Question

Evaluate x^2-6x+13=0 and tell whether or not it is extraneous

Answer 1

We can solve this equation by completing the square since it doesn't factor.

1) x^2-6x =13
2) x^2-6x+ 9 =13+9
3) (x-3)^2=22
4) x-3=+/-sq rt( 22)
5) x=3 +/- sq rt(22)

Answer 2

@mackenzie_willa

Answer 3

To determine if it has real or imaginary roots, you take a part of the quadratic formula:
Square Root (b^2 -4*a*c)
In this quadratic a=1 b=-6 and c=13
Square Root (6^2 -4*1*13)
Square Root (36 -52)
Square Root (-16)
You must take a negative square root and therefore the roots are complex (or imaginary).

Answer 4

geny55's solution is wrong
To complete the square the equation should be set up as
1) x^2-6x =MINUS 13
2) x^2-6x+ 9 = -13+9
and then continue from there

Answer 5

hey @wolf1728 can you help me with like 12 math questions please? :)

Answer 6

TWELVE math questions? I could help you with a couple

Answer 7

but they are easy math questions :) please help with 12 :)

Answer 8

Well at least a few.
By the way, why not start a new topic?

Answer 9

or how about science questions? are you good in science?

Answer 10

I know science pretty well too

Answer 11

ok awesome.. ill tag you :)

Answer 12

Okay