
2
1
)+1
\bf P( \frac{1}{2} ) = \frac{6}{8} + \frac{1}{4} - \frac{4}{2} + 1P(
2
1
)=
8
6
+
4
1
−
2
4
+1
\bf P(\frac{1}{2}) =\frac{6+2-16+8}{8}P(
2
1
)=
8
6+2−16+8
\bf P(\frac{1}{2})=\frac{0}{8}P(
2
1
)=
8
0
\bf P(\frac{1}{2})=0P(
2
1
)=0
P(1/2) is also a zero of P(x).
\rule{200}2
\bf P( \frac{1}{3} ) = 6(\frac{1}{3} )^3 + ( \frac{1}{3} )^2 - 4 ( \frac{1}{3} ) + 1P(
3
1
)=6(
3
1
)
3
+(
3
1
)
2
−4(
3
1
)+1
\bf P(\frac{1}{3}) =\frac{6}{27}+\frac{1}{9}-\frac{4}{3}+1P(
3
1
)=
27
6
+
9
1
−
3
4
+1
\bf P(\frac{1}{3}) = \frac{2}{9}+\frac{1}{9}-\frac{4}{3} +1P(
3
1
)=
9
2
+
9
1
−
3
4
+1
\bf P(\frac{1}{3}) =\frac{2+1}{9}-\frac{4}{3}+1P(
3
1
)=
9
2+1
−
3
4
+1
\bf P(\frac{1}{3}) = \frac{3}{9}- \frac{4}{3}+1P(
3
1
)=
9
3
−
3
4
+1
\bf P(\frac{1}{3}) = \frac{1}{3}-\frac{4}{3}+1P(
3
1
)=
3
1
−
3
4
+1
\bf P(\frac{1}{3})=\frac{1-4}{3} +1P(
3
1
)=
3
1−4
+1
\bf P(\frac{1}{3})=\frac{-3}{3}+1P(
3
1
)=
3
−3
+1
\bf P(\frac{1}{3})=-1+1P(
3
1
)=−1+1
\bf P(\frac{1}{3})=0P(
3
1
)=0
So,
(1/3) is also a zero.
\rule{200}1
Now,
To verify the relationship between the zeroes and the coefficients.
From the polynomial, 6x³ + x² – 4x + 1
a = 6, b = 1, c = ( - 4 ), d = 1
And, we verified that (-1), (1/2), and (1/3) are zeroes so,
Let,
\boxed{\red{\alpha = -1} ,\ \pink{\beta = \frac{1}{2}} ,\ \green{\gamma = \frac{1}{3}}}
α=−1, β=
2
1
, γ=
3
1
\begin{gathered}\bf If\ \alpha,\beta,\gamma\ \bf are\ the\ zeroes\ of\ a\ cubic\ polynomial,\\ \bf ax^3+bx^2+cx+d,\ then\end{gathered}
If α,β,γ are the zeroes of a cubic polynomial,
ax
3
+bx
2
+cx+d, then
\boxed{\sf \alpha+\beta+\gamma =\frac{-b}{a}}
α+β+γ=
a
−b
, \boxed{\sf \alpha \beta+\beta \gamma + \gamma \alpha=\frac{c}{a}}
αβ+βγ+γα=
a
c
, \boxed{\sf \alpha \beta \gamma =\frac{-d}{a}}
αβγ=
a
−d
So,
\begin{gathered}\sf \dfrac{-b}{a}= \dfrac{-1}{6}\\ \\ \\ \sf \dfrac{c}{a} = \dfrac{-4}{6} = \dfrac{-2}{3} \\ \\ \\ \sf \dfrac{-d}{a} = \dfrac{-1}{6}\end{gathered}
a
−b
=
6
−1
a
c
=
6
−4
=
3
−2
a
−d
=
6
−1
\star \rule{50}2 \star \rule{50}2 \star
\begin{gathered}\boxed{\sf \alpha+\beta+\gamma =\frac{-b}{a}} \\ \\ \\ \sf =(-1)+(\frac{1}{2})+\frac{1}{3} \\ \\ \\ \sf = \dfrac{-6+3+2}{6} \\ \\ \\ \sf = \dfrac{-6+5}{6}\\ \\ \\ = \boxed{\frac{-1}{6}} = \dfrac{-b}{a}\end{gathered}
α+β+γ=
a
−b
=(−1)+(
2
1
)+
3
1
=
6
−6+3+2
=
6
−6+5
=
6
−1
=
a
−b
\rule{150}2
\begin{gathered}\boxed{\sf \alpha \beta+\beta \gamma + \gamma \alpha=\frac{c}{a}} \\ \\ \\ \sf= \bigg[(-1)\times (\frac{1}{2})\bigg] + \bigg[(\frac{1}{2}) \times (\frac{1}{3}) \bigg]+ \bigg[(\frac{1}{3})\times(-1)\bigg] \\ \\ \\ \sf = \frac{-1}{2} + \frac{1}{6} + \frac{-1}{3} \\ \\ \\ \sf = \dfrac{-3+1-2}{6} \\ \\ \\ \sf = \dfrac{-4}{6} \\ \\ \\ \sf =\boxed{ \frac{-2}{3}} = \dfrac{c}{a}\end{gathered}
αβ+βγ+γα=
a
c
=[(−1)×(
2
1
)]+[(
2
1
)×(
3
1
)]+[(
3
1
)×(−1)]
=
2
−1
+
6
1
+
3
−1
=
6
−3+1−2
=
6
−4
=
3
−2
=
a
c
\rule{150}2
\begin{gathered}\boxed{\sf \alpha \beta \gamma =\frac{-d}{a}} \\ \\ \\ \sf = (-1) \times (\frac{1}{2}) \times \frac{1}{3} \\ \\ \\ \sf = \boxed{\frac{-1}{6}}=\dfrac{-d}{a}\end{gathered}
αβγ=
a
−d
=(−1)×(
2
1
)×
3
1
=
6
−1
=
a
−d
Hence,
The relationship between the zeroes and the coefficients. is verified.