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If a^2 + 4a + 21 = 0 and a^4 - a^3 + a^2 = na, find n.

 Jun 11, 2020
 #1
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+1

Multiply the first equation by a^2:

\(a^4+4a^3+21a^2=0\)

We still have to clean up the a^3 term, so we multiply the first equation by -5a:

\(-5a^3-20a^2-105a=0\)

Add the two equations to get:

\(a^4-a^3+a^2-105a=0\)

Rearrange:

\(a^4-a^3+a^2=105a\)

The value of n is 105.

 

Hope this helps!

 Jun 11, 2020
edited by Guest  Jun 11, 2020
edited by Guest  Jun 11, 2020
 #7
avatar+110235 
-1

This first answer is correct.

Nicely done guest.   laugh

 

 

Jimkey, maybe you can give him/her back the point you took   wink

Melody  Jun 11, 2020
 #2
avatar+1093 
+3

 simplfy the fist equation to get a^2+4a=-21 and it can also be (a+4)a=-21 so we get a=-3 since 3+4=7 and -3*7=-21 then we plug a=-3 into the second equation to get 81-(-27)+9=3n which is 117=3n so n=39

 Jun 11, 2020
edited by jimkey17  Jun 11, 2020
 #3
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+1

a should actually be -7, you forgot the negative sign I think

Guest Jun 11, 2020
 #4
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+1

There is actually no real value associated with a

Guest Jun 11, 2020
 #5
avatar+1093 
+1

@Guest

There is a value of a

 Jun 11, 2020
edited by jimkey17  Jun 11, 2020
 #6
avatar+110235 
0

Your answer is incorrect Jimkey17 AND there is no real value of a.

 

I certainly do give you credit for having a go though  laugh

 Jun 11, 2020

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