Leonhard Euler · 1770 (1828 English ed.) · Internet Archive
Elements of Algebra
All 627 leaves · leaf 309
Transcription
272 ELEMENTS SECT. IV.
CHAP. XIII.
Of the Resolution of Equations of the Fourth Degree.
750. When the highest power of the quantity $x$ rises to the fourth degree, we have equations of the fourth degree; the general form of which is
$x^4 + ax^3 + bx^2 + cx + d = 0$.
We shall, in the first place, consider pure equations of the fourth degree; the expression for which is simply $x^4=f$; the root of which is immediately found by extracting the biquadrate root of both sides, since we obtain $x = \sqrt[4]{f}$.
751. As $x^4$ is the square of $x^2$, the calculation is greatly facilitated by beginning with the extraction of the square root; for we shall then have $x^2 = \sqrt{f}$; and, taking the square root again, we have $x = \sqrt{\sqrt{f}}$; so that $\sqrt[4]{f}$ is nothing but the square root of the square root of $f$.
For example, if we had the equation $x^4=2401$, we should immediately have $x^2 = 49$, and then $x = 7$.
752. It is true this is only one root; and since there are always three roots in an equation of the third degree, so also there are four roots in an equation of the fourth degree: but the method which we have explained will actually give those four roots. For, in the above example, we have not only $x^2 = 49$, but also $x^2 = -49$; now, the first value gives the two roots $x = 7$, and $x = -7$, and the second value gives $x = \sqrt{-49}, = 7 \sqrt{-1}$, and $x = -\sqrt{-49} = -7 \sqrt{-1}$; which are the four biquadrate roots of 2401. The same also is true with respect to other numbers.
753. Next to these pure equations, we shall consider others, in which the second and fourth terms are wanting, and which have the form $x^4 + fx^2 + g = 0$. These may be resolved by the rule for equations of the second degree; for if we make $x^2 = y$, we have $y^2 + fy + g = 0$, or $y^2 = -fy - g$, whence we deduce
$y = -\frac{1}{2}f \pm \sqrt{(\frac{1}{4}f^2 - g)} = (\frac{-f \pm \sqrt{(f^2-4g)}}{2})$.
Now, $x^2 = y$; so that $x = \pm \sqrt{(\frac{-f \pm \sqrt{(f^2-4g)}}{2})}$, in which the double signs $\pm$ indicate all the four roots.
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