plus the double product of this first term x by a second, which must

Ρ

p

Ρ p2

be

2

since pr=2-
pa=2; therefore, if the square of

or

be

2

4

added to x2+px, the first member of the equation will become the

Ρ
2

square of x+. ; but in order that the equality may not be destroy.

p2

ed

must be added to the second member.

4

By this transformation, the equation x2+px=q becomes

From this we derive, for the resolution of complete equations of the second degree, the following general

RULE.

After reducing the equation to the form x2+px=q, add the square of half of the co-efficient of x, or of the second term, to both members; then extract the square root of both members, giving the double sign to the second member; then find the value of x from the resulting equation.

This formula for the value of x may be thus enunciated.

The value of the unknown quantity is equal to half the co-efficient

of x, taken with a contrary sign, plus or minus the square root of the known term increased by the square of half the co-efficient of x.

Take, for an example, the equation

Clearing the fractions, we have

10x2-6x+9=96-8x-12x2+273,

which agrees with the enunciation given above for the double value

of x.

It remains to perform the numerical operations. In the first

360

12

place, + must be reduced to a single number, having

22

extracting the square root of 7921, we find it to be 89; therefore,

Therefore, one of the two values which will satisfy the proposed equation, is a positive whole number, and the other a negative fraction.

For another example, take the equation

therefore, it is only necessary to multiply 57 by 24, then 37 by itself, and divide the difference of the two products by (12). Now, 37×37=1369; 57×24=1368;

This example is remarkable, as both of the values are positive, and answer directly to the enunciation of the question, of which the proposed equation is the algebraic translation.

Let us now take the literal equation

4a2-2x2+2ax=18ab—18b2.

By transposing, changing the signs, and dividing by 2, it becomes

These two values will be positive at the same time, if 2a>3b, and 36>a, that is if the numerical value of b is greater than

1. Find a number such, that twice its square, increased by three

times this number, shall give 65.

Let a be the unknown number, the equation of the problem will be

For,

Both these values satisfy the question in its algebraic sense.

2X(5)2+3×5=2×25+15=65.

But, if we wish to restrict the enunciation to its arithmetical

sense, we will first observe, that when a is replaced by —x, in the