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(a + b√3)^2 = 7 + 4√3 then find the value of a and b

(a + b√3)^2 = 7 + 4√3
then find the value of a and b

Grade:9

2 Answers

Phanindra
226 Points
5 years ago
(a+b)^2 = a^2 + b^2 + 2ab....Here.... (a+b3^1/2)^2 = a^2 + 3b^2 + 2ab3^1/2... Here...On comparing....a^2 +3b^2 = 7 and 2ab3^1/2 = 4 × 3^1/2 ab = 2 so....a = 2/b..(1)equation4/b^2 + 3b^2 = 7....4 + 3b^4 = 7b^2...Put b^2 = t.... 4 + 3t^2 - 7t = 0... 3t^2 - 4t -3t + 4 = 0..3t(t-1) - 4(t-1) = 0..t = 1 or t = 4/ 3...So b = 1 or b= -1 or b = 2/3^1/2... or b = -2/3^1/2...Now put values of b in 1st equation to get values of a...Hope you have understood this.....
Samyak Jain
333 Points
5 years ago
(a + b√3)2 = 7 + 4√3
LHS = (a + b√3)2  = a2 + 2ab√3 + (b√3)2 
         = a2 + 3b2 + 2ab√3
RHS = 7 + 4√3
Comparing LHS & RHS, we get 
a2 + 3b2 = 7  …........(1)     &    2ab√3 = 4√3       ….........(2)
From (2), a = 2/b  Put this value of a in (1)
(4/b2) + 3b2 = 7     i.e.    4 + 3b4 = 7b2    or    3b4 – 7b2 + 4 = 0
3b4 – 3b2 – 7b2 + 4 = 0
(3b2 – 1) (3b2 – 1) = 0
Solving we get  b = 1 or –1 or 2/\sqrt{}3 or –2/\sqrt{}3
Respective value of a = 2 or –2 or \sqrt{}3 or – \sqrt{}3       [\because a = 2/b]
 

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