Posted by Amanda on Monday, February 4, 2008 at 4:58pm.
Approximate to the nearest tenth the positive real zeroes of f(x)=x^2+1. I got .6
Solve over the set of complex numbers.
x^413x^2+36=0 i got 3,2,2,3 but i don't know if i even did it right.
Solve x^2+4x9=0 by completing the square. I don't even know where to begin on this one. So help me please!!!!! I have a test on this stuff Wed.

Algebra 2  Damon, Monday, February 4, 2008 at 5:18pm
Huh? x^2 + 1 = 0 has no real zeros
x = + or  sqrt (1) = +i or i
let y = x^2
then
y^2  13 y + 36 = 0
I will do this by completing the square so you see how
y^2  13 y = 36
take half of 13, sqare it, add to both sides
y^2  13 y +169/4 = 36 + 169/4
(y6.5)^2 = 6.25
y  6.25 = +2.5 or y  6.25 = 2.5
so
y = 8.75 or y = 3.75
BUT y = x ^2
so
x = + sqrt(8.75)
or
x = sqrt (8.75)
or
x = + sqrt (3.75)
or
x = sqrt(3.75)
x^2+4x9=0 This has a typo but I already showed you how to complete square above.
One thing is that coef of x^2 must be one. In your problems it already is, but if it were not, you would have to divide both sides by the nonzero coefficient of x^2 before starting.