Solving Quadratics: When to Use Each Quadratic Equation?

[Lavace]

I've been having trouble understanding why my answers are always different to the solutions and I found that there's two equations:

{-b ± SQRT(b^2 - 4ac)}/2a
and
{-b ± SQRT(b^2 -ac)}/2a

I don't know when to use either equation, always having used the first, and for someone at my level to have just encountered this is fairly embarressing (undergraduate on Physics)!

Any quick answers is REALLY appreciated, or an explanation of why etc, or even a website describing this would be very helpful!

 

[mgb_phys]

Where did you find "{-b ± SQRT(b^2 -ac)}/2a" ?

The first (correct) equation is fairly easy to derive

 

[Lavace]

Is it something to do with complex numbers?

 

[Lavace]

Mgb, I'm revising for an exam and answering excerise questions, take a look at this:

http://www.ph.qmul.ac.uk/mt2/Homework/mt2HW1.pdf

Quesions 1) a and b

http://www.ph.qmul.ac.uk/mt2/Homework/hw11.pdf

The solutions are -ac instead of -4ac which I don't understand!

 

[Lavace]

I've just realized he's divided the sqrt to make it simpler, I'm checking it out now!

 

[Lavace]

Actualy I'm still lost if you can be of any help!

 

[Lavace]

Yes it seems he's divided the SQRT by 4 to make it simpler, is this correct?

If so I feel pretty silly!

 

[Lavace]

Seems I jumped ahead of myself.

If I answer the quadratic without looking at the answers, I get the same answer.

Why has he done that to the SQRT?

 

[HallsofIvy]

Yes, that is correct.
The "standard" quadratic formula is
$$\frac{-b\pm\sqrt{b^2- 4ac}}{2a}$$
Of course, you can separate that into two fractions:
$$\frac{-b}{2a}\pm\frac{\sqrt{b^2- 4ac}}{2}$$
If you take that second denominator, 2, inside the square root, it becomes 4:
$$\frac{-b}{2a}\pm\sqrt{\frac{b^2- 4ac}{4}}$$
$$= \frac{-b}{2a}\pm\sqrt{\frac{b^}{4}- ac}$$

Presumably, he did that in order to simplify the arithmetic!

 

[statdad]

Yes, that is correct.
The "standard" quadratic formula is
$$\frac{-b\pm\sqrt{b^2- 4ac}}{2a}$$
Of course, you can separate that into two fractions:
$$\frac{-b}{2a}\pm\frac{\sqrt{b^2- 4ac}}{2}$$
If you take that second denominator, 2, inside the square root, it becomes 4:
$$\frac{-b}{2a}\pm\sqrt{\frac{b^2- 4ac}{4}}$$
$$= \frac{-b}{2a}\pm\sqrt{\frac{b^}{4}- ac}$$

Presumably, he did that in order to simplify the arithmetic!

Er, Halls, what happened to the 'a' in the second denominator (I'm looking at at your work immediately following "Of course...")

I'm on a little medication after oral surgery, so if I missed something obvious I will apologize and blame that for my problem.