How can I verify the statement A * B = c/a for quadratic equations?

[mathdad]

Let A and B be roots of the quadratic equation
ax^2 + bx + c = 0. Verify the statement.

A * B = c/a

What are the steps to verify this statement?

I can let A = [-b + sqrt{b^2 - 4ac}]/2a and, of course, let
B = [-b - sqrt{b^2 - 4ac}]/2a. If I multiply the left side, the statement A * B becomes c/a, right?

 

[MarkFL]

I think what I would do is write:

\(\displaystyle ax^2+bc+c=k(x-A)(x-B)=kx^2-k(A+B)x+kAB\)

Equating coefficients, we then find:

\(\displaystyle k=a\)

\(\displaystyle kAB=c\implies AB=\frac{c}{a}\)

 

[mathdad]

I think what I would do is write:

\(\displaystyle ax^2+bc+c=k(x-A)(x-B)=kx^2-k(A+B)x+kAB\)

Equating coefficients, we then find:

\(\displaystyle k=a\)

\(\displaystyle kAB=c\implies AB=\frac{c}{a}\)

Ok but can it be done as expressed in my post?

 

[Greg]

Hint:

$$(a-b)(a+b)=a^2-b^2$$

 

[mathdad]

Hint:

$$(a-b)(a+b)=a^2-b^2$$

Can you be more specific?

 

[Greg]

$$\frac{-b+\sqrt{b^2-4ac}}{2a}\cdot\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{b^2-(\sqrt{b^2-4ac})^2}{4a^2}=\frac{4ac}{4a^2}=\frac ca$$

 

[mathdad]

$$\frac{-b+\sqrt{b^2-4ac}}{2a}\cdot\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{b^2-(\sqrt{b^2-4ac})^2}{4a^2}=\frac{4ac}{4a^2}=\frac ca$$

This is exactly what I thought should be done.