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Algebraic equation with square root and power

Posted 11 years ago
(1-t) / (sqrt(1+t^2))*(sqrt(2)) = 0.5

the result is t=2-sqrt(3), but i'm not able to have it !
POSTED BY: Charles Vey
4 Replies
Hi, 
Using proper (capitalized) function names, and If the sqrt(2) is in the denominator as i MISREAD  the first time, then I am able to get the solution presented with the question.
(* The following form has the proposed solution. *)
Solve[ (1-t)/(Sqrt[(1+t^2)] Sqrt[2])==(1/2), t]
{{t -> 2 - Sqrt[3]}}

(* The question as presented has the solutions shown by Thiel. *)
I think its easy to mis-read the brackets when the equation is in the in-line form.
POSTED BY: Isaac Abraham
Dear Charles,

please do type it into Wolfram alpha or click on this link:

http://www.wolframalpha.com/input/?i=%281-t%29+%2F+%28sqrt%281%2Bt%5E2%29%29*%28sqrt%282%29%29+%3D%3D+0.5

Once you see the result click on "step-by-step" solution. For your convenience I have attached the notebook that wolfram alpha generates.

M.
Attachments:
POSTED BY: Marco Thiel
Posted 11 years ago
I'm trying to understand where i made a mistake, do you have a step by step solution to propose ? 
POSTED BY: Charles Vey
Hi,

there are some parenthesis that need to be changed and Sqrt is capitalised.
Solve[(1 - t)/(Sqrt[1 + t^2])*(Sqrt[2]) == 1/2, t]

gives
{{t -> 1/7 (8 - Sqrt[15])}}
or 
{{t -> 0.589574}}
which is different from your solution. 

In fact, your suggested solution does not appear to solve the equation:
In[90]:= (1 - t)/(Sqrt[1 + t^2])*(Sqrt[2]) == 1/2 /. t -> 2 - Sqrt[3]

Out[90]= False

In[93]:= (1 - t)/(Sqrt[1 + t^2])*(Sqrt[2]) == 1/2 /.
  t -> 1/7 (8 - Sqrt[15]) // N

Out[93]= True

M.
POSTED BY: Marco Thiel
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