SOLUTION: How do I solve: {{{ sqrt( s+25 ) +5 = s}}} Thank you.

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Question 336199: How do I solve:
+sqrt%28+s%2B25+%29++%2B5+=+s

Thank you.
Found 2 solutions by jim_thompson5910, galactus:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
+sqrt%28+s%2B25+%29++%2B5+=+s Start with the given equation.


+sqrt%28+s%2B25+%29+=+s-5 Subtract 5 from both sides.


s%2B25+=+%28s-5%29%5E2 Square both sides.


s%2B25+=+s%5E2-10s%2B25 FOIL


0+=+s%5E2-10s%2B25-s-25 Get every term to one side.


0=s%5E2-11s%2B0 Combine like terms.


Notice that the quadratic s%5E2-11s%2B0 is in the form of As%5E2%2BBs%2BC where A=1, B=-11, and C=0


Let's use the quadratic formula to solve for "s":


s+=+%28-B+%2B-+sqrt%28+B%5E2-4AC+%29%29%2F%282A%29 Start with the quadratic formula


s+=+%28-%28-11%29+%2B-+sqrt%28+%28-11%29%5E2-4%281%29%280%29+%29%29%2F%282%281%29%29 Plug in A=1, B=-11, and C=0


s+=+%2811+%2B-+sqrt%28+%28-11%29%5E2-4%281%29%280%29+%29%29%2F%282%281%29%29 Negate -11 to get 11.


s+=+%2811+%2B-+sqrt%28+121-4%281%29%280%29+%29%29%2F%282%281%29%29 Square -11 to get 121.


s+=+%2811+%2B-+sqrt%28+121-0+%29%29%2F%282%281%29%29 Multiply 4%281%29%280%29 to get 0


s+=+%2811+%2B-+sqrt%28+121+%29%29%2F%282%281%29%29 Subtract 0 from 121 to get 121


s+=+%2811+%2B-+sqrt%28+121+%29%29%2F%282%29 Multiply 2 and 1 to get 2.


s+=+%2811+%2B-+11%29%2F%282%29 Take the square root of 121 to get 11.


s+=+%2811+%2B+11%29%2F%282%29 or s+=+%2811+-+11%29%2F%282%29 Break up the expression.


s+=+%2822%29%2F%282%29 or s+=++%280%29%2F%282%29 Combine like terms.


s+=+11 or s+=+0 Simplify.


So the possible solutions are s+=+11 or s+=+0


However, if you plug in s=0 into the original equation, you'll get a contradiction. So s=0 is NOT a solution.


So the only true solution is s=11


If you need more help, email me at jim_thompson5910@hotmail.com

Also, feel free to check out my website.

Jim

Answer by galactus(183) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%28s%2B25%29%2B5=s
Subtract 5 from both sides:
sqrt%28s%2B25%29=s-5
Square both sides to shed the radical:
s%2B25=%28s-5%29%5E2
Expand to form the quadratic:
s%2B25=s%5E2-10s%2B25
Set to 0 by bringing everything to one side:
s%5E2-11s=0
Factor out s:
s%28s-11%29=0
Now, it is easy to see that the two solutions are s=0 and s=11.
But, one of these may be extraneous. Check to see if they work:
For s=0:
sqrt%280%2B25%29%2B5=0........clearly untrue.
Check s=11:
sqrt%2811%2B25%29%2B5=11......clearly true.
s=11 is the solution we want.