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avatar+1347 

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

 Jan 13, 2024
 #1
avatar+289 
+1

Let's simplify:

 

\(\frac{a}{3}+1=1+\frac{3}{a}-a+\frac{2}{a}\)

 

\(a-\frac{5}{a}+\frac{a}{3}=0\)

 

I just put this on a calculator and got a = \(\frac{\sqrt 15}{2}\)\(\frac{-\sqrt 15}{2}\)

 

I didn't know what to do.

 

Answer: (sqrt 15)/2, -(sqrt 15)/2

 Jan 13, 2024
 #2
avatar+128732 
+1

Simplify as

 

(a + 3) / 3 =    [ -a^2 + a + 1 ] /  a             cross-multiply

 

a ( a + 3)  =  3 (-a^2 + a + 1)

 

a^2 + 3a   = -3a^2 + 3a + 3

 

4a^2  = 3

 

a^2  = 3/4                 take both roots

 

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

 

cool cool cool

 Jan 13, 2024

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