Question

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

Answer 1

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 ,

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

Answer 2

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