solve this equation?

Question

2x^4+9x^3+8x^2+9x+2=0
what is the value of x?

Answer 1

2x^4 + 9x^3 + 8x^2 + 9x + 2 = 0

Divide the whole thing by x^2,
2x^2 + 9x + 8 + (9/x) + (2/x^2) = 0
2x^2 + (2/x^2) + 9x + (9/x) + 8 = 0 (Rearranging)
2(x^2 + 1/x^2) + 9(x + 1/x) + 8 = 0

Put x + 1/x = y ..... (1)

2(y^2 - 2) + 9y + 8 = 0
2y^2 - 4 + 9y + 8 = 0
2y^2 + 9y + 4 = 0
2y^2 + y + 8y + 4 = 0
y(2y + 1) + 4(2y + 1) = 0
(y + 4)(2y + 1) = 0

y = -4 (or) -1/2

Putting value of y in (1),

x + 1/x = -4 (or) x + 1/x = -1/2
x^2 + 1 = -4x (or) x^2 + 1 = -x/2
x^2 + 4x + 1 = 0 (or) 2x^2 + x + 2 = 0
x = [-4 +/- sqrt 12]/2 (or) x = [-1 +/- sqrt (1 - 16)]/4
x = -2 +/- sqrt 3 (or) x = (-1 +/- sqrt -15)/4

The real values of x are:
-2 + sqrt 3
-2 - sqrt 3

We can't involve the others as roots as they involve square root of a negative number, which is imaginary.

The total list of roots:
-2 + sqrt 3
-2 - sqrt 3
(-1 + sqrt -15)/4
(-1 - sqrt -15)/4

Answer 2

2x^4+9x^3+8x^2+9x+2=0
(divide by x^2)
2x^2+9x+8+9/x+2/x^2=0
(2x^2+2/x^2)+(9x+9/x)+8=0
2*(x^2+1/x^2)+9*(x+1/x)+8=0 ---------------------(1)
let's now consider x+1/x=y
then(x+1/x)^2=x^2+1/x^2+2.
hence x^2+1/x^2=y^2-2
sub. the values in (1) , we have,
2*(y^2-2)+9*(y)+8=0
2y^2-4+9y+8=0
2y^2+9y+4=0
2y^2+y+8y+4=0(splitting the middle terms)
y*(2y+1)+4*(2y+1)=0
(y+4)(2y+1)=0
y= -4 (or) y= -1/2
x+1/x= -4 (or) x+1/x = -1/2
x^2+1= -4x (or) x^2+1 = -1/2x (multiply by x)
x^2+4x+1=0 (or) x^2+1/2x+1=0
w.k.t. {-b+(/)-#(b^2-4ac)}/2a (-b + or - isa given as +(/)- and sq. root is given by #))
x= {-4+(/)-#([4]^2-4*1*1)}/2*1 (or)
x= {-1/2+(/)-#([1/2]^2-4*1*1)}/2*1
x= {-4+(/)-#(16-4)}/2 (or) x={-1/2+(/)-#(1/4-4)}/2
x={-4+(/)-#(12)}/2 (or) x={-1/2+(/)-#((1-16)/4)}/2
x={-4+(/)-2#3}/2 (or) x={-1/2+(/)-i/4#15}/2
x= -2+(/)-#3 (or) x= -1+(/)-i#15
=> x= -2+#3, x= -2-#3, x= -1+i#15, x= -1-i#15

Answer 3

According to Matlab:

-1/4 + 1/4i*sqrt(15)
-1/4 - 1/4i*sqrt(15)
-2 + sqrt(3)
-2 - sqrt(3)

There should be four answers since it's a 4th degree equation, 2 of which are imaginary.

Answer 4

x is simply 0 becoz o to the power anything is 0 and 0*anything is also 0

therefore over here 2*0^4+9*0^3+8*0^2+9*0^2===
0+0+0+0=0
there's yur ans

Answer 5

You have one solution between 0 and -1. As for the others, crank it into your TI whatever and see what it looks like.

Answer 6

well theres a couple of answers x=(-3.732051.....,0) and (-(.2679492),0)