Elements of Algebra

CHAP. XVI. OF ALGEBRA. 291 required to find by approximation cube roots, biquadrate roots, &c. Let there be given the equation of the third degree, $x^3 = a$; or let it be proposed to find the value of $\sqrt[3]{a}$. Knowing that it is nearly $n$, we shall suppose $x = n + p$; neglecting $p^2$ and $p^3$, we shall have $x^3 = n^3 + 3n^2p = a$; so that $3n^2p = a - n^3$, and $p = \frac{a-n^3}{3n^2}$; whence $x = (n+p) = \frac{2n^3+a}{3n^2}$. If, therefore, $n$ is nearly $\sqrt[3]{a}$, the quantity which we have now found will be much nearer it. But for still greater exactness, we may again substitute this new value for $n$, and so on. For example, let $x^3 = a = 2$; and let it be required to determine $\sqrt[3]{2}$. Here, if $n$ is nearly the value of the number sought, the formula $\frac{2n^3+2}{3n^2}$ will express that number still more nearly; let us therefore make 1. $n=1$, and we shall have $x = \frac{4}{3}$, 2. $n=\frac{4}{3}$, and we shall have $x = \frac{91}{72}$, 3. $n=\frac{91}{72}$, and we shall have $x = \frac{162130596}{128634294}$. 789. This method of approximation may be employed, with the same success, in finding the roots of all equations. To shew this, suppose we have the general equation of the third degree, $x^3 + ax^2 + bx + c = 0$, in which $n$ is very nearly the value of one of the roots. Let us make $x = n - p$; and, since $p$ will be a fraction, neglecting the powers of this letter, which are higher than the first degree, we shall have $x^2 = n^2 - 2np$, and $x^3 = n^3 - 3n^2p$; whence we have the equation $n^3 - 3n^2p + an^2 - 2anp + bn - bp + c = 0$, or $n^3 + an^2 + bn + c = 3n^2p + 2anp + bp = (3n^2 + 2an + b)p$; so that $p = \frac{n^3+an^2+bn+c}{3n^2+2an+b}$, and $x = n - (\frac{n^3+an^2+bn+c}{3n^2+2an+b}) = \frac{2n^3+an^2-c}{3n^2+2an+b}$. This value, which is more exact than the first, being substituted for $n$, will furnish a new value still more accurate. 790. In order to apply this operation to an example, let $x^3 + 2x^2 + 3x - 50 = 0$, in which $a=2, b=3$, and $c=-50$. If $n$ is supposed to be nearly the value of one of the roots, $x = \frac{2n^3+2n^2+50}{3n^2+4n+3}$, will be a value still nearer the truth. U 2