Multiply by (x-5) (x − 3), the L. C. M. of the denominators; then we have

Multiply by x2-1, the L.C.M. of the denominators; then we have x2 - 3x+2(x2-1)+x+1=0,

which has only one root, namely x= -.

From the above example it will be seen that when an equation has been made integral by multiplication, some of the roots of the resulting equation may have to be rejected.

In this case it is best not to multiply at once by the L.C.M. of the denominators of the fractions; much labour is often saved by a judicious arrangement and grouping of the terms.

Now multiply by the L.C.M. of the denominators; then

2(x-7)(x-6)-2(x-2) (x-1)=0,

From II. we have on multiplication by the L.C.M.

that is

(ca − x) (ab — x) + (ab − x) (bc − x) + (bc − x) (ca − x)=0,

3x2 - 2x (bc+ca+ab) + abc (a+b+c)=0,

whence x={bc+ca+ab±√√b2c2 + c2a2 + a2b2 − abc (a+b+c)}.

125. Irrational Equations. An irrational equation is one in which square or other roots of expressions containing the unknown quantity occur.

In order to rationalize an equation it is first written with one of the irrational terms standing by itself on one side of the sign of equality: both sides are then raised to the lowest power necessary to rationalize the isolated term; and the process is repeated as often as may be

necessary.

Ex. 1. Solve the equation √x+4+ √√x+20 − 2 √x+11=0.

Ex. 2. Solve the equation √2x+8-2x+5=2.

Square both members: then

2x+8+4(x+5)-4√2x+8√√x+5=4;

.. 3x+12=2√2x+8 √x+5.

Ex. 3. Solve the equation ax+a+ √bx+B+√cx+y=0.

We have

√ax+a+ √bx+B= − √cx+y.

Square both members: then we have after transposition

(a+b−c)x+a+ß − y = −2√ax+a√/bx+ß.

Squaring again, we have

that is

{(a+b−c)x+a+ß − y}2 = 4 (ax+ a) (bx+ß),

x2 (a2+b2 + c2 − 2bc - 2ca - 2ab)

+2x(aa+bB+cy - by - cẞ - ca - ay - aß - ba)

+ a2+B2 + y2 - 2ẞy - 2ya - 2aß=0.

Thus the given equation is equivalent to a quadratic equation.

It should be observed that it is quite immaterial what sign is put before a radical in the above examples; for there are two square roots of every algebraical expression and we have no symbol

which represents one only to the exclusion of the other; so that + √√x+1 and √x+1 are alike equivalent to ±√√x+1; also x+1 has the same two values as x±√√x+1.

x+

126. By squaring both members of the rational equation A = B, we obtain the equation A2 = B2; and the equation A2=B', or A2 -B2=0, is not only satisfied when A−B=0, but also when A + B = 0. Hence an equation is not in general equivalent to that obtained by squaring both its members; for the latter equation has the same roots as the original equation together with other roots which are not roots of the original equation. Additional roots are not however always introduced by squaring both sides of an irrational equation. For example, the equation x+1 = √x + 13 is really two equations since the radical may have either of two values; and by squaring both members we obtain the equation (x+1)=x+13, which is equivalent to the two.

127. A quadratic equation can only have two roots. We have already proved that an expression of the nth degree in x cannot vanish for more than n values of x, unless it vanishes for all values of x. This shews that an equation of the nth degree cannot have more than n roots, and in particular that a quadratic equation cannot have more than two roots.

The following is another proof that a quadratic equation can only have two roots.

We have to prove that ax2 + bx + c cannot vanish for a, B, y three unequal values of x. That is we have to prove that

cannot be simultaneously true, unless a, b, c are all zero.