| T O P I C R E V I E W |
| cavanbas |
Posted - 07/23/2004 : 20:54:53
1. Given that 3-x1/2=(2x-7)1/2, what is x?
(i.e., "1/2" means square root.)
2. Given that x+y=11 and x1/2=y-5, solve for x and y.
1. Ans. x=4 or 64.
2. Ans. x=4, y=7; x=9, y=2.
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| 7 L A T E S T R E P L I E S (Newest First) |
| kaushalasp |
Posted - 07/30/2010 : 06:50:12
rightly said...all the best...
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| y82benji |
Posted - 10/06/2004 : 17:56:38
Gotta love math. It's just a language with more grammar than vocabulary! |
| liebenator |
Posted - 10/06/2004 : 11:41:57
I should not have read this thread. Now I have a headache. |
| y82benji |
Posted - 09/26/2004 : 22:28:44
No, 64 is still correct because the square root of 64 can be -8 or +8. If you take the square root of 64 to be -8 then you have 3 - -8 = 11 = sqrt(121) |
| jello |
Posted - 09/26/2004 : 08:35:04
For the 1st question, the answer should be ONLY 4, not 64.
3-sqrt(64)=sqrt(2*64-7)
3-8=sqrt(121)
-5=11 which is incorrect.
Hence, the answer is only 4
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| y82benji |
Posted - 07/25/2004 : 01:01:11
sqrt[x] = y - 5
square both sides
x = y^2 - 10y + 25
then
x + y = 11
subtract y from both sides
x = 11 - y
substitute "x = 11 - y" in x = y^2 - 10y + 25
11 - y = y^2 - 10y + 25
subtract 11 and add y to both sides
0 = y^2 - 9y + 14
use quadratic to solve and get
y = 2 or 7
go back to "x + y = 11"
x = 9 or 4 |
| y82benji |
Posted - 07/25/2004 : 00:53:29
3 - sqrt[x] = sqrt[2x-7]
square both sides
(3-sqrt[x])(3-sqrt[x]) = ([2x-7]^0.5)^2
multiply out on left, mulitply exponents on the right
3*3 - 3*sqrt[x] - 3*sqrt[x] + sqrt[x]*sqrt[x] = [2x-7]^1
simplify
9 - 6*sqrt[x] + x = 2x - 7
subtract 9 and x from both sides
-6*sqrt[x] = x - 16
square both sides
36x = x^2 - 32x + 256
subtract 36x from both sides
x^2 - 68x + 256
use quadratic
get x = 4 or 64 |
