2x^2+4x-1 ; x is greater than or equal to one
i can't solve for x in terms of y... help! tnx
2006-12-30 19:06:20 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Put the equation in standard form and use the quadratic formula:
2x²+4x-1=y
2x²+4x-1-y=0
x=(-4±√(16+8(y+1)))/4
x=-1±√(y+3)/√2
Since x≥1, we must take the positive solution, and we have that y≥5. So f^(-1)(x) = -1+√(x+3)/√2, x≥5
2006-12-30 19:18:34 · answer #1 · answered by Pascal 7 · 0⤊ 0⤋
y = 2(x+1)^2 - 3
y+3 = 2(x+1)^2
(y+3)/2 = (x+1)^2
x+1 = sqrt((y+3)/2)
x = sqrt((y+3)/2) - 1
2006-12-31 03:15:17 · answer #2 · answered by Johnny Handsome 2 · 0⤊ 0⤋
+/-sq.rt[(x+3)/2]-1
2006-12-31 03:13:57 · answer #3 · answered by agarwalsankalp 2 · 0⤊ 0⤋
imma have to go with raj on this one... i give you both thumbs up!
2006-12-31 03:14:04 · answer #4 · answered by Jordan 3 · 0⤊ 0⤋
y=2x^2+4x-1
x=2y^2+4y-1
=2(y^2+2y+1-1)-1
=2(y^2+2y+2)-3
x+3=2(y+1)^2
(x+3)/2=(y+1)^2
y+1=+/-sq.rt.[(x+3)/2]
y=+/-sq.rt[(x+3)/2]-1
2006-12-31 03:11:42 · answer #5 · answered by raj 7 · 1⤊ 0⤋