SOLUTION: (1+sqroot50)*(1+sqroot50)= 1+2sqroot50+50= 51+2sqroot50= 51+7sqroot2 Is this correct ?

Algebra ->  Square-cubic-other-roots -> SOLUTION: (1+sqroot50)*(1+sqroot50)= 1+2sqroot50+50= 51+2sqroot50= 51+7sqroot2 Is this correct ?      Log On


   

Question 404371: (1+sqroot50)*(1+sqroot50)= 1+2sqroot50+50= 51+2sqroot50= 51+7sqroot2
Is this correct ?
Answer by Tatiana_Stebko(1539) About Me  (Show Source):
You can put this solution on YOUR website!
%28a%2Bb%29%2A%28a%2Bb%29=%28a%2Bb%29%5E2
Formula %28a%2Bb%29%5E2=a%5E2%2B2ab%2Bb%5E2
Thats why %281%2Bsqrt%2850%29%29%2A%281%2Bsqrt%2850%29%29=1%2B2sqrt%2850%29%2B50 is correct.
1%2B2sqrt%2850%29%2B50=51%2B2sqrt%2850%29 is correct,
but
So 51%2B2sqrt%2850%29=51%2B10sqrt%282%29