If a+b=5 and a ^2+b ^2 =13 find the value of (a-b )
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Answer:
a+b=5
a^2+b^2=13
Now
(a+b)^2=5x5=25
a^2+b^2+2ab=25
13+2ab=25
2ab=12
Therefore
(a-b)^2=a^2+b^2-2ab
=13-12=1
hence a-b= 1
Answered by
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(a+b)^2=a^2+b^2+2ab
25=13+2ab
2ab=12
ab=6
a(5-a)=6
5a-a^2=6
a^2-5a+6=0
a^2-3a-2a+6=0
a(a-3)-2(a-3)=0
(a-2)(a-3)=0
a=2,3
2+b=5 or 3+b=5
b=3,2
a-b=-1,1
25=13+2ab
2ab=12
ab=6
a(5-a)=6
5a-a^2=6
a^2-5a+6=0
a^2-3a-2a+6=0
a(a-3)-2(a-3)=0
(a-2)(a-3)=0
a=2,3
2+b=5 or 3+b=5
b=3,2
a-b=-1,1
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