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• July 24th 2011, 12:40 PM
David Green

3x^2 - 2x +3

OK this is a learning curve for me which I don't profess to understand!

I have the above and I think the first step is to find the sum of the two numbers 'a' and 'c', so ac = 9. The sum is b

for this quadratic expression, ac = 3 x 3 = 9 and b = -2

I think I now need to find the factors of 9!

I see 1, 3 and 9

The problem I now see is there is no sum of these to make -2, so I am stuck here at this point?

Anyone know what I am doing wrong?
• July 24th 2011, 12:49 PM
Siron
Do you know the discriminant is:
$D=b^2-4ac$
?
You have: a,b and c so you can calculate it directly.
• July 24th 2011, 12:50 PM
Plato
Quote:

Originally Posted by David Green
3x^2 - 2x +3
to calaculate the discriminate of the quadratic expression

$\Delta=b^2-4ac$ is the discriminate.

$b=-2,~a=3,~\&~c=3$ so $\Delta=(-2)^2-4(3)(3)=-32$
• July 24th 2011, 12:55 PM
David Green
Quote:

Originally Posted by Siron
Do you know the discriminant is:
$D=b^2-4ac$
?
You have: a,b and c so you can calculate it directly.

How does - 49 sound?

I didn't know D = b^2 - 4ac, although it does look very similar to the quadratic equation b sqrt b^2 -4ac /2a
• July 24th 2011, 01:00 PM
Plato
Quote:

Originally Posted by David Green
How does - 49 sound?

I didn't know D = b^2 - 4ac, although it does look very similar to the quadratic equation b sqrt b^2 -4ac /2a

Read reply #3 carefully. It is correct. The answer is $-32$
• July 24th 2011, 01:05 PM
David Green
Quote:

Originally Posted by Plato
Read reply #3 carefully. It is correct. The answer is $-32$

Thanks all for your help, I just worked it out. b^2 first, the - 2 becomes a positive when squared, then the product rule, then subtract from the squared value to = - 32.
• July 24th 2011, 01:08 PM
Siron
Maybe something important, if D<0, do you know what that means? Does the parabola intersect the x-axis in certain point(s)? ...
• July 24th 2011, 01:16 PM
David Green
Quote:

Originally Posted by Siron
Maybe something important, if D<0, do you know what that means? Does the parabola intersect the x-axis in certain point(s)? ...

I might be wrong but it could be referring to a negative solution?

On this example; x^2 + x - 12 = 0

can I use the quadratic formula?

If so is it only the integers on the top line under the square root that should be square rooted, then divide then either add or subract the remaining b?
• July 24th 2011, 01:25 PM
Siron
Quote:

Originally Posted by David Green
I might be wrong but it could be referring to a negative solution?

On this example; x^2 + x - 12 = 0

can I use the quadratic formula?

If so is it only the integers on the top line under the square root that should be square rooted, then divide then either add or subract the remaining b?

Now, it can be also important to know why you're calculating the discriminant. Probably you'll lear more about that.
But if you calculate: $0=ax^2+bx+c$ which can be solved with the quadratic formula, you're calculing the point(s) where the parabola intersects the x-axis, because there $y=0$, now you can distinguish tree cases:
- $D<0$, the parabola doesn't intersects the x-axis at all, so there are no intersect points and so no solution.
- $D=0$, there parabola intersects the x-axis, there are two equal solutions.
- $D>0$, the parabola intersects the x-axis, there are two different intersect points and so two different solutions.
• July 24th 2011, 01:35 PM
David Green
Quote:

Originally Posted by Siron
Now, it can be also important to know why you're calculating the discriminant. Probably you'll lear more about that.
But if you calculate: $0=ax^2+bx+c$ which can be solved with the quadratic formula, you're calculing the point(s) where the parabola intersects the x-axis, because there $y=0$, now you can distinguish tree cases:
- $D<0$, the parabola doesn't intersects the x-axis at all, so there are no intersect points and so no solution.
- $D=0$, there parabola intersects the x-axis, there are two equal solutions.
- $D>0$, the parabola intersects the x-axis, there are two different intersect points and so two different solutions.

Interesting stuff which I am learning, I have worked out that I have I think two routes, the positive route being 3 using the equation.

This is what I did;

-1 + or - square root 1^2 - 4 x 1 - 12
2 x 1

= 3

Hope I got that right?
• July 24th 2011, 01:51 PM
Siron
You have found one solution, because $D>0$, there are two different solutions.
You have found:
$x_1=\frac{-1+\sqrt{D}}{2}=3$
$x_2=\frac{-1-\sqrt{D}}{2}=...$?
• July 24th 2011, 02:00 PM
David Green
Quote:

Originally Posted by Siron
You have found one solution, because $D>0$, there are two different solutions.
You have found:
$x_1=\frac{-1+\sqrt{D}}{2}=3$
$x_2=\frac{-1-\sqrt{D}}{2}=...$?

There are two solutions, one positive and one negative, I only required the positive one on this example. Thanks David
• July 24th 2011, 02:11 PM
Siron
Quote:

Originally Posted by David Green
There are two solutions, one positive and one negative, I only required the positive one on this example. Thanks David

Fine! :D
But why do you need only the positive solution?
• July 24th 2011, 02:15 PM
David Green