Quadratic Equation

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November 23rd 2008, 01:21 PM
Larianne

Quadratic Equation

Hi,

I don't think I'm getting the correct roots for this equation...

3x2(squared)- 2x- 6 = 0

That should read 3xsquared.

I had to use the b +/- formula and I'm getting x= 1.79 or x= -1.12

That can't be right???
November 23rd 2008, 01:30 PM
euclid2

Quote:

Originally Posted by Larianne

Hi,

I don't think I'm getting the correct roots for this equation...

3x2(squared)- 2x- 6 = 0

That should read 3xsquared.

I had to use the b +/- formula and I'm getting x= 1.79 or x= -1.12

That can't be right???

$3x^2 - 2x - 6 = 0$

Use the quadratic equation $-b+/- square root of \frac{b^2 -4ac}{2a}$

Where a is 3, b is -2 and c is -6.
November 23rd 2008, 01:34 PM
o_O

Let's format that a little better: $x = \frac{b \pm \sqrt{b^2 - 4ac}}{2a}$

Why don't you think that's right? You rounded off to a few decimal places which will throw off your answer a little bit but it is correct (again, rounded of course)
November 23rd 2008, 01:35 PM
Larianne

Quote:

Originally Posted by euclid2

$3x^2 - 2x - 6 = 0$

Use the quadratic equation $-b+/- square root of \frac{b^2 -4ac}{2a}$

Where a is 3, b is -2 and c is -6.

Yes, I've done that but I put my answers back into the equation to check and I'm not getting zero.
November 23rd 2008, 01:43 PM
euclid2

Quote:

Originally Posted by Larianne

Yes, I've done that but I put my answers back into the equation to check and I'm not getting zero.

$x = \frac{2 \pm \sqrt{-2^2 - 4(3)(-6)}}{2(3)}$
$x = \frac{2 \pm \sqrt{76}}{6}$
$x = \frac{2 + \sqrt{76}}{6}$ +
$x = \frac{2 - \sqrt{76}}{6}$
Therefore, $x =1.78 + x=-1.12$
November 23rd 2008, 01:43 PM
Larianne

Quote:

Originally Posted by o_O

Let's format that a little better: $x = \frac{b \pm \sqrt{b^2 - 4ac}}{2a}$

Why don't you think that's right? You rounded off to a few decimal places which will throw off your answer a little bit but it is correct (again, rounded of course)

Should it not work out evenly???
November 23rd 2008, 01:44 PM
Larianne

Quote:

Originally Posted by euclid2

$x = \frac{2 \pm \sqrt{-2^2 - 4(3)(-6)}}{2(3)}$
$x = \frac{2 \pm \sqrt{76}}{6}$
$x = \frac{2 + \sqrt{76}}{6}$ +
$x = \frac{2 - \sqrt{76}}{6}$
Therefore, $x =1.78 and x=-1.11$

So my answer is okay or should it be your answer???
November 23rd 2008, 01:45 PM
euclid2

Quote:

Originally Posted by Larianne

So my answer is okay or should it be your answer???

Your answer is fine. Good job.