If a+b=21 ,ab=108 find a square+ b square
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(a+b)^2=a^2+b^2+2ab
21^2=a^2+b^2+2×108
441=a^2+b^2+216
441-216
225
21^2=a^2+b^2+2×108
441=a^2+b^2+216
441-216
225
SumanthRayudu:
answer is 333
is it so???
sorry my answer is wrong
i took ab instead of 2(ab)
ok fine
ok
Answered by
0
By using formulae:
(a+b)^2= a^2 + b^2 + 2ab
By substituting values:
➡️ (21)^2 = a^2 + b^2 + 2*108
➡️441= a^2 + b^2 + 216
➡️441-216=a^2 + b^2
➡️ a^2 + b^2= 225.
(a+b)^2= a^2 + b^2 + 2ab
By substituting values:
➡️ (21)^2 = a^2 + b^2 + 2*108
➡️441= a^2 + b^2 + 216
➡️441-216=a^2 + b^2
➡️ a^2 + b^2= 225.
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