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^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.

- 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

(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.

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