Question

If alpha and beta are the zeros of a polynomial x^2-3x-1 then find a quadratic polynomial whose zeroes are 1/alpha^2 &1/beta^2​

Answer 1

Solution

Given :-

• Polynomial , x² - 3x - 1 = 0
• α & β are zeroes .

Find :-

• Equation of polynomial, whose zeroes are 1/α² & 1/ β²

Explanation

Using Formula

☛ Sum of zeroes = -(Coefficient of x)/(coefficient of x²)

☛ product of zeroes = constant part /(coefficient of x²)

So,

==> Sum of zeroes = - (-3)/1

==> α + β = 3 -----------(1)

And,

==> product of zeroes = -1/1

==> α . β = -1 -----------(2)

Now, squaring both side of equ(1)

==>( α + β)² = 3²

==> α² + β² + 2 α β = 9

Keep Value by equ(2)

==> α² + β² = 9 - 2 * -1

==> α² + β² = 9 + 2

==> α² + β² = 11 ----------(3)

Now, calculate equation , whose zeroes are (1/α² & 1/ β²)

For This,

☛ Sum of zeroes = (1/α² + 1/ β²)

==> Sum of zeroes = ( α² + β²)/(α² . β²)

Keep Value by equ(2) & (3)

==> Sum of zeroes = 11/(-1)²

==> Sum of zeroes = 11

And,

☛ Product of zeroes = ((1/α² ) * ( 1/ β²)

==> Product of zeroes = 1/ (α² *β²)

Keep Value by equ(2)

==> Product of zeroes = 1/(-1)²

==> Product of zeroes = 1

Formula of equation,

☛ x² - (Sum of zeroes)x + product of zeroes = 0

Keep above values ,

==> x² - 11x + 1 = 0

Hence, Required formula

• x² - 11x + 1 = 0

__________________

Answer 2

$\implies x^2-3x-1\\ a=1\\b=-3\\ c=-1$

→ Finding the sum of the zeroes of equation

$\implies \alpha+\beta=\frac{-b}{a}=\frac{-3}{1}=3 \\ \implies \alpha\beta=\frac{-1}{1}=1$

Now,

the root are $\frac{1}{\alpha^2} \:and\:\frac{1}{\beta^2}$

Then,

The sum "S" is written as

$\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\alpha^2+\beta^2/( \alpha^2)(\beta^2)------(1)$

$(\alpha+\beta)^2-2\alpha\beta=\alpha^2\beta^2$

Putting value of $\alpha^2+\beta^2$in eq--------(1)

$(\alpha+\beta)^2-2\alpha\beta)/(\alpha^2)(\beta^2)\\ \alpha+\beta^2+2\alpha\beta/(\alpha\beta)^2$

put value of $\alpha+\beta \:and\:\alpha\beta$as derived earlier.

$[(3)^2-2(-1)]/(-1)^2\\9+2/1\\11/1$

Now product "P" is as follows

$1/ \alpha×\frac{1}{\beta}=\frac{1}{\alpha\beta}$

put the value of $\alpha\beta$

$\frac{1}{\alpha\beta}=\frac{1}{-1}=-1$

so,

equation is written as

$x^2-Sx+p=0$

put the value of S&P

$x^2-11x+(-1)=0\\ x^2-11x-1=0$

$\rule{300}2$

Hence

The required polynomial is ↓

$x^2-11x-1$