Question

Quadratic polynomial 2x2 – 3x + 1 has zeroes as α and β. Now form a quadratic polynomial whose zeroes are 3α and 3β

Answer 1

Answer:

There are two ways to do it.

One is factorize 2x²-3x+1 as (2x-1)(x-1) and get the zeroes as x=1 and x=1/2.

And then for the new polynomial, the zeroes are x=3 and x=3/2, i.e the new polynomial is k(x-3)(x-3/2), where k is any non zero integer.

Let's take k as 2. Therefore we get the polynomial as 2(x-3)(x-3/2) = (x-3)(2x-3) = 2x²-9x+9.

(You could also you the form k[x² - (α'+β')x + α'β'] to get the new polynomial. Where α' and β' are the new roots i.e 3 and 3/2 and k is any non zero integer)

Other method is to transform the roots of the equation.

i.e use y = 3x and replace x=y/3 in the original equation to get a new polynomial in y.

i.e 2(y/3)² - 3(y/3) + 1 = 2y²/9 - y + 1. We can get any other polynomial with same roots by multiplying a non zero number such as say, 9. So we get 2y² - 9y + 9. Replacing y with x, we get, 2x²-9x+9 as the required polynomial.

Please note: The first method is most common everywhere and most preferred.