1911 Encyclopædia Britannica/Equation/Biquadratic Equations

IV. Biquadratic Equations.

1. When a biquadratic equation contains all its terms, it has this form,

x⁴ + Ax³ + Bx² + Cx + D = 0,

where A, B, C, D denote known quantities.

We shall first consider pure biquadratics, or such as contain only the first and last terms, and therefore are of this form, x⁴ = b⁴. In this case it is evident that x may be readily had by two extractions of the square root; by the first we find x² = b², and by the second x = b. This, however, is only one of the values which x may have; for since x⁴ = b⁴, therefore x⁴ − b⁴ = 0; but x⁴ − b⁴ may be resolved into two factors x² − b² and x² + b², each of which admits of a similar resolution; for x² − b² = (x − b)(x + b) and x² + b² = (x − b√−1)(x + b√−1). Hence it appears that the equation x⁴ − b⁴ = 0 may also be expressed thus,

(x − b) (x + b) (x − b√−1) (x + b√−1) = 0;

so that x may have these four values,

+b,    −b,    +b√−1,    −b√−1,

two of which are real, and the others imaginary.

2. Next to pure biquadratic equations, in respect of easiness of resolution, are such as want the second and fourth terms, and therefore have this form,

x⁴ + qx² + s = 0.

These may be resolved in the manner of quadratic equations; for if we put y = x², we have

y² + qy + s = 0,

from which we find y = 1/2 {−q ± √(q² − 4s) }, and therefore

x = ±√1/2 {−q ± √(q² − 4s) }.

3. When a biquadratic equation has all its terms, its resolution may be always reduced to that of a cubic equation. There are various methods by which such a reduction may be effected. The following was first given by Leonhard Euler in the Petersburg Commentaries, and afterwards explained more fully in his Elements of Algebra.

We have already explained how an equation which is complete in its terms may be transformed into another of the same degree, but which wants the second term; therefore any biquadratic equation may be reduced to this form,

y⁴ + py² + qy + r = 0,

where the second term is wanting, and where p, q, r denote any known quantities whatever.

That we may form an equation similar to the above, let us assume y = √a + √b + √c, and also suppose that the letters a, b, c denote the roots of the cubic equation

z³ + Pz² + Qz − R = 0;

then, from the theory of equations we have

a + b + c = −P,    ab + ac + bc = Q,    abc = R.

We square the assumed formula

y = √a + √b + √c,

and obtain y² = a + b + c + 2(√ab + √ac + √bc);

or, substituting −P for a + b + c, and transposing,

y² + P = 2(√ab + √ac + √bc).

Let this equation be also squared, and we have

y⁴ + 2Py² + P² = 4 (ab + ac + bc) + 8 (√a²bc + √ab²c + √abc²);

and since ab + ac + bc = Q,

and   √a²bc + √ab²c + √abc² = √abc (√a + √b + √c) = √R·y,

the same equation may be expressed thus:

y⁴ + 2Py² + P² = 4Q + 8√R·y.

Thus we have the biquadratic equation

y⁴ + 2Py² − 8√R·y + P² − 4Q = 0,

one of the roots of which is y = √a + √b + √c, while a, b, c are the roots of the cubic equation z³ + Pz² + Qz − R = 0.

4. In order to apply this resolution to the proposed equation y⁴ + py² + qy + r = 0, we must express the assumed coefficients P, Q, R by means of p, q, r, the coefficients of that equation. For this purpose let us compare the equations

y⁴ + py² + qy + r = 0,
y⁴ + 2Py² − 8√Ry + P² − 4Q = 0,

and it immediately appears that

2P = p,    −8√R = q,    P² − 4Q = r;

and from these equations we find

P = 1/2 p,   Q = 1/16 (p² − 4r),   R = 1/64 q².

Hence it follows that the roots of the proposed equation are generally expressed by the formula

y = √a + √b + √c;

where a, b, c denote the roots of this cubic equation,

z3 +	p	z2 +	p2 − 4r	z −	q2	= 0.
	2		16		64

But to find each particular root, we must consider, that as the square root of a number may be either positive or negative, so each of the quantities √a, √b, √c may have either the sign + or − prefixed to it; and hence our formula will give eight different expressions for the root. It is, however, to be observed, that as the product of the three quantities √a, √b, √c must be equal to √R or to −1/8 q; when q is positive, their product must be a negative quantity, and this can only be effected by making either one or three of them negative; again, when q is negative, their product must be a positive quantity; so that in this case they must either be all positive, or two of them must be negative. These considerations enable us to determine that four of the eight expressions for the root belong to the case in which q is positive, and the other four to that in which it is negative.

5. We shall now give the result of the preceding investigation in the form of a practical rule; and as the coefficients of the cubic equation which has been found involve fractions, we shall transform it into another, in which the coefficients are integers, by supposing z = 1/4 v. Thus the equation

z3 +	p	z2 +	p2 − 4r	z −	q2	= 0
	2		16		64

becomes, after reduction,

v³ + 2pv² + (p² − 4r) v − q² = 0;

it also follows, that if the roots of the latter equation are a, b, c, the roots of the former are 1/4 a, 1/4 b, 1/4 c, so that our rule may now be expressed thus:

Let y⁴ + py² + qy + r = 0 be any biquadratic equation wanting its second term. Form this cubic equation

v³ + 2pv² + (p² − 4r) v − q² = 0,

and find its roots, which let us denote by a, b, c.

Then the roots of the proposed biquadratic equation are,

when q is negative,	when q is positive,
y = 1/2 (√a + √b + √c),	y = 1/2 (−√a − √b − √c),
y = 1/2 (√a − √b − √c),	y = 1/2 (−√a + √b + √c),
y = 1/2 (−√a + √b − √c),	y = 1/2 (√a − √b + √c),
y = 1/2 (−√a − √b + √c),	y = 1/2 (√a + √b − √c).

See also below, Theory of Equations, § 17 et seq.  (X.)