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# :O why won't it solve??? Watch

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1. hi,

need to find x, but i cant seem to solve it using quadratic formula, any other method?

x^2 - 2x + 3 = 0

help!! thanks
2. No real solutions, if you use the quadratic formula you will see that you get the square root of -8 which is about 2.82i. Perhaps you have the quadratic wrong if you could not see that?
3. to see if something has a double root, 2 unique roots, or no real roots
use the discriminant!
if it only solves for the complex numbers, ensure that your solutions match up with the function of definition!!

D= b^2 - 4ac
4. (Original post by mir3a)
which are two numbers which multiply to give +3, but add t give -2....
i've tried the ones i can think of :/
-1 and 3 , no
-3 and 1 , no
-3 and -1 , no
-2 and 1 , no
-1 and 2 , no
3 and -5 , no

:?
5. you can also try completing the square?
that helped me alot with solving quadratics
6. If you put it into a calculator or program for the formula, it will probably tell you you are giving it an error due to the root part being negative (b^2 - 4ac). However, it actually comes to the square root of -8, which, when divided by 2 as in the formula, gives root 2 i, so the overall solution is 1 plus or minus iroot2. If in doubt, put it into wolfram alpha.
7. Ive got the quadratic from this, it is correct? x^2 - 2x + 3 = 0

8. you will still get complex answers tho becoz the discriminant is less than 0!
9. lol no real rooots!!!! my friend you need some i's!
10. (Original post by dream123)
Ive got the quadratic from this, it is correct? x^2 - 2x + 3 = 0

no that isn't correct

you must work out (~0.38)

then add that into the equation to get x^2 - 2x + 0.38 = 0

then use quadratic formula to solve for x
11. (Original post by dream123)
Ive got the quadratic from this, it is correct? x^2 - 2x + 3 = 0

Use a substitution of to get:

Solve this for y and then solve for x. This has a real solution.

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Updated: December 9, 2010
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