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

3 Answers
Narad T.
Aug 2, 2017

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))

Charlie B.
Aug 2, 2017

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)

Alan P.
Aug 2, 2017

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)

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