Question

If alpha and beta are zeros of a polynomial ax2+bx+c then evaluate alpha-beta.

Answer 1

We know that α+β=−ba and α×β=ca, by Vieta’s formulas.

First note that (a−b)2=a2−2ab+b2. If we can find this value, we can find the square root to get our desired quantity.

Let’s find (a+b)2:

(a+b)2=a2+2ab+b2

Substituting a and b with α and β and a+b and ab with −ba and ca, respectively, we get:

(−ba)2=α2+2(ca)+β2

=b2a2−2ca=α2+β2

Now that we know the value of α2+β2, we can solve for α−β.

(α−β)2=α2+β2−2αβ

Substituting:

(α−β)2=b2a2−2ca−2ca

(α−β)2=b2a2−4ca

(α−β)2=b2a2−4aca2

(α−β)2=b2−4aca2

Taking the square root of both sides:

(α−β)=±√b²-4ac/a²

Thus we get the result

α−β=±√b²4ac/a

Hope u get the answer........