# Chapter 8. "Of the Extraction of the Square Roots of Binomials."

### Part I. Section IV. Chapter 8. “Of the Extraction of the Square Roots of Binomials.”

669 By a *binomials* we mean a quantity composed of
two parts, which are either both affected by the sign of the
square root, or of which one, at least, contains that sign.

For this reason 3+√5 is a binomial, and likewise √8 + √3; and it is indifferent whether the two terms be joined by the sign + or by the sign - . So that 3 - √5 and 3 + √5 are both binomials.

670 The reason that these binomials deserve particular attention is, that in the resolution of quadratic equations we are always brought to quantities of this form, when the resolution cannot be performed. For example, the equation

\[x^2=6x-4\]gives \(x = 3 + \surd 5\).

It is evident, therefore, that such quantities must often occur in algebraic calculations; for which reason, we have already carefully shewn how they are to be treated in the ordinary operations of addition, subtraction, multiplication, and division, but we have not been able till now to show how their square roots are to be extracted; that is, so far as that extraction is possible; for when it is not, we must be satisfied with affixing to the quantity another radical sign. Thus, the square root of 3 + √2 is written \(\sqrt{3 + \surd 2}\); \(\surd(3+\surd 2)\).

671 It must here be observed, in the first place, that the squares of such binomials are also binomials of the same kind; in which also one of the terms is always rational.

For, if we take the square of \(a+\surd b\). we shall obtain \((a^2+b)+2a\surd b\). If therefore it were required reciprocally to take the root of the quantity \((a^2+b)+2a\surd b\), we should find it to be \(a+\surd b\), and it is undoubtedly much easier to form an idea of it in this manner, than if we had only put the sign √ before the quantity.

672 It is chiefly required, therefore, to assign a character, which may, in all cases, point out whether such a square root exists or not; for which purpose we shall begin with an easy quantity, requiring whether we can assign, in the sense that we have explained, the square root of the binomial 5 + 2√6.

Suppose, therefore, that this root is \(\surd x+\surd y\); the square of it is \((x+y)+2\surd xy\), which must be equal to the quantity 5 + 2√6. Consequently, the rational part \(x+y\) must be equal to 5, and the irrational part \(2\surd xy\) must be equal to 2√6; which last equality gives \(\surd xy=\surd 6\). Now, since \(x+y=5\), we have \(y=5-x\), and this value substituted in the equation \(xy=6\), produces \(5x-x^2=6\), or \(x^2=5x-6\); therefore

\[x=\frac{5}{2}+\sqrt{\vphantom{\left( \frac{25}{4}-\frac{24}{4}\right)}}\left( \frac{25}{4}-\frac{24}{4}\right)=\frac{5}{2}+\frac{1}{2}=3.\]So that \(x=3\) and \(y=2\); whence we conclude, that the square root of 5 + 2√6 is √3 + √2.

673 As we have here found the two equations, \(x + y = 5\), and \(xy = 6\), we shall give a particular method for obtaining the values of \(x\) and \(y\).

Since \(x+y = 5\), by squaring, \(x^2+2xy+y^2= 25\); and as we know that \(x^2- 2xy + y^2\) is the square of \(x - y\), let us subtract from \(x^2+2xy+y^2= 25\), the equation \(xy=6\), taken four times, or \(4xy=24\), in order to have \(x^2-2xy+y^2=1\); whence by extraction we have \(x-y=1\); and as \(x+y=5\), we shall easily find \(x=2\) and \(y=3\): wherefore, the square root of 5 + 2√6 is √3 + √2.

674 Let us now consider the general binomial \(a+\surd b\), and supposing its square root to be \(\surd x+\surd y\), we shall have the equation

\[(x+y)+2\surd xy=a+\surd b;\]so that \(x+y=a\), and \(2\surd xy=\surd b\), or \(4xy=b\); subtracting this square from the square of the equation \(x+y=a\), that is, from \(x^2+2xy+y^2=a^2\), there remains \(x^2-2xy+y^2=a^2-b\), the square root of which is \(x-y=\surd(a^2-b)\). Now, \(x+y=a\); we have therefore \(x=\dfrac{a+\surd(a^2-b)}{2}\) and \(y=\dfrac{a-\surd(a^2-b)}{2}\); consequently, the square root required of \(a+\surd b\) is

