example: x^2 - 13x +36 = 0

one question says Write a quadratic equation that has the given solutions.
12 and 5

how would I go about doing this?

2. Originally Posted by trippym

example: x^2 - 13x +36 = 0

one question says Write a quadratic equation that has the given solutions.
12 and 5

how would I go about doing this?
When you solve a quadratic that has whole number solutions, what does the second to last line usually look like?

The solutions are: x = 12 and x = 5
=> x - 12 = 0 and x - 5 = 0
so (x - 12)(x - 5) = 0
=> x^2 - 17x + 60 = 0

3. Originally Posted by trippym

example: x^2 - 13x +36 = 0
Here's one way, the "ac" method.
x^2 - 13x + 36 = 0

Multiply the coefficient of x^2 by the constant term: 1*36 = 36
Now write down all the ways to factor 36:
36 = 1*36, -1*-36
36 = 2*18, -2*-18
etc.
Now look at your list and find the two factors that add up to the coefficient of the x term -13:
36 = -4*-9; -4 + -9 = -13

So this is what we do:
x^2 - 13x + 36 = 0

x^2 + (-4x + -9x) + 36 = 0 <--- We could have used (-9x + -4x) also.

(x^2 - 4x) + (-9x + 36) = 0 <--- This part is called "factoring by grouping"

x(x - 4) + -9(x - 4) = 0

Note that each term has a factor of x - 4 in it, so factor an x - 4 out on the right:
(x - 9)(x - 4) = 0

Thus either x - 9 = 0 or x - 4 = 0. Thus x = 4 or x = 9.

This method will always work unless the quadratic cannot be factored over the rational numbers, that is to say that it won't work if the solutions are (non-rational) real or complex numbers.

-Dan