# Solve the equation

• Jan 6th 2010, 12:54 PM
Mukilab
Solve the equation
How would I go about solving this?

$6x^2+17x-39=0$

EDIT: Factorised into $(-13+-3x)(3+-2x)=0$But that didn't help :(
• Jan 6th 2010, 12:57 PM
pomp
Quote:

Originally Posted by Mukilab
How would I go about solving this?

$6x^2+17x-39=0$

EDIT: Factorised into $(-13+-3x)(3+-2x)=0$But that didn't help :(

You have that (-13 - 3x)(3 - 2x) = 0 . Therefore either (3 - 2x) = 0 , or (-13 - 3x) = 0.

Can you solve from here?
• Jan 6th 2010, 12:59 PM
Mukilab
(3 - 2x) = 0 , or (-13 - 3x) = 0.

therefore

$x=1.5$ or $x=\frac{-13}{3}$

It has to be one definite answer though. It tells me from this I should be able to calculate a side in a polygon (I need to find x). Do I try subbing both into the formula and see which works?
• Jan 6th 2010, 01:02 PM
e^(i*pi)
Quote:

Originally Posted by Mukilab
(3 - 2x) = 0 , or (-13 - 3x) = 0.

therefore

$x=1.5$ or $x=\frac{-13}{3}$

It has to be one definite answer though. It tells me from this I should be able to calculate a side in a polygon (I need to find x). Do I try subbing both into the formula and see which works?

If you're measuring a side on a polygon is follows that $x>0$ as you're measuring a length. So discard the negative root
• Jan 6th 2010, 01:02 PM
pomp
Quote:

Originally Posted by Mukilab
(3 - 2x) = 0 , or (-13 - 3x) = 0.

therefore

$x=1.5$ or $x=\frac{-13}{3}$

It has to be one definite answer though. It tells me from this I should be able to calculate a side in a polygon (I need to find x). Do I try subbing both into the formula and see which works?

Both are correct answers to the equation, but only one is the answer you seek for the given problem. Only one of the answers makes sense as the length of a polygon's side, can you see which one and why?
• Jan 6th 2010, 01:03 PM
pickslides
Quote:

Originally Posted by Mukilab

Do I try subbing both into the formula and see which works?

Not really, you can do it as a check, both should give 0.
• Jan 6th 2010, 01:07 PM
Mukilab
Quote:

Originally Posted by e^(i*pi)
If you're measuring a side on a polygon is follows that $x>0$ as you're measuring a length. So discard the negative root

Ok

Quote:

Originally Posted by pomp
Both are correct answers to the equation, but only one is the answer you seek for the given problem. Only one of the answers makes sense as the length of a polygon's side, can you see which one and why?

Quote:

Originally Posted by pickslides
Not really, you can do it as a check, both should give 0.

Thank you for all the posts, most considerate.