Question

Solve for x using the quadratic equation.

x^2-10x+50=0

The solution set is: { , }

Note: If someone can tell me how to simplify the quadratic equation, that would be great since I can't seem to simplify correctly after I plug a, b, and c in.

Answer 1

Apparently the answer is {5+5i, 5-5i} but I don't know why...

Answer 2

Compare your quadratic equation with \(ax^2+bx+c=0\)
find a,b,c
then the two rots of x are:
\(\huge{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)

Answer 3

what are your a,b,c ?

Answer 4

a=1, b=-10, c=50

Answer 5

so what is b^2 - 4ac ?

Answer 6

-300

Answer 7

this is of the form, ax^2 +bx+c =0 a=1 b=-10 c=50

and the roots are x= +10+/-root{100-4.1.50}/2.1

x= 10+/- root{100-200}/2

10+/- root{-100}/2

10+/- 10i/2

5+/-5i becoz rooot{-1} is treated as i

Answer 8

and yes, the solution is 5+/-5i
100-200 = -100
not -300

Answer 9

so @Jalberra your b^2-4ac was incorrect, did u get your mistake ?
now can u solve on your own ?

Answer 10

Yes. I got it now. I didn't make b^2 positive which messed everything up. Thanks!

Answer 11

welcome :)

Answer 12

I was looking at this problem again and b^2 is (-10)^2 which is -100 so how did it become positive?

Answer 13

(-10)^2 =(-10)(-10) = +100
minus * minus = plus

Answer 14

No wonder, I was entering it into my calculator without the parenthesis... Figures, thanks again... so many minor mistakes... -.-

Answer 15

happens with all of us sometimes....