How does one get y^2 - 2y = x^3 + 2x^2 + 2x + 3 to turn into: y = 1+- sqrt(x^3 + 2x^2 + 2x +4) ? Thanks for the help.
2007-07-21 18:16:15 · 2 answers · asked by Charlie4590 2 in Science & Mathematics ➔ Mathematics
Looking at left side figures y^2-2y carefully, you will find the figures of (y-1)^2 where constant figure is missing which is 1.
So adding 1one to both sides of you equation renders L.H.S. a perfect square.
y^2-2y+1=(y-1)^2=x^3+2x^2+2x+3+1
or (y-1)^2 = x^3+2x^2+2x+4
Taking sq.rt. of both sides
y-1=(+/-)sq.rt.(x^3+2x^2+2x+4)..............(sq.rt.a^2=+/- a)
or y = 1 +/- sq.rt.(x^3+2x^2+2x+4)
which the required result.
2007-07-21 18:37:31 · answer #1 · answered by Indian Primrose 6 · 0⤊ 0⤋
If you add 1 to both sides, you have y^2-2y+1 on the left, which is the perfect square that factors to
(y-1)^2. Taking the square root of both sides and then adding 1 to both sides yeilds the result you have.
2007-07-21 18:20:35 · answer #2 · answered by cattbarf 7 · 0⤊ 1⤋