Solve 7x^4=-28x^2-21?

Answer 1

We have that 7x^4 + 28x^2 + 21 = 0

Let X = x^2, then 7X² + 28X + 21 = 0 or X² + 4X + 3 = 0

Factorising gives (X + 3)(X + 1) so X = -3 or X = -1

since X = x² then x = √ X

Therefore x = ±√ (-1) or x = ±√ (-3) i.e. -3i, -i, i, 3i.

Answer 2

7x^4=-28x^2-21
7x^4+28x^2+21=0
(7x^2+7)(x^2+3)=0

x^2=-1
x= -1^(1/2)=1i or -1i
or
x^2=-3
x= -3^(1/2)= square root of 3i or -square root of 3i

note: I checked all my answer on the calculator on CMPLX mode.

Answer 3

7x^4=-28x^2-21
=>7x^4+28x^2+21=0
=>7(x^4+4x^2+3)=0
=>x^4+4x^2+3=0
=>x^4+3x^2+x^2+3=0
=>x^2(x^2+3)+1(x^2+3)=0
=>(x^2+1)(x^2+3)=0
Therefore either x^2+1=0 or x^2+3=0
x=sqrt -1 or sqrt -3

Answer 4

set up your perfect square
7x^4+28x^2+21=0
factor out 7
7(x^4+4x^2+3)=0
7((x^2+3)(x^2+1))=0
solve for x
x=sqrt(-3) and x=sqrt(-1)

I think that these two answers are irrational... meaning that the graph never crosses the x axis... no value of x will give you zero.

Answer 5

7x^4 + 28x^2 +21 = 0 solve for zero
x^4 + 4x^2 +3 = 0 factor out 7

think carefully about possible factors...

you can get this...

(hint: x^2 * x^2 = x^4)


...got it?


okay good, the factors are

(x^2 +3)(x^2 +1) = 0

so x = sqrt(-3) & x = sqrt(-1)

remember the sqrt(-1) = i (the imaginary number)

so x = i*sqrt(3) & x = i

Answer 6

7x^4+28x^2+21=0

7(x^4+4x^2+3)=0
7(x^2+1)(x^2+3)=0
x=sqrt of -1 or i
x= sqrt of -3=1.70205i

Answer 7

x^4 = - 4x^2 - 3
x^4 + 4 x^2 + 3 = 0
(x^2 -3 )(x^2 -1 ) = 0
so x^2 = 3 or x^2 =1
x = 3^(1/2) or - 3^(1/2) or 1 or - 1

Don't you use Answers again to do your homework! I've been watching you!