How do you solve x^4-2x^2-3=0 ?

Answer 1

First: multiply the 1st & 3rd coefficient to get "-3." Find two numbers that give you "-3" when multiplied & "-2" (2nd/middle coefficient) when added/subtracted. The numbers are: "-3 and 1"

Sec: rewrite the expression with the new middle coefficients...

x^4 - 3x^2 + x^2 - 3 = 0

*When you factor an express with 4 terms, group "like" terms...

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

Third: solve the two x-variables > set them to equal "0"...

a. x^2 + 1 = 0
*Subtract "1" from both sides...

x^2 + 1 - 1 = 0 - 1
x^2 = - 1

*Find the square root of both sides...

V`(x^2) = +/- V`(-1)

*You can't find the square root of a negative number > the negative sign becomes an imaginary number >"i"

x = +/- i V`1
x = i and - i

b. x^2 - 3 = 0
*Add "3" to both sides...

x^2 - 3 + 3 = 0 + 3
x^2 = 3

*Find the square root of both sides...

V`(x^2) = +/- V`(3)
x = +/- V`(3)

Solutions: +/- V`(3) and i, - i

Answer 2

x^4 - 2x^2 - 3 = 0

The first thing you need to do is treat this like a quadratic. That is, should this factor conveniently, we should have something in the form

(x^2 + ?) (x^2 + ?) = 0

In fact, we can obtain two numbers that multiply to get -3 and add up to get -2.

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

Now, we equate each factor to 0.

x^2 + 1 = 0
x^2 - 3 = 0

The first equation, x^2 + 1 = 0, will not have any real solutions; the complex solutions will be x = {i, -i}.

The second equation, x^2 - 3 = 0 can be solved for x. Keep in mind that when taking the square root of both sides of an equation, you have to add a "+/-", or "plus or minus".

x^2 = 3
x = +/- sqrt(3)

Therefore, our solutions are

x = {i, -i, sqrt(3), -sqrt(3)}

Answer 3

you just have to substitute another variable instead of x^4, like this: a^2=x^4 , then the equation will be like this:
a^2-2a-3=0 => (a-3)(a+1)=0 that means "a" is either 3 or -1 and bcuz a=x^2 and square root of any number is bigger than 0, then a is not -1 so it is 3 and you'll have:
a=3 and 1=x^2 => x^2=3 then you have x= +√3 or -√3

Answer 4

x^4-2x^2-3=0 is the same as
x^4- 3x^2 + x^2 -3=0 ie
x^2(x^2-3)+1(x^2-3)=0
(x^2+1)(x^2-3)=0
x^2= -1, 3
x= sqrt(-1) which is unreal, -sqrt(3), +sqrt(3)

Answer 5

it is insolvable. no method of factoring could ever solve a fourth degree equation! that's unheard of.

Answer 6

by making x the subject of the equation hope that helps

Answer 7

x⁴- 2x² - 3

x⁴-x² + 3x² - 3

x²(x² - 1) + 3(x² - 1)

(x² + 3)(x - 1)

- - - - - - - - - - - - - s-

Answer 8

(x^2+1)(x^2-3)=0
x=+/-i,+/-sq.rt(3)