Given that a+b=3 , a^2+b^2=8 and; b should be greater than a. Then the value of b/a?

Question

Answer 1

Hi
We have
a+b = 3
a^2+b^2 = 8
Now a^2 = 8-b^2
8-b^2 = 9+b^2-6b
we get 2b^2-6b+1 =0
b = (6+rt28)/4
we get a = (6-rt28)/4
taking b/a we get (6+rt28/6-rt28)

Answer 2

Hi
We have
a+b = 3
a^2+b^2 = 8
Now a^2 = 8-b^2
8-b^2 = 9+b^2-6b
we get 2b^2-6b+1 =0
b = (6+rt28)/4
we get a = (6-rt28)/4
taking b/a we get (6+rt28/6-rt28)

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Given that a+b=3 , a^2+b^2=8 and; b should be greater than a. Then the value of b/a?