Question

If α and β are the zeros of the quadratic polynomial f(x) = x² – 1, find a quadratic polynomial whose zeros are `(2alpha)/beta` and `(2beta)/alpha`.

Answer 1

Since α and β are the zeros of the quadratic polynomial f(x) = x² − 1

The roots are α and β

`alpha+beta="-coefficient of x"/("coefficient of "x^2)`

`alpha+beta=0/1`

`alpha+beta=0`

`alphabeta="constant term"/("coefficient of "x^2)`

`alphabeta=(-1)/1`

`alphabeta=-1`

Let S and P denote respectively the sum and product of zeros of the required polynomial. Then,

`S=(2alpha)/beta+(2beta)/alpha`

Taking least common factor we get,

`S=(2alpha^2+2beta^2)/(alphabeta)`

`S=(2(alpha^2+beta^2))/(alphabeta)`

`S=(2[(alpha+beta)-2alphabeta])/(alphabeta)`

`S=(2[(0)-2(-1)])/-1`

`S=(2[-2(-1)])/-1`

`S=(2xx2)/-1`

`S=4/-1`

S = -4

`P=(2alpha)/betaxx(2beta)/alpha`

P = 4

Hence, the required polynomial f(x) is given by,

f(x) = k(x² - Sx + P)

f(x) = k(x² -(-4)x + 4)

f(x) = k(x² +4x +4)

Hence, required equation is f(x) = k(x² + 4x + 4) Where k is any non zero real number.