Algebra 2

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.

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asked by Amanda
  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.

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    posted by Damon

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