Question

x^4+x^2-6=0

Answer 1

directly factor out it, becomes
x^4+x^2-6=0
(x^2 + 3)(x^2 - 2) = 0
set each factor be equal zero then solve for x

Answer 2

thanks i really dont remember how to do it

Answer 3

first, take x^2 + 3 = 0
this cant be factored, so use the quadratic formula :
x = (-b +- sqrt(b^2 - 4ac))/(2a)

here, known a = 1, b = 0 and c = 3

Answer 4

thanks

Answer 5

for the 2nd case, to solve of equation :
x^2 - 2 = 0, it can be factored
(x + sqrt(2)) (x - sqrt(2)) = 0
set all factor equal zero, then solve for x.
x + sqrt(2) = 0
x = - sqrt(2)

(x - sqrt(2) = 0
x = sqrt(2)

if the domain of x is real nmber, so the solution for x = { - sqrt(2), sqrt(2)}

Answer 6

so there are 4 answers

Answer 7

yeah, if the imaginer number allow here

Answer 8

is this the equation for the first one -0+-√0^2-4(1)(3)/2(1)

Answer 9

yes, just simplify again then

Answer 10

would it end up being √-4(1)(3)/2

Answer 11

because u cant square root 0

Answer 12

yup, or it can simplied becomes
+- √-4(1)(3)/2
= +- (√-12)/2
= +- (√4 * 3 * -1)/2
= +- (√4 * √3 * √-1)/2
= +- 2(√-3)i/2
= +- 2i√3
(dont forget about the sign of +- infront of the square root, and we knowed that i = sqrt(-1))

Answer 13

can i also ask u another question

Answer 14

step by step answer
http://www.mathskey.com/question2answer/