Question

If someone can, I need someone to solve this so I can compare my answer with their's on the quadratic formula.

Answer 1

\[2x^2-x-2=0\]

My answer was:
\[x+/-1i \sqrt{x^2}/2\]

Answer 2

If you can't read that end piece, it's radical x^2 over 2

Answer 3

how did you get that answer?
quadratic formula doesn't involve x's in it

Answer 4

Heh... I miraculously just caught that. Give me a bit.

Answer 5

For the b in this equation, it would be -1 correct?

Answer 6

yes

Answer 7

and a=2, c=-2 and you correctly identified the value of b

Answer 8

So when I substituted b^2-4ac with \[-1^2-4(2)(-2)\], I should get \[\sqrt{1+16}\] Correct?

Answer 9

As well as the 2(2) being 4 on the denominator.

Answer 10

yep. (tho i recommend using parentheses around the -1 when squaring it, since the exponent applies to the 1 in -1)

Answer 11

I cannot simply this equation any further can I?

Answer 12

\[1\pm \sqrt{1+16}\over4\]

Answer 13

Since the 17 in the radical can't be rooted, that should be the solution?

Answer 14

Unless that 4 divides into the 16.

Answer 15

yes, you cannot simplify it any further

Answer 16

\[1\pm \sqrt{17}\over 4\]

Answer 17

Don't forget the "-b" in the very beginning of the numerator.

Answer 18

Thank you kindly.

Answer 19

@jpigott, did you get my post regarding the -b in the numerator??

Answer 20

Well i'm not necessarily sure what is supposed to happen to it.
If I were to take a crack at it would it be
1+17/4 and 1-17/4?

Answer 21

The final answer could look like:\[1\pm \sqrt{17}\over 4\]

Answer 22

or 1/4 +/- (1/4)sqrt 17

Answer 23

I just didn't want you to just drop it, it is part of the solution.

Answer 24

Your last post is partially correct, but the 17 should be sqrt 17, or shown in a radical.

Answer 25

The equation editor is handy for expressing quadratics and their solution. All you have to do is click on the left lower button below your typing box.

Answer 26

The one labeled "Equation"\[\Sigma"Equation"\]

Answer 27

I do that. My apologies for the confusion.
If i am reading what you typed correctly, I have to do
(1+√17)/4 and vise vera?

Answer 28

By Jove, I believe you have it. Good luck with them.

Answer 29

Oh now I see what you were trying to get at.
The actual answer is supposed to include both the + and the 1 equation as the answer, right?

Answer 30

Yes,\[1+\sqrt{17}\over 4\] and \[1-\sqrt{17}\over 4\]

Answer 31

Whoop! Thank you!

Answer 32

U r welcome.