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Given that a(a + 2b) = 104/3 and b^2 = 7/9 then find |a + b|.

 Nov 23, 2020
 #1
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+1

Given that a(a + 2b) = 104/3 and b^2 = 7/9 then find |a + b|.

 

b^2 =7/9

b =sqrt(7/9)

 

a(a + 2*sqrt(7/9) =104/3, solve for a

a^2 + 2*sqrt(7/9)a =104/3

 

Solve for a:
a^2 + (2 sqrt(7) a)/3 = 104/3

Add 7/9 to both sides:
a^2 + (2 sqrt(7) a)/3 + 7/9 = 319/9

Write the left hand side as a square:
(a + sqrt(7)/3)^2 = 319/9

Take the square root of both sides:
a + sqrt(7)/3 = sqrt(319)/3 or a + sqrt(7)/3 = -sqrt(319)/3

Subtract sqrt(7)/3 from both sides:
a = sqrt(319)/3 - sqrt(7)/3 or a + sqrt(7)/3 = -sqrt(319)/3

Subtract sqrt(7)/3 from both sides:
 
a = sqrt(319)/3 - sqrt(7)/3      or        a = -sqrt(7)/3 - sqrt(319)/3

 

Abs[a + b] =sqrt(319) / 3

 Nov 23, 2020
 #2
avatar+128053 
+1

a(a + 2b)  = 104/3

 

a^2  + 2ab    = 104/3      add b^2  to both sides and factor the  left side

 

a^2  + 2ab  + b^2 =    104/3  + b^2

 

(a + b)^2  =  104/3 + 7/9

 

( a + b)^2  =  313/9

 

sqrt [  (a + b)^2  ]  =   l a + b l

 

sqrt  [ 313/9 ]  =  l a + b l  =  sqrt (313) / 3

 

 

cool cool cool

 Nov 23, 2020

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