Solve the Equation:
(x^2-3)^1/2 - (x+3)^1/2 = 0
2006-10-18 21:20:13 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
This sum is just a matter of algebraic manipulation.
(x^2-3)^1/2 - (x+3)^1/2 = 0
Squaring the entire equation will result,
(x^2-3) - (x+3) = 0
x^2 - x - 6 = 0
(x + 2)(x - 3) = 0
Thus, letting individual terms equate to 0, we have,
x + 2 = 0
x = -2
OR
x - 3 = 0
x = 3
Thus, the solution is x = -2 OR x = 3.
Hope this helps.
2006-10-18 23:08:38 · answer #1 · answered by xxmizuraxx 2 · 0⤊ 0⤋
Not too hard:
move the minus term to the other side
(x^2 - 3)^1/2 = (x + 3)^1/2
Square each side:
x^2 - 3 = x + 3
Solve quadratic
x^2 - x - 6 = 0 : (x - 3)(x + 2) = 0, so x = 3 or -2
BUT...
Because we squared earlier we may have imntroduced extraneous roots, so you *must* test these (usually you only *should* test answers)
(3^2 - 3)^1/2 = 6^1/2 = (3 + 3)^1/2 ... check
((-2)^2 - 3)^1/2 = 1 = (-2 + 3)^1/2 ... check
So in this case both answers -2 and 3 are correct.
2006-10-19 04:33:29 · answer #2 · answered by sofarsogood 5 · 0⤊ 0⤋
Yes it's right x=-2 or x=3
2006-10-19 07:47:21 · answer #3 · answered by mms 3 · 0⤊ 0⤋
(x^2-3)^1/2-(x+3)^1/2=0
(x^2)^1/2 .(-3)^1/2 - (x)^1/2.(3)^1/2=0
COntiuneu
2006-10-19 07:02:53 · answer #4 · answered by mzamo n 1 · 0⤊ 0⤋
Try moving one term to the right side of the equation and squaring both terms.
Use the quadractic equation after that.
2006-10-19 04:32:17 · answer #5 · answered by Phillip 3 · 0⤊ 0⤋
could be wrong but my guess is that x=0, 1 but i would wait for someone else's answer before you go on.
2006-10-19 04:30:58 · answer #6 · answered by Anonymous · 0⤊ 0⤋
look very very well and you will see what the answer is....
3 and -2
2006-10-19 07:10:17 · answer #7 · answered by gjmb1960 7 · 0⤊ 0⤋