# Thread: quadratic equation

1. ## quadratic equation

have i done this correctly? was unsure about the 18/2 part?

2. ## Re: quadratic equation

Given an equation of the form

$ax^2 + bx +c=0,$

we can find solutions for $x$ using the formula

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$

Check that you have used this formula correctly. An easy way to tell if you have the right answer is to substitute your values for $x$ back into the original equation to see if you get the right answer i.e try and work out whether $(4.45)^2+2(4.45)-3.5$ really equals zero. If not then you need to check your calculations carefully.

3. ## Re: quadratic equation

You have to be more careful; you should get:

x = [-2 +- SQRT(18)] / 2

4. ## Re: quadratic equation

I always seem to make careless mistakes, so it should have been 2.12 rather than 2.45 !!!

5. ## Re: quadratic equation

there are two problems with your solution:

1) EVERYTHING goes over the 2, not just "the square root part"

2)you should wind up with:

$\frac{-2 \pm \sqrt{18}}{2}$

it's ok to leave it like this, but if you want to simplify:

$\sqrt{18} = \sqrt{2}\sqrt{9} = 3\sqrt{2}$.

neither:

$\frac{-2 + 3\sqrt{2}}{2}$ nor $\frac{-2 - 3\sqrt{2}}{2}$ are even close to either one of your "proposed" answers 2.12 or 2.45. there is no √6 involved in the correct solution.

6. ## Re: quadratic equation

Yeh that's what I mean about making careless mistakes, so really it would be 0.12 and -4.12 ? Thanks!

7. ## Re: quadratic equation

NO! Try again ..... CAREFULLY!

8. ## Re: quadratic equation

1.12 and -3.12?

Eeek, hope this is right now :/

YES.

10. ## Re: quadratic equation

Woooo haha thanks Wilmer!