+0

0
116
7

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

Jun 11, 2020

#1
+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
+111111
-1

Nicely done guest.

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

Melody  Jun 11, 2020
#2
+1130
+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
+1

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

Guest Jun 11, 2020
#4
+1

There is actually no real value associated with a

Guest Jun 11, 2020
#5
+1130
+1

@Guest

There is a value of a

Jun 11, 2020
edited by jimkey17  Jun 11, 2020
#6
+111111
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

Jun 11, 2020