\[\dfrac{a+\surd(a^2-b)}{2}+\dfrac{a-\surd(a^2-b)}{2}.\]675 We admit that this expression is more complicated than if we had simply put the radical sign √ before the given binomial \(a+\surd b\), and written it \(\surd(a+\surd b)\): but the above expression may be greatly simplified when the numbers \(a\) and \(b\) are such, that \(a^2-b\) is a square; since then the sign √, which is under the radical, disappears. We see also, at the same time, that the square root of the binomial \(a+\surd b\) cannot be conveniently extracted, except when \(a^2-b=c^2\); for in this case the square root required is

\[\sqrt{\vphantom{\left(\frac{a+c}{2}\right)}} \left(\frac{a+c}{2}\right) + \sqrt{\vphantom{\surd\left(\frac{a-c}{2}\right)}}\left(\frac{a-c}{2}\right):\]but if \(a^2-b\) is not a perfect square, we cannot express the square root of \(a+\surd b\) more simply, than by putting the radical sign √ before it.

676 The condition, therefore, which is requisite, in order that we may express the square root of a binomial \(a+\surd b\) in a more convenient form, is, that \(a^2-b\) be a square; and if we represent that square by \(c^2\), we shall have for the square root in question \(\sqrt{\vphantom{\left(\frac{a+c}{2}\right)}} \left(\frac{a+c}{2}\right) + \sqrt{\vphantom{\surd\left(\frac{a-c}{2}\right)}}\left(\frac{a-c}{2}\right)\); for, by squaring this quantity, we get

\[a-2\sqrt{\vphantom{\left(\frac{a^2-c^2}{4}\right)}}\left(\frac{a^2-c^2}{4}\right);\]now, since \(c^2=a^2-b\), and consequently \(a^2-c^2=b\), the same square is found equal to

\[a-2\sqrt{\vphantom{\frac{b}{4}}} \frac{b}{4} = a-\dfrac{2\surd b}{2} = a-\surd b.\]677 When it is required, therefore, to extract the square root of a binomial, as \(a \pm \surd b\), the rule is, to subtract from the square \(a^2\) of the rational part the square \(b\) of the irrational part, to take the square root of the remainder, and calling that root \(c\), to write for the root required

\[\sqrt{\vphantom{\left(\frac{a+c}{2}\right)}} \left(\frac{a+c}{2}\right) \pm \sqrt{\vphantom{\surd\left(\frac{a-c}{2}\right)}}\left(\frac{a-c}{2}\right):\]678 If the square root of 2+√3 were required, we should have \(a=2\) and \(\surd b=\surd 3\); wherefore \(a^2-b=c=1\); so that by the formula just given, the root sought equals √³⁄₂ + √½.

Let it be required to find the square root of the binomial 11 + 6√2. Here we shall have \(a=11\), and \(\surd b=6\surd 2\); consequently, \(b=36\cdot 2=72\), and \(a^2-b=49\), which gives \(c=7\); and hence we conclude, that the square root of 11 + 6√2 is √9 + √2, or 3 + √2.

Required the square root of 11+2√30. Here \(a=11\), \(\surd b = 2\surd 30\); consequently, \(b=4\cdot 30=120\), \(a^2-b=1\), and \(c=1\); therefore the root required is √6 + √5.

679 This rule also applies, even when the binomial contains imaginary, or impossible quantities.

Let there be proposed, for example, the binomial 1 + 4√-3. First, we shall have \(a=1\) and \(\surd b = 4\surd -3\), that is to say, \(b=-48\), and \(a^2-b=49\); therefore \(c=7\), and consequently the square root required is √4 + √-3 = 2 + √-3.

Again, let there be given -½+½√-3. First, we have \(a=-\frac{1}{2}\); \(\surd b = \frac{1}{2}\surd -3\), and \(b=\frac{1}{4}\cdot -3=-\frac{3}{4}\); whence \(a^2-b=\frac{1}{4}+\frac{3}{4}=1\), and \(c=1\); and the result required is \(\surd \frac{1}{4}+\surd -\frac{3}{4}=\frac{1}{2}+\frac{\surd -3}{2}\), or \(\frac{1}{2}+\frac{1}{2}\surd -3\).

