Question

What is the product of the roots of `ax^2+bx+c = 0" "` where `a != 0` ?

Answer 1

`c/a`

Explanation:

Method 1 - Direct multiplication

`(-b-sqrt(b^2-4ac))/(2a) * (-b+sqrt(b^2-4ac))/(2a)`

`=((-b)^2-(sqrt(b^2-4ac))^2)/(4a^2)`

`=(b^2-(b^2-4ac))/(4a^2)`

`=(4ac)/(4a^2)`

`=c/a`

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Method 2 - Think about the roots

Call the two values of this expression `r_1` and `r_2` - the roots of the quadratic equation `ax^2+bx+c = 0`.

Then we have:

`ax^2+bx+c`

`= a(x-r_1)(x-r_2)`

`= a(x^2-(r_1+r_2)x+r_1 r_2)`

`=ax^2-a(r_1 + r_2)x + a r_1 r_2`

Equating coefficients we find:

`c = a r_1 r_2`

So dividing both sides by `a` we find:

`r_1 r_2 = c/a`