How many REAL solutions for y are there in this equation
sqrt y+3 - sqrt y = 3?
0, 1, 2, or 3?
2006-07-22 18:34:06 · 9 answers · asked by samadhi97 1 in Science & Mathematics ➔ Mathematics
I assume your problem is
[1] ... sqrt (y+3) - sqrt(y) = 3
Rewrite:
[2] ... sqrt (y+3) = 3 + sqrt (y)
Take squares:
[3] ... y + 3 = 9 + 6 sqrt(y) + y
Simplify:
[4] ... 6 + 6 sqrt(y) = 0
It follows that sqrt(y) = -1, so there are no real solutions for y.
================
Others find the value y = 1. This is not a solution, as is checked easily:
sqrt (1 + 3) - sqrt (1) = 2 - 1 = 1, not 3
The problem is that squaring an equation with sqrt's can generate more solutions than there really are. When using that method, always check the outcome with the original formula.
2006-07-22 20:55:13 · answer #1 · answered by dutch_prof 4 · 1⤊ 0⤋
3
2006-07-23 02:50:41 · answer #2 · answered by Kianna A 3 · 0⤊ 0⤋
3
2006-07-23 01:37:17 · answer #3 · answered by Anonymous · 0⤊ 0⤋
sqrt(y+3) -sqrt(y) =3,
Squaring,
(y+3) +y - 2 * sqrt( y^2 + 3*y) = 9;
=> 2*y -6 = 2 * sqrt(y^2+ 3*y)
=> y -3 = sqrt( y^2+ 3*y)
=> y^2 -6*y + 9 = y^2 +3*y
=> 9*y = 9
=> y =1
Therefore, Just one real solution
2006-07-23 01:43:22 · answer #4 · answered by adi007boy 2 · 0⤊ 0⤋
sqrt y + 3 - sqrt y = 3, simplifies to :
sqrt y = sqrt y,
an equation which has an infinite number of solutions.
2006-07-23 04:03:28 · answer #5 · answered by coolzadar 2 · 0⤊ 0⤋
there is just 1 real solution and that is y=1
2006-07-23 02:42:55 · answer #6 · answered by abhinav 2 · 0⤊ 0⤋
sqrt y+3-sqrt y=3
=>sqrt y=sqrt y .so it is not a equn.
2006-07-23 09:43:43 · answer #7 · answered by Sayom 2 · 0⤊ 0⤋
simple
square root y - square root y will be cancel and +3 and -3 will be cancel
nothing left
so 1
2006-07-23 10:31:54 · answer #8 · answered by Anonymous · 0⤊ 0⤋
quick help > quick guess : 2
2006-07-23 02:02:15 · answer #9 · answered by Chie 5 · 0⤊ 0⤋