Question

extremely confused on this one:

solve using quadratic formula
-2t^2+4t+3=0

Answer 1

What exactly confuses you about this problem in particular?

Answer 2

i have it equated down to \[-4\pm \sqrt{40}/-4\] and somehow that makes \[2\pm \sqrt{10}/2\] and im not understanding the math involved... others have tried to help but im not getting it

Answer 3

the math im doing is\[-4\pm \sqrt{40}/4\]\[-4\pm \sqrt{4*10}/-4\]\[-4\pm 2\sqrt{10}/-4\]

Answer 4

Hm... This is the property you're using between the second last and last lines: \(\sqrt{ab} = \sqrt{a}\sqrt{b} \).
Basically you can do this: \(\sqrt{4* 10} = \sqrt{4}\sqrt{10} = 2\sqrt{10}\)

Or is it something else that you're unsure of?

Answer 5

im just not sure how it equates to \[2\pm \sqrt{10}/2\] when the division doesnt make sense... you divide by -4 which does not equal 2.... someone said you divide by -2 but i cannot see how

Answer 6

Oh, this is what you're referring to then:
\[ \frac{-4 \pm 2\sqrt{10}}{-4} \\
= \frac{2(-2 \pm \sqrt{10})}{-(2 * 2} \\
= \frac{\cancel{2}(-2 \pm \sqrt{10})}{-\cancel{2}* 2} \]

Answer 7

Which then leaves,
\[ \frac{-2 \pm \sqrt{10}}{-2} \]
And then you can move the negative to the numerator:
\[ \frac{-(-2\pm\sqrt{10})}{2} \\
=\frac{2 \pm \sqrt{10}}{2} \qquad \quad \text{Plus-minus is the same two expressions multiplied by -1.} \]

Answer 8

im not completely sure of what you did there and that is no where in my book but thank you very much

Answer 9

I factored out a 2 from the numerator and denominator and cancelled them by the property: \[ \frac{a}{a} = 1 \qquad a \ne 0 \]

Answer 10

ok i understand now.... i never would have found that on my own...thank you again

Answer 11

You're welcome!
Yeah, you have to be be aware of all the properties that you can apply. The power of this, however, is quite strong. Some of these properties can follow you very far into the future in Math.
I recall a Calculus question that involved just subtracting this complicated expression from both sides to solve. I never even thought of it because its just not obvious when you're looking at all these complicated things. :P

Answer 12

thanks for the warning Calculus is next lol

Answer 13

Oh, I found it too. After some repetitive integration work, we get to a point that's like this:
\[ \int e^x \cos x \ dx = e^x \sin x + e^x \cos x - \int e^x \cos x \ dx \]
The next step actually being to add the "integral of e^x cos x dx" to both sides to get
\[ 2\int e^x \cos x \ dx = e^x \sin x + e^x \cos x \]
And then you divide off the 2 and you're done.

This kind of problem doesn't come up early in Calculus though, so no worries if it looks daunting. :P