# solving for x

• Jan 25th 2010, 12:33 AM
jmedsy
solving for x
My algebra is rusty:

I'm solving for x in the following equation:

(x)^(1/2) + (x-(1-x)^(1/2))^(1/2) = 1

After several steps (in some of which I square both sides of the equality), I end up with the equality

x(25x-16) = 0

At this point I'm afraid to eliminate either term of the product in fear of dividing by 0. So I solve for x. I get x=16/25 , 0.

Maple says the only solution is x = 16/25.

Am I supposed to solve this way, but then check the solutions with the original equation in order to compensate for any misleading algebra brought on by me multiplying both sides by a variable?
• Jan 25th 2010, 01:01 AM
furor celtica
it is always helpful to plug in the values after as some answers can be illogical.
• Jan 25th 2010, 04:15 AM
HallsofIvy
You should always check answers. In this case, 0 is not a valid solution as x= 0 gives the square root of -1.
• Jan 25th 2010, 04:30 AM
Pulock2009
x^1/2 +(x-(1-x)^1/2)^1/2=1
=>x-(1-x)^1/2=1+x-2x^1/2
=>2x^1/2-1=(1-x)^1/2
=>5x=4x^1/2
=>x^1/2(5x^1/2-4)=0
=>x=0,2/sqrt(5)

wonder what solutions will come if we cosider x=r(cos(theta)+isin(theta))
ofcourse that would give the actual solution set.
• Jan 25th 2010, 04:31 AM
Pulock2009
x^1/2 +(x-(1-x)^1/2)^1/2=1
=>x-(1-x)^1/2=1+x-2x^1/2
=>2x^1/2-1=(1-x)^1/2
=>5x=4x^1/2
=>x^1/2(5x^1/2-4)=0
=>x=0,2/sqrt(5)

i wonder what solutions will come if we cosider x=r(cos(theta)+isin(theta))
ofcourse that would give the actual solution set.
• Jan 25th 2010, 04:37 AM
Pulock2009