Question

How do you solve `x^4 - 3x^2 + 4 = 0` using the quadratic formula?

Answer 1

The solutions are `S={+-sqrt((3+-isqrt7)/(2))}`

Explanation:

Our equation is

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

This is a biquadratic equation

Let `X=x^2`

Then,

`X^2-3X+4=0`

We solve this like quadratic equation

The discriminant is `Delta=b^2-4ac=(-3)^2-4(1)(4)=-7`

As `Delta <0`, the solutions are in `CC`

`X=(-b+-sqrtDelta)/(2a)=(3+-isqrt7)/(2)`

Therefore,

`x=+-sqrtX=+-sqrt((3+-isqrt7)/(2))`

Answer 2

Substitute `p` for `x^2`, so `p = x^2`. The quadratic equation we get is then:

`p^2 - 3p + 4 = 0`

Note that this equation has complex roots. Then, using these roots (`p_1` and `p_2`), calculate the values of `x` using the fact that `p = x^2`.

so:

`x_1 = sqrt(p_1)`
`x_2 = -sqrt(p_1)`
`x_3 = sqrt(p_2)`
`x_4 = -sqrt(p_2)`

Answer 3

`x=+-sqrt((3+-sqrt(5)i)/2)`

Explanation:

Let `r=x^2`

`x^4-3x^2+4=0color(white)("xx")harrcolor(white)("xxx")r^2-3r+4=0`

We can now apply the quadratic formula to the equation in term of `r`
`color(white)("XXX")r=(-(-3)+-sqrt((-3)^2-4 * (1) * (4)))/(2 * (1))`

`color(white)("XXXX")=(3+-sqrt(-5))/2`

`color(white)("XXXX")=(3+-sqrt(5)i)/2`

and since `r=x^2color(white)("xx")rarrcolor(white)("xx")x=+-sqrt(r)`
we have
`color(white)("XXX")x=+-sqrt((3+-sqrt(5)i)/2)`