If (a^2-b^2)=21 and (a^2+b^2)=29 which of the following can be the value of ab? (a.)-10 (b.)5 root2 (c.)10 it is a multi-correct question. Please give the solution.

given,a²-b²=21------------(1) and a²+b²=29-----------------(2) by adding (1) and (2) we get 2a²=50 then a=5 by substiting a in (1) we get b=2 then ab=10 c is correct option.

a²-b²= 21 a²+b² = 29

Approved

let a2=A b2=B A+B=29 A-B=21 SOLVING A=25 B=4 OR a2=25 which follows a= 5 or a= -5 b2=4 which follows b=2 or b = -2 so ab = 5x2=10 or -5x2=-10 or 5x-2=-10 or -5x-2=10 hence correct options= a,c

a^2-b^2=21,a^2+b^2=29Adding both sides:a^2+a^2-b^2+b^2=21+292a^2=50a^2=25a=√25a=+/-5Put value of a in a^2+b^2=29(+/-5)^2+b^2=29b^2=29-25b^2=4b=√4b=+/-2Now ab=(+/-5X+/-2) ab=+/-10