x^2+2x+5=0
x-3-sqrt(x+1)=0
x^2+10x+k
2007-01-04 16:49:24 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
x=[-2+/-rt(4-20)]/2
=[-2+/-rt-16]/2
=-2+/-4i]/2
=-1+/-2i
2.x-3=rt(x+1)
squaring
x^2-6x+9=x+1
x^2-7x+8=0
x=[7+/-rt(49-32)]/2
=[7+/-rt17]/2
3.x^2+10x+k
x=[-10+/-rt(100-4k]/2
2007-01-04 16:54:59 · answer #1 · answered by raj 7 · 0⤊ 0⤋
x^2 + 2x + 5 = 0
x = (- 2 ± â(4 + 20))/2
x = - 1 ± â(1 + 5)
x = - 1 ± â6
x - 3 - sqrt(x + 1) = 0
- sqrt(x + 1) = x - 3
x + 1 = x^2 - 6x + 9
x^2 - 7x + 8 = 0
(x - 8)(x + 1) = 0
x = -1, 8
x^2 + 10x + k =
x^2 + 10x + 25 + k - 25 =
(x + 5)^2 - (25-k) =
(x + 5 + â(25-k))(x + 5 - â(25-k))
2007-01-05 01:11:36 · answer #2 · answered by Helmut 7 · 0⤊ 0⤋
x=[-2+/-rt(4-20)]/2
=[-2+/-rt-16]/2
=-2+/-4i]/2
=-1+/-2i
2.x-3=rt(x+1)
squaring
x^2-6x+9=x+1
x^2-7x+8=0
x=[7+/-rt(49-32)]/2
=[7+/-rt17]/2
3.x^2+10x+k
x=[-10+/-rt(100-4k]/2
2007-01-05 01:06:08 · answer #3 · answered by LIL Mike 1 · 0⤊ 0⤋