solve: (x^4)+(10x^2)=-9?

If i answer this choose me a best answer
let -9 turn to the left so
x^4+10x^2+9=0

let y=x^2

so subt.

y^2+10y+9=0

(y+1)(y+9)=0

y=-1 and -9

since y=x^2


x^2=-9 and -1

the answer is x=+-3i and +-1i

imaginary

u can search more about like trigonometric imagineries in complex number

x^4 + 10x^2 + 9 = 0
let n = x^2 so
n^2 + 10n + 9 = 0
(n + 9)(n + 1) = 0
(x^2 + 9)(x^2 + 1) = 0
so there aren't any real answers
complex answers: x = +-3i, +-i

x^4 is always positive and also x^2 is always positive. So, their sum cant be a negative no. so, the given equation doesnt have ne real solution.

(x^4)+(10x^2)=-9
or x^4 + 10x^2 + 9 = 0
or x^4 + 9x^2 + x^2 + 9 = 0
or x^2(x^2 + 9) + 1(x^2 + 9) = 0
or (x^2 + 9)(x^2 + 1) = 0
giving x = 3i, -- 3i, i, -- i.

x^4 - 10x^2 + 9 = 0...factoring (x^2 - 9)(x^2 - a million) = 0 (x-3)(x+3)(x-a million)(x+a million) = 0 x = ±a million, ±3 i might could say not one of the above. Is there some thing else provided that limits x, like x>0? Or does C particularly say ±a million, ±3 ?

let y = x²
y² + 10y + 9 = 0
(y + 9)(y + 1) = 0
y = - 9 , y = - 1
x² = - 9 , x² = - 1
x² = 9 i² , x = i²
x = ± 3 i , x = ± i

(x^4) + (10x^2) = -9
x^4 + 10x^2 + 9 = 0
(x^2 + 9)(x^2 + 1) = 0

x^2 + 9 = 0
x^2 = -9
x = √-9 (imaginary number)

x^2 + 1 = 0
x^2 = -1
x = √-1 (imaginary number)

(this equation cannot be solved)

x^4 + 10 x ^2 = -9
10x^6 = -9
x^6 = -9/10
x = 6square root -9/10


that's what i think anyway

There are no real solutions, since x^4 and x^2 are always positive. There are however complex solutions:


(x^2)^2 + 10(x^2) + 9 = 0

(x^2) = [ -b +/- sqrt(b^2 - 4ac) ] / 2a
(x^2) = [ -10 +/- sqrt(100 - 36) ] / 2
= [ -10 +/- 8 ] / 2
= -1 or -9


For -1:

x^2 = -1
x = +/- i


For -9:

x^2 = -9
x = +/- 3i




Therefore:

x = -3i or x = -i or x = i or x = 3i

x^2 = y
y^2 + 10y +9 = 0
d = 10^ - 4*9 = 64
y1,2 = (-10 +- sqrt(d)) / 2
y1 = (-10 - 8) / 2 = -9
y2 = (-10 +8)/2 = -1
x^2 >= 0 whatever x => the equation has no real solutions, but complex solutions

x^2 = -9 => x1,2 = +-3i
x^2 = -1 => x3,4 = +-i