Question

If alpha and beta are zeros of polynomial x2-3.Then a form a quadratic polynomial whose zeros are 1/alpha and 1/beta

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

The quadratic equation whose roots are   $\frac{1}{\alpha }$ and $\frac{1}{\beta }$ = 3x² - 1

Step-by-step explanation:

Given,

α and β are zeros of the polynomial x²-3

To find,

A quadratic polynomial whose zeros are $\frac{1}{\alpha }$ and $\frac{1}{\beta }$

Recall the concepts

If α and β are zeros of the polynomial ax²+bx+c, then

Sum of roots  = α+β = $\frac{-b}{a}$

Product of roots =αβ = $\frac{c}{a}$

If the roots are given, then the equation of a quadratic polynomial is given by   x²-(Sum of zeros)x+ Product of zeros ----------------(A)

Solution:

Since α and β are zeros of the polynomial x²-3

Comparing this equation with ax²+bx+c, we get

a = 1 , b = 0 and c = -3,

Then we have  α+β = $\frac{-b}{a}$  = 0

αβ  = $\frac{c}{a}$ = $\frac{-3}{1}$ = -3

α+β  = 0 and αβ  = -3 -----------------(1)

Required to find the quadratic equation whose zeros are $\frac{1}{\alpha }$ and $\frac{1}{\beta }$

From equation (A),

the quadratic equation whose zeros are  $\frac{1}{\alpha }$ and $\frac{1}{\beta }$ is given by

x²-(Sum of zeros)x+ Product of zeros

Sum of zeros =  $\frac{1}{\alpha }$ +  $\frac{1}{\beta }$ = $\frac{\alpha +\beta }{\alpha \beta }$

Product of zeros =  $\frac{1}{\alpha }$ × $\frac{1}{\beta }$  = $\frac{1}{\alpha \beta }$

Substituting the values of α+β and αβ   from equation (1) we get,

Sum of zeros = $\frac{0}{-3}$ = 0

Product of zeros = $\frac{1}{-3}$ = $\frac{-1}{3}$

the quadratic equation whose zeros are $\frac{1}{\alpha }$ and $\frac{1}{\beta }$ is given by

x²-(0)x+ $\frac{-1}{3}$

$\frac{1}{3}(3x^2 -1)$

Ignoring the constant term,

The quadratic equation whose roots are   $\frac{1}{\alpha }$ and $\frac{1}{\beta }$ = 3x² - 1

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