I'm supposed to solve:

3X^2+17X+20=0

This is my working:

-> 3X^2-3X+20X+20=0

-> 3X(X-1)-20(X-1)=0

-> (X-1)(3X-20)=0

(X-1) -> X=1

(3X-20) -> 3X=20 -> X=20/3

X= 1 or 20/3

However, the anser to that question is

-4 or -5/3

What did I do wrong?

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- September 6th 2008, 11:52 PMjuliakWhy is this incorrect (quadratics)?
I'm supposed to solve:

3X^2+17X+20=0

This is my working:

-> 3X^2-3X+20X+20=0

-> 3X(X-1)-20(X-1)=0

-> (X-1)(3X-20)=0

(X-1) -> X=1

(3X-20) -> 3X=20 -> X=20/3

X= 1 or 20/3

However, the anser to that question is

-4 or -5/3

What did I do wrong? - September 6th 2008, 11:58 PMo_O
In general, can be factored by finding two numbers that multiply to get and add up to , call them . Then you split your quadratic: and you should be able to factor from there (assuming it was factorable in the first place)

For your quadratic, you have to find 2 numbers that multiply to (3)(20) = 60 that add up to 17. 12 and 5 should come to mind. So:

- September 6th 2008, 11:58 PMMatt Westwood
What's with all the negative signs already?

... and your factorisation falls over.

What I do in this situation when I can't find the numbers to factor by is use the quadratic formula. Then when I know what the answer is (or I've looked it up in the answers in the back of the book) I go back and use those numbers to factor it properly and I don't tell the teacher I cheated. - September 7th 2008, 12:07 AM11rdc11

Use the quad formula

- September 7th 2008, 12:07 AMjuliak
Oh okay I get it now.

Heh, thank you :D.