find the value of (a+b) if (a-b)= 3 and ab=40
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Answered by
5
(a-b)^2=9
(a+b)^2=(a-b)^2+4ab
(a+b)^2= 9+4*40
(a+b)^2= 9+160
(a+b)^2=169
a+b=+13 or -13
but since a-b is positive 3 so a+b must be positive thirteen
a+b=13
a-b=3
2a=16
a=8
so b =5
a=8, b=5
(a+b)^2=(a-b)^2+4ab
(a+b)^2= 9+4*40
(a+b)^2= 9+160
(a+b)^2=169
a+b=+13 or -13
but since a-b is positive 3 so a+b must be positive thirteen
a+b=13
a-b=3
2a=16
a=8
so b =5
a=8, b=5
shailaahmed:
thanx
Answered by
7
(a+b)^2=(a-b)^2+4ab
(a+b)^2= 3^2+4(40)
(a+b)^2=9+160
(a+b)^2=169
(a+b)= √169
(a+b)=13
(a+b)^2= 3^2+4(40)
(a+b)^2=9+160
(a+b)^2=169
(a+b)= √169
(a+b)=13
given (a-b)=3
given (a-b)=3, ab=40 (a+b)^2=3^2+4(40) (a+b)^2=9+160 (a+b)^2=169 (a+b)=sqrt169 (a+b)=13
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