Can we find the solutions of *x*^{4 }+ *x*^{2 }- 1 = 0? Defining *y *= *x*^{2}, we get *y*^{2 }+ *y *- 1 = 0, whose solutions are -1/2 ± √ 5/2. Then the four solutions of the original equation stem from *x*^{2 }= -1/2 ± √ 5/2; they are $$\pm\sqrt{-1+\sqrt{5}\over 2}, \quad \pm i\sqrt{1+\sqrt{5}\over 2}.$$ Problems of this sort were discussed in section 2. It is clear that, given the equation *x*^{4 }+ *ax*^{2 }+ *b *= 0, with real coefficients, the transformation *y *= *x*^{2} works well whenever *a*^{2 }- 4*b* > 0. We use the phrase "works well" to mean that we do not have to worry about finding the square root of a complex number. From a slightly different perspective we might observe that *x*^{4 }+ *ax*^{2 }= -*b* implies (*x*^{2}+*a*/2)^{2 }= *a*^{2}/4 - *b*; thus, (*x*^{2 }+ *a*/2)^{2 }= (√(*a*^{2}-4*b*)/2)^{2}, an equality that allows us to conclude that *x*^{2 }+ *a*/2=(1/2)√(*a*^{2 }- 4*b)* or *x*^{2 }+ *a*/2 = -(1/2)√(*a*^{2}-4*b)*. From these quadratic equations the four solutions (real or complex) do follow.

At the beginning of section 4 we analyzed the equation *x*^{4 }+ *x*^{2 }+ 1 = 0 after writing *x*^{4 }+ 1 = -*x*^{2} and "completing squares" in the sense that (*x*^{2 }+ 1)^{2 }= -*x*^{2 }+ 2*x*^{2}. Is it possible to extend this idea to all biquadratics whenever the transformation cannot be applied with ease? In his quest of a proof of the Fundamental Theorem of Algebra, Leonhard Euler (1707-1783) considered the equation *x*^{4 }+ *ax*^{2 }+ *b *= 0, where *a*^{2 }- 4*b* < 0 ([5], p. 364). The inequality *a*^{2} < 4*b* implies *a* <= |*a*| < 2√*b*, hence 2√*b *- *a* > 0. We have *x*^{4 }+ *b *= -*ax*^{2}, which is equivalent to (*x*^{2 }+ √*b*)^{2 }= -*ax*^{2 }+ 2√*b**x*^{2}. Therefore (*x*^{2 }+ √*b*)^{2 }= (*x*√(2√*b*-*a*)^{2}, which in turn leads to *x*^{2 }+ √*b *= *x*√(2√*b*-*a)* or *x*^{2 }+ √*b *= -*x*√(2√*b*-*a)*. The two quadratics provide the four solutions, real or complex, of *x*^{4 }+ *ax*^{2 }+ *b *= 0. We note that the expression *a*^{2 }- 4*b*, called the "discriminant" of the biquadratic, determines the path to be followed. We have succeeded in solving the most general biquadratic with real coefficients.

To illustrate the procedure let us analyze with care the equation *x*^{4 }+ 12 = 6*x*^{2}, one of Cardano's original equations that we mentioned in section 2. We note that its discriminant is negative, thus we write (*x*^{2}+√12)^{2 }= 6*x*^{2 }+ 2(√12)*x*^{2}, that is to say (*x*^{2 }+ √12)^{2 }= (√(6 + 2√12)*x*)^{2}. Hence *x*^{2 }+ √12 = √(6 + 2√12)*x* or *x*^{2 }+ √12 = -√(6+2√12)*x*. The first equation has the two complex solutions $${1\over 2}\sqrt{6+2\sqrt{12}}\pm i{1\over 2}\sqrt{2\sqrt{12}-6}$$ while the second equation has the two complex solutions $$-{1\over 2}\sqrt{6+2\sqrt{12}} \pm i{1\over 2}\sqrt{2\sqrt{12}-6}.$$ These are the four solutions of the given biquadratic in the complex field. Cardano was right in claiming that *x*^{4 }+ 12 = 6*x*^{2} has no solutions in the "real" realm, the only type acceptable to him and his contemporaries.