Another remarkable example is that in which it is required to find the square root of 2√-1. As there is here no rational part, we shall have \(a=0\). Now, \(\surd b=2\surd -1\), and \(b=-4\); wherefore \(a^2-b=4\) and \(c=2\); consequently, the square root required is √1 + √-1 = 1 + √-1, and the square of this quantity is found to be 1 + 2√-1 - 1 = 2√-1.

680 Suppose now we have such an equation as \(x^2=a^2 \pm b\), and that \(a^2-b=c^2\); we conclude from this, that the value of

\[x=\sqrt{\vphantom{\left(\frac{a+c}{2}\right)}} \left(\frac{a+c}{2}\right) \pm \sqrt{\vphantom{\surd\left(\frac{a-c}{2}\right)}}\left(\frac{a-c}{2}\right),\]which may be useful in many cases.

For example, if \(x^2=17+12\surd 2\), and we shall have \(x=3+\surd 8=3+2\surd 2\).

681 This case occurs most frequently in the resolution of equations of the fourth degree, such as \(x^4=2ax^2+d\). For, if we suppose \(x^2=y\), we have \(x^4=y^2\), which reduces the given equation to \(y^2=2ay+d\), and from this we find \(y=a \pm \surd(a^2+d)\), therefore, \(x^2=a \pm \surd(a^2+d)\), and consequently we have anoth.er evolution to perform. Now, since \(\surd b=\surd(a^2+d)\), we have \(b=a^2+d\), and \(a^2-b=-d\); if, therefore, \(-d\) is a square, as \(c^2\), that is to say, \(d=-c^2\), we may assign the root required.

Suppose, in reality, that \(d=-c^2\); or that the proposed equation of the fourth degree is \(x^4=2ax^2-c^2\), we shall then find that

\[x=\sqrt{\vphantom{\left(\frac{a+c}{2}\right)}} \left(\frac{a+c}{2}\right) \pm \sqrt{\vphantom{\surd\left(\frac{a-c}{2}\right)}}\left(\frac{a-c}{2}\right).\]682 We shall illustrate what we have just said by some examples.

*Example 1.* Required two numbers, whose product may be 105,
and whose squares may together make 274.

Let us represent those two numbers by \(x\) and \(y\); we shall then have the two equations,

\(xy=105\)

\(x^2+y^2=274.\)

The first gives \(y=\dfrac{105}{x}\), and this value of \(y\) being substituted in the second equation, we have

\[x^2+\dfrac{105^2}{x^2}=274.\]Wherefore \(x^4+105^2=274x^2\), or \(x^4=274x^2-105^2\).

If we now compare this equation with that in the preceding article, we have \(2a=274\), and \(-c^2=-105^2\); consequently, \(c=105\), and \(a=137\). We therefore find

\[x=\sqrt{\vphantom{\left(\frac{137+105}{2}\right)}} \left(\frac{137+105}{2}\right) \pm \sqrt{\vphantom{\surd\left(\frac{137-105}{2}\right)}}\left(\frac{137-105}{2}\right) =11 \pm 4.\]Whence \(x = 15\), or \(x - 7\). In the first case, \(y=7\), and in the second case, \(y=15\); whence the two numbers sought are 15 and 7.

683 It is proper, however, to observe, that this calculation may be performed much more easily in another way. For, since \(x^2+2xy+y^2\) and \(x^2-2xy+y^2\) are squares, and since the values of \(x^2+y^2\) and \(xy\) are given, we have only to take the double of this last quantity, and then to add and subtract it from the first, as follows: \(x^2+y^2=274\); to which if we add \(2xy=210\), we have \(x^2+2xy+y^2=484\), which gives \(x+y=22\).

But subtracting \(2xy\), there remains \(x^2-2xy+y^2=64\), whence we find \(x-y=8\).

So that \(2x=30\), and \(2y=14\); consequently, \(x = 15\) and \(y=7\).

The following general question is resolved by the same method.

*Example 2.* Required two numbers, whose product may be \(m\), and
the sum of the squares \(n\).

