or, clearing of fractions—

24x2 + 148 x + 224 = 25 x2 + 150 x + 200;

=

24 + 1

or, completing the square, x2 + 2 x + 12
Taking the square root of each side, we have-

Ex. 3. Solve the equation x2 + 6x + 25 = 0.

As the quantity

=

± √√ 16

3 ± √ 16.

16 has no exact or approximate value, the given equation has no real roots. The roots are therefore said to be imaginary.

4. Solution by breaking into factors.

We have seen (Stage I., Alg., Art. 30) that it is often easy to find by inspection the factors of quadratic expressions. We may make use of this knowledge to solve quadratic equations.

Ex. 1. Solve the equation x2 + 5x = 66.
Transposing all to the first side, we have—

And, resolving the first side into its elementary factors, we get

(x6) (x+11) = 0.

Now, if either of these factors is put equal to 0, the equation is satisfied.

Ex. 2. Given z2

(a + b) x + ab = 0, find x.

We have (x-a) (x

Now this equation is satisfied by making either of the factors = 0.

Equations which may be Solved like Quadratics.

5. Certain equations of a degree higher than the second may be solved like quadratics. It will be seen that, although it is impossible to lay down definite rules for the treatment of every such equation, the object to be attained is either (1.) To throw the equation into the form

X2 + pX = 9,

when I is an expression containing the unknown quantity, and solving when possible this equation for X; or (2.), To strike out from each side a factor containing the unknown quantity, thus reducing the dimensions of the equation, and obtaining a value or values of the unknown quantity by equating this factor to zero.

Ex. 1. Solve the equation x + 144

=

25 x2.

144;

* See Remark, page 304,

The given equation has therefore four roots, being as many roots as the degree of the equation.

We have, (2) + 3 (23)= 88; or, transposing,

=

0;

(23)2 + 3 (203) 88
or, (1⁄23 — 8) (23 + 11) = 0.

Hence, the given equation is satisfied by

2) (° + x•2 + 2)

=

∞ • VII + ( ~11)? } = 0.

From the first of these equations, we get

0, and

0, from which two other roots may be

and x-x11 +

121

=

0, which gives two other roots. We have therefore shown how to obtain the six roots of

Changing the sign of each side, and transposing, we have

6 x√4x2 9x+9= 23 x2 75x + 40;

adding to each side 13-9x + 9, then

13x9x+9 + 6 x√4x2 - 9x+9= 36 x2 - 84x+49; or, (4x2 - 9x+9) + 2 (3x) √4 x2 - 9 x + 9 + (3 x)2 =(6x)2 - 2 (6x)· 7 + 73;