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If a^2 - 3a + 9 = 0, then find a^3.

 Nov 16, 2019
 #1
avatar+470 
0

The first step in this is to separate \(a^2\) from the rest of the function.

 

\(a^2-3a+9=0\) becomes \(a^2=3a-9\).

 

Now there is just one step to find \(a^3\). Can you take it from here?

 Nov 16, 2019
 #3
avatar+106885 
+1

Withdrawn.

Melody  Nov 16, 2019
edited by Melody  Nov 16, 2019
 #4
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0

I think that this is what zzz had in mind.

 

.\(\displaystyle \text{Given that }a^{2}-3a+9=0,\dots(1),\\a^{2}=3a-9,\\\text{so}\\a^{3}=3a^{2}-9a.\\ \text{But from (1), }\\a^{2}-3a=-9,\\ \text{so}\\3(a^{2}-3a)=a^{3}=-27.\)

Guest Nov 16, 2019
 #5
avatar+106885 
0

ok thanks.

Maybe you are right.

In which case I offer my appologies to ZZZZZZZZZZ.

Sorry ZZZZZZ

Melody  Nov 16, 2019
 #2
avatar+106515 
+1

a^2  -3a  + 9   = 0   subtract  9 from both sides

 

a^2  - 3a   =   -9        complete the square on a

 

a^2  - 3a  + 9/4  =  -9  + 9/4

 

(a - 3/2)^2  =  -27/4       take both roots

 

a - 3/2  =  ±√ [-27] / 2

 

a  =  [3 ± 3i√3 ]

       _________

               2

 

When a  = [ 3 + 3i√ 3]

                __________   then   a^3  =

                         2

 

( [ 3 + 3i√ 3] / 2)^2  ( [3 + 3i√ 3 ] / 2 )  =

 

[ 9 + 18i√3 + 9i^2 (3) ] 

_______________        *   (  [ 3 + 3i√3] / 2)  =

             4

 

[ -18 + 18i√3 ]

_____________     (  [ 3 + 3i√3 ] / 2 )  =

         4

 

[ -9 + 9i√3]            [  3 + 3i√3]

__________  *    __________   =

      2                          2

 

[-27 + 27i√3 + 27i√3 - 27*3 ]         [ -27 * 4 ]

______________________    =  __________   =                -27

                4                                          4

 

The same result is obtained for a^3 when  a  =  ( 3 - 3i√3  )  / 2  

 

 

 

cool cool cool

 Nov 16, 2019

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