Therefore, taking the square root of the square of the coefficient of the second term increased by four times the coefficient of the first term into the term independent of x, we get

±√2025+576=±√2601=±51.

This added to the coefficient of the second term with the sign changed, gives

45+51.

which must be divided by twice the coefficient of the first term. Hence,

Either of which values of x, will verify the equation.

This, when reduced to the general form, becomes

Squaring 18, we get

x2-18x=-72.

(18)2=-324.

Four times the first coefficient multiplied into -72,

gives

4x-72-288,

which added to 324, gives 36, the square root of which

This, cleared of fractions, becomes

3x2-5x=7x+36.

Therefore,

4,or-31

can be solved by the above rule, which indeed will agree with the form under consideration in the particular case of n=1.

If, in the above equation, we write y for x", and consequently y2 for x2", it will become

ay2+by=c,

which is precisely of the form of (A), Art. 117. Consequently,

This value of x must hold for all values of the con

stants n, a, b, and c, whether positive or negative, integral or fractional.

EXAMPLES.

1. Given x+ax2=b, to find x.

This becomes y2+ay=b, when for x2 we write y.

1

3. Given 2(1+x—x2)—√1+x—x2———, to find x.

If, for 1+x—x2, we put y2, our equation will become

Resubstituting 1+x-x2, for y2, we have, when we

When we take the other value of y2, we have

4. Given x1-25x2——144, to find the four values

of x.

Let x2-y, and the above equation will become

Consequently, x=+4, or -4; or else x=+3, or -3.

5. Given 7-8, to find the four values of x.