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

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Answer 1 · 2017-08-02T12:35:56

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

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 · 2017-08-02T12:37:44

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 · 2017-08-02T12:38:21

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

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)#