Question

If alpha and beta are the zeroes of polynomial ax2+bx+c, then what is the value of alpha-beta?

Answer 1

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

First note that (a−b)2=a2−2ab+b2(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:

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

Substituting aa and bb with αα and ββ and a+ba+b and abab with −−baba and caca, respectively, we get:

(−ba)2=α2+2(ca)+β2(−ba)2=α2+2(ca)+β2

=b2a2−2ca=α2+β2=b2a2−2ca=α2+β2

Now that we know the value of α2+β2α2+β2, we can solve for α−βα−β.

(α−β)2=α2+β2−2αβ(α−β)2=α2+β2−2αβ

Substituting:

(α−β)2=b2a2−2ca−2ca(α−β)2=b2a2−2ca−2ca

(α−β)2=b2a2−4ca(α−β)2=b2a2−4ca

(α−β)2=b2a2−4aca2(α−β)2=b2a2−4aca2

(α−β)2=b2−4aca2(α−β)2=b2−4aca2

Taking the square root of both sides:

(α−β)=±b2−4aca2−−−−−−−√(α−β)=±b2−4aca2

Thus we get the result

α−β=±b2−4ac−−−−−−−√aα−β=±b2−4aca

Answer 2

i m representing here alpha as A and beeta as B,

then

A+B=-b/a,

A.B=c/a,

hence

(A-B)²=(A+B)²-4AB,

(A-B)²=(-b/a)²-4×c/a,

(A-B)²=b²/a² - 4c/a,

(A-B)²=(b²-4ac)/a²,

then

(A-B)=√(b²-4ac)/a