(75.) If we take the equation a2 + bx = c, and multiply both members by 44, it becomes,

4a2x2+4abx=4ac;

then adding b2 to each member, we have,

4a2x2+4abx+b2 = 4ac +b2.

The first member of this last equation is a perfect square,. the middle term being twice the product of the square roots of the extreme terms (41). We can, therefore, extract the root of both members, and the equation will be reduced to

Now, as any equation may be reduced to the form ax2 + bx = c, we have another rule for the reduction of complete equations of the second degree.

RULE II.

Bring the equation to the form ax2 + bxc; that is, so that the first member has two terms, one of which contains x2, and the other x. Then multiply both members by four times the coefficient of x2, and add the square of the coefficient of x to both; and extract the square root. The resulting equation may be reduced by the rules for simple equations.

EXAMPLES.

1. Given 2x2-6x80, to find x.

Multiplying by 8, which is four times the coefficient of x2, we have, 16.x2-48x640.

Adding the square of the coefficient of x,

16x248x+36 640 +36;

whence, 4x-6= ± √640+36=± √⁄676 = ± 26,

4x62632, or -20,

2. Given x2+3x=72, to find x. Multiplying by 4, 4x2+12x288. Adding the square of the coefficient of a,

4x2+12x+9=288+9=297.

33x-3x2-72-12x+66= x2-5x+6.

Transposing and reducing,

16x104x+169-48 +169 121.

Extracting the root, 4x-13=±11,

=

4. Given V+5× √x+12= 12, to find a.

Squaring both members,

or,

(x+5) × (x+12) = 144;.
x2+17x=84;

Then by Rule II,

4x2+68x+(17) 336+ 289625.

Extracting the root,

=

2x+17=±√625 = 25;

therefore, 2x=-17+25=8; or-42;

NOTE. The rules given for reducing equations of the form x2+paq, will apply to all equations which contain in two powers, one of which is double the other. The following

examples will show their application to a variety of such equa

Adding the square of p to both members, we have,

Extracting the square root,

+ ==

+ 4

'By Rule II, 42" + 4px" + p2 = 4q +p2.

Extracting the root,

2x" + p = ± √4q +p2 ;

4

n

Ρ

2

4

7. Given Vx+a+b√x + a = 2b2, to find x.

(77.) Divide 34 into two such parts, that their product shall be 225.

Let x=one of the parts; then will the other be 34 and taking their product, we have,

Completing the square,

x2-34x+289-225+28964;

In this case, the two values of x give the two parts into which the given number is to be divided, for, 25×9=225. 2. There are two numbers whose difference is 7, and half their product plus 30, is equal to the square of the lesser number. What are the numbers?

Let x be the less; then x+7 will be the greater; and by the conditions of the question.