Math, asked by reem51, 5 months ago

if a and b are zeroes of polynomial pX= x²-3x+2. then find 1/a+1/a​

Answers

Answered by Anonymous
43

Given :

\sf\:p(x)=x^2-3x+2

To Find :

\sf\dfrac{1}{\alpha}+\dfrac{1}{\beta}

Theory :

If \sf \alpha \: and \: \beta are zeroes of quadratic polynomial

\sf  f(x) = {x}^{2} + bx + c

Then ,\sf  \alpha + \beta = \dfrac{ - cofficient \: of \: x}{cofficient \: of \: x {}^{2} }

\sf \: and \: \alpha \beta = \dfrac{constant}{cofficient \: of \: x {}^{2} }

Solution

We have ,

\rm\:p(x)=x^2-3x+2

We know that

\sf  \alpha + \beta = \dfrac{ - cofficient \: of \: x}{cofficient \: of \: x {}^{2} }

\sf  \implies \alpha + \beta = \dfrac{ - ( - 3)}{ 1} = 3...(1)

And

\sf \: and \: \alpha \beta = \dfrac{constant}{cofficient \: of \: x {}^{2} }

\sf\implies\alpha\times\beta=\dfrac{2}{1}=2..(2)

We have to find the value of

\sf\dfrac{1}{\alpha}+\dfrac{1}{\beta}

\sf=\dfrac{\alpha+\beta}{\alpha\times\beta}

Now put the values of equation(1)and(2)

\sf=\dfrac{3}{2}

Therefore ,

\sf\dfrac{1}{\alpha}+\dfrac{1}{\beta}=\dfrac{3}{2}


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Answered by Anonymous
17

Step-by-step explanation:

Given :

\sf\:p(x)=x^2-3x+2p(x)=x

2

−3x+2

To Find :

\sf\dfrac{1}{\alpha}+\dfrac{1}{\beta}

α

1

+

β

1

Theory :

If \sf \alpha \: and \: \betaαandβ are zeroes of quadratic polynomial

\sf f(x) = {x}^{2} + bx + cf(x)=x

2

+bx+c

Then ,\sf \alpha + \beta = \dfrac{ - cofficient \: of \: x}{cofficient \: of \: x {}^{2} }α+β=

cofficientofx

2

−cofficientofx

\sf \: and \: \alpha \beta = \dfrac{constant}{cofficient \: of \: x {}^{2} }andαβ=

cofficientofx

2

constant

Solution

We have ,

\rm\:p(x)=x^2-3x+2p(x)=x

2

−3x+2

We know that

\sf \alpha + \beta = \dfrac{ - cofficient \: of \: x}{cofficient \: of \: x {}^{2} }α+β=

cofficientofx

2

−cofficientofx

\sf \implies \alpha + \beta = \dfrac{ - ( - 3)}{ 1} = 3...(1)⟹α+β=

1

−(−3)

=3...(1)

And

\sf \: and \: \alpha \beta = \dfrac{constant}{cofficient \: of \: x {}^{2} }andαβ=

cofficientofx

2

constant

\sf\implies\alpha\times\beta=\dfrac{2}{1}=2..(2)⟹α×β=

1

2

=2..(2)

We have to find the value of

\sf\dfrac{1}{\alpha}+\dfrac{1}{\beta}

α

1

+

β

1

\sf=\dfrac{\alpha+\beta}{\alpha\times\beta}=

α×β

α+β

Now put the values of equation(1)and(2)

\sf=\dfrac{3}{2}=

2

3

Therefore ,

\sf\dfrac{1}{\alpha}+\dfrac{1}{\beta}=\dfrac{3}{2}

α

1

+

β

1

=

2

3

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