9x+ 9 + 3x)2 = (6 x — 7)2. √4x29x+9+3x= ±(6x-7),.(1.);

or, taking the upper sign

Again, taking the lower sign of (1), we have—

Hence, the roots of the given equation are 5, 8, 1, 49.

REMARK.-If we proceed to verify these values of x, we shall find that the last two values-viz., 1 and 4-will not satisfy the given equation unless we obtain the value of √4x2 9x9 by means of the equation from which these last roots were found.

-

Thus from (2) we find, on putting 1 and 4 successively for x

and on substituting either of these values of √4x2 - 9x + 9 along with the corresponding value of x, the given equation is satisfied.

Ex. 4. Solve

=

(x + b + c) (x + b
(a + b + c) (a + b

c) (b + c
c) (b + c − a).

x)

We shall

By inspection we see that a is one of the roots.

therefore so arrange the equation as to be able to strike out - a as a factor of each side (Art. 5).

or, (abx) – 4 a2b3 (abx" "r") = (a - b)2;

* We shall assume in this example that the laws of multiplication and division proved in Algebra, Stage I,, Arts. 25, 27, hold for frac tional indices,

Hence, extracting the square root

abx2 - 2 ab = ± (a + b).

. abx = ±(a + b) + 2 a1b¥ = (a¥ + b1)2 or — (a1 — b1)2;

Then, raising each side to the (2 pq)th power, we have→

Hence, taking the (p-q)th root, we get

We have also, since the factor x has been struck outXP = 0, and :: x = O as another solution,

6. The following worked examples are given as specimens of the methods to be employed, but it must be understood that practice alone will give the student complete mastery over equations of this class.

As we have given the sum of the unknown quantities, we shall work for the difference.

From (2), multiplying each side by 2, we have—

2 xy + y

=

4;

=

and, taking the square root, we have xy +2........(3).

4 or 8, and.. y

=

NOTE. Having found that x = 4, y = 2, we might have told by inspection that the values x = 2, y = 4, would also satisfy the given equations, for x and y are similarly involved in both equations.

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As we have given here the difference of the squares of the unknown quantities, it will be convenient to work for the sum of the squares,