Math, asked by tanmaysingh1021, 4 hours ago

If ava+bvb=108 and avb+b√a= 36 then find the value of a² + 2? (a, b are positive real numbers]​

Answers

Answered by minakshideka1117
0

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

Answered by mguna90
0

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.

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