• Lavace
In summary, the student is having trouble understanding why their answers are always different to the solutions and has found that there are two equations- one easier to use and the other for more advanced users. They are still lost and would appreciate any help.

#### 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!

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

The first (correct) equation is fairly easy to derive

Is it something to do with complex numbers?

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 [Broken]

Quesions 1) a and b

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

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

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

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

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

If so I feel pretty silly!

Seems I jumped ahead of myself.

Why has he done that to the SQRT?

Yes, that is correct.
$$\frac{-b\pm\sqrt{b^2- 4ac}}{2a}$$
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!

HallsofIvy said:
Yes, that is correct.
$$\frac{-b\pm\sqrt{b^2- 4ac}}{2a}$$
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.