Hi,

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

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

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

That can't be right???

2. Originally Posted by Larianne
Hi,

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

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

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.

3. 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)

4. 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.

5. 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, $x =1.78 + x=-1.12$

6. 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???

7. 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, $x =1.78 and x=-1.11$