If those numbers are represented by \(x\) and \(y\), we have the two following equations:

\(xy=m\)

\(x^2+y^2=n.\)

Now, \(2xy=2m\), being added to \(x^2+y^2-n\), we have \(x+2+2xy+y^2=n+2m\), and consequently,

\[x+y=\surd(n+2m).\]But subtracting \(2xy\), there remains \(x^2-2xy+y^2=n-2m\), whence we get \(x-y=\surd(n-2m)\); we have, therefore,

\[x=\frac{1}{2}\surd(n+2m)+\frac{1}{2}\surd(n-2m);\]and

\[y=\frac{1}{2}\surd(n+2m)+\frac{1}{2}\surd(n-2m).\]684 *Example 3.* Required two numbers, such, that their product
may be 35, and the difference of their squares 24.

Let the greater of the two numbers be \(x\), and the less \(y\): then we shall have the two equations

\(xy=35,\)

\(x^2-y^2=24;\)

and as we have not the same advantages here, we shall proceed in the usual manner. Here, the first equation gives \(y=\dfrac{35}{x}\), and, substituting this value of \(y\) in the second, we have \(x^2-\dfrac{1225}{x^2}=24\); or \(x^4=24x^2+1225\). Now, the second member of this equation being affected by the sign +, we cannot make use of the formula already given, because having \(c^2=-1225\), \(c\) would become imaginary.

Let us therefore make \(x^2=z\); we shall then have \(z^2=24z+1225\), whence we obtain

\[z=12 \pm \surd(144+1225), \: \textrm{or,} \: z=12 \pm 37;\]consequently, \(x^2=12 \pm 37\); that is to say, either \(x^2=49\) or \(x^2=-25\).

If we adopt the first value, we have \(x=7\) and \(y=5\).

The second value gives \(x=\surd -25\); and, since \(xy=35\), we have

\[y=\frac{35}{\surd -25} = \sqrt{\vphantom{\frac{1225}{-25}}} \frac{1225}{-25}=\surd -49.\]685 We shall conclude this chapter with the following question.

*Example 4.* Required two numbers, such, that their sum, their
product, and the difference of their squares, may be all
equal.

Let \(x\) be the greater of the two numbers, and \(y\) the less; then the three following expressions must be equal to one another: namely, (I) the sum, \(x+y\); (II) the product, \(xy\); and (III) the difference of the squares, \(x^2-y^2\). If we compare the first with the second, we have \(x+y=xy\); which will give a value of \(x\): for \(y = xy - x = x(y-1)\), and \(x=\dfrac{y}{y-1}\). Consequently,

\[x+y=\dfrac{y}{y-1}+y=\dfrac{y^2}{y-1},\]and \(xy=\frac{y^2}{y-1}\), that is to say, the sum is equal to the product; and to this also the difference of the squares ought to be equal. Now, we have

\[x^2-y^2=\dfrac{y^2}{y^2-2y+1} - y^2 = \dfrac{-y^4+2y^3}{y^2-2y+1};\]so that making this equal to the quantity found \(\dfrac{y^2}{y-1}\), we have

\[\dfrac{y^2}{y-1}=\dfrac{-y^4+2y^3}{y^2-2y+1};\]dividing by \(y^2\), we have

\[\dfrac{1}{y-1}=\dfrac{-y^2+2y}{y^2-2y+1};\]and multiplying by \(y^2-2y+1\), or \((y-1)^2\), we have \(y-1=-y^2+2y\); consequently, \(y^2=y+1\); which gives

\[y=\frac{1}{2} \pm \sqrt{\vphantom{\frac{1}{4}+1}} \left(\frac{1}{4}+1\right) =\frac{1}{2} \pm \sqrt{\vphantom{\frac{5}{4}}} \frac{5}{4};\]or \(y=\dfrac{1 \pm \surd 5}{2}\), and since \(x=\dfrac{y}{y-1}\), we shall have, by substitution, and using the sign +, \(x=\dfrac{\surd 5+1}{\surd 5-1}\).

In order to remove the surd quantity from the denominator, multiply both terms by √5+1, and we obtain

\[x=\dfrac{6+2\surd 5}{4} = \dfrac{3+\surd 5}{2}.\]Therefore the greater of the numbers sought, or \(x\), equals \(\dfrac{3+\surd 5}{2}\); and the less, or \(y\), equals \(\dfrac{1 + \surd 5}{2}\).

Hence their sum \(x+y=2+\surd 5\); their product \(xy=2+\surd 5\); and since \(x^2=\dfrac{7+3\surd 5}{2}\), and \(y^2=\dfrac{3+\surd 5}{2}\), we have also the difference of the squares \(x^2-y^2=2+\surd 5\), being all the same quantity.

