Question

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^4-13x^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.

Answer 1

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
(y-6.5)^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 non-zero coefficient of x^2 before starting.