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avatar+1420 

Find all values of $a$ that satisfy the equation
\[\frac{a}{3} + 1 = \frac{a + 3}{a} - \frac{a^2 + 2}{a}.\]

 Apr 4, 2024
 #1
avatar+118 
+1

a/3 + 1 = (( a+3) - (a^2 + 2))/a

 

a/3 + 1 =( a^2 - a - 1 ) / a 

 

a ( a/3 + 1 ) = a^2 - a - 1 

 

a^2/3 + a = a^2 - a -1 

 

a^2/3 = a^2 - 2a -1 

 

0 = (2a^2 - 2a) / 3  - 1 

 

then we can plug this equation into the quadratic equation, 

 

this gives us the answer 

 

(3+sqrt 15 )/2 or (3-sqrt15)/2 

 

please if anyone could check my answer that would be a fire smiley

 Apr 4, 2024
 #2
avatar+129741 
+1

Assuming "a" not equal to  0, multiply through by a

 

a^2/3 + a  = (a + 3) - (a^2 + 2)

 

(1/3)a^2  + a  = -a^2 + a + 1          simplify as

 

(4/3)a^2  =  1             multiply both sides by 3/4

 

a^2  = 3/4                  take both roots

 

a =  sqrt (3)  / 2      or     a =  -sqrt (3)  / 2

 

cool cool cool

 Apr 4, 2024
 #3
avatar+118 
+1

Srry mb gang 

 Apr 4, 2024
 #4
avatar+129741 
+1

I still gave you a point  for a good  effort !!!!

 

cool cool cool

CPhill  Apr 4, 2024

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