686 As this solution is very long, it is proper to remark that it may be abridged. In order to which, let us begin with making the sum \(x+y\) equal to the difference of the squares \(x^2-y^2\); we shall then have \(x+y=x^2-y^2\); and dividing by \(x+y\), because \(x^2-y^2=(x+y)(x-y)\), we find \(1=x-y\) and \(x=y+1\). Consequently, \(x+y=2y+1\), and \(x^2-y^2=2y+1\); farther, as the product \(xy\), or \(y^2+y\), must be equal to the same quantity, we have \(y^2+y=2y+1\), or \(y^2=y+1\), which gives, as before, \(y=\dfrac{1+\surd 5}{2}\).

687 The preceding question leads also to the solution of the following.

*Example 5.* To find two numbers, such, that their sum, their product, and the sum of their squares, may be all equal.

Let the numbers sought be represented by \(x\) and \(y\); then there must be an equality between \(x+y\), \(xy\), and \(x^2+y^2\).

Comparing the first and second quantities, we have \(x+y=xy\), whence \(x=\dfrac{y}{y-1}\); consequently, \(xy\) and \(x+y\) equal \(\dfrac{y}{y-1}\). Now, the same quantity is equal to \(x^2+y^2\); so that we have

\[\dfrac{y^2}{y^2-2y+1} + y^2 = \dfrac{y^2}{y-1}.\]Multiplying by \(y^2-2y+1\), the product is \(y^4-2y^3+2y^2=y^3-y^2\), or \(y^4=3y^3-3y^2\); and dividing by \(y^2\), we have \(y^2=3y-3\); which gives

\[y=\frac{3}{2}\sqrt{\vphantom{\frac{9}{4}-3}} \frac{9}{4}-3=\dfrac{3+\surd -3}{2};\]consequently, \(y-1=\dfrac{1+\surd -3}{2}\), whence results \(x=\dfrac{3+\surd -3}{1+\surd -3}\); and multiplying both terms by 1-√-3, the result is \(x=\dfrac{6-2\surd -3}{4}\), or \(x=\dfrac{3-\surd -3}{2}\).

Therefore the numbers sought are \(x=\dfrac{3-\surd -3}{2}\), and \(\dfrac{3+\surd -3}{2}\), the sum of which is \(x+y=3\), their product \(xy=3\); and lastly, since \(x^2=\dfrac{3-3\surd -3}{2}\), and \(y^2=\dfrac{3+3\surd -3}{2}\), the sum of the squares \(x^2+y^2=3\), all the same quantity as required.

688 We may greatly abridge this calculation by a particular artifice, which is applicable likewise to other cases; and which consists in expressing the numbers sought by the sum and the difference of two letters, instead of representing them by distinct letters.

In our last question, let us suppose one of the numbers sought to be \(p+q\), and the other \(p-q\), then their sum will be \(2p\), their product will be \(p^2-q^2\), and the sum of their squares will be \(2p^2+2q^2\); which three quantities must be equal to each other; therefore making the first equal to the second, we have \(2p=p^2-q^2\), which gives \(q^2=p^2-2p\).

Substituting this value of \(q^2\) in the tliird quantity \((2p^2+2q^2)\), and comparing the result \(4p^2-4p\) with the first, we have \(2p=4p^2-4p\), whence \(p=\frac{3}{2}\).

Consequently, \(q^2=p^2-2p=-\frac{3}{4}\), and \(q=\dfrac{\surd -3}{2}\); so that the numbers sought are \(p+q=\dfrac{3+\surd -3}{2}\), and \(p-q=\dfrac{3-\surd -3}{2}\), as before.

#### Editions

- Leonhard Euler.
*Elements of Algebra*. Translated by Rev. John Hewlett. Third Edition. Longmans, Hurst, Rees, Orme, and Co. London. 1822. - Leonhard Euler.
*Vollständige Anleitung zur Algebra. Mit den Zusätzen von Joseph Louis Lagrange.*Herausgegeben von Heinrich Weber. B. G. Teubner. Leipzig and Berlin. 1911. Leonhardi Euleri Opera omnia. Series prima. Opera mathematica. Volumen primum.