We have seen in Art. 412 that the square of a binomial (a + b) equals a2 + 2 ab + b2 - a trinomial - the first and last terms of which are each the square of one of the two quantities, while the second term contains the second quantity multiplied by twice the first quantity -
In analytical investigations it frequently occurs that an expression will be obtained which may be reduced to this form:
a2 + m a b = f, (118.)
in which m is the coefficient of the second term, and a and b are two quantities represented by a and b or any other two symbols.
A comparison of this expression with the square of a binomial (112.) contained in Art. 412, shows that the member at the left comprises two out of the three terms of the square of a binomial; as thus -
a2 + 2 a b 4. b2,
but with a coefficient m instead of 2. It is desirable, as will be seen, to ascertain a proper third term for the given expression; or, as it is termed, " to complete the square." The method by which this is done will now be shown.
A consideration of the above trinomial shows that the third term is equal to the square of the quotient obtained by dividing the second term by twice the square root of the first; or -
(2 ab/2 a)2 = b2.
Now a third term to the above binomial, equation (118.), may be obtained by this same rule. For example -
(mab/2 a)2 = (mb/2)2.
The rule for the third term then is: Divide the second term by twice the squdre root of the first, and square the quotient.
As an example, let it; be required to find the third term required to complete the square in the expression -
6 n x + 4x2 = f
in which n and f are known quantities and x unknown. Putting it in this form -
4 x3 + 6 n x = f, and dividing by 4, we have -
x2+ 6/4 nx = f/4,
which reduces to -
x2 + 3/2 nx = f/4
Now applying the above rule for finding the third term, we have -
(3/2nx/2x)3 = (3/4n)3
which is the required third term. To complete the square we add this third term to both members of the above reduced expression, and have -
x2 + 3/2 nx + (3/4n)2 = f/4 + (3/4n)2
The member of this expression at the left is the completed square of a binomial, the two quantities constituting which are the square roots of the first and third terms respectively; or x and 3/4 n, and we therefore have -
and now taking the square root of both sides of the expression, we have -
and, by transferring the second quantity to the right member, we have -
an expression in which x, the unknown quantity, is made to stand alone and equal to known quantities.
The process of completing the square is useful, as has been shown, in developing the value of an unknown quantity where it enters into an expression in two forms, one as the square of the other.
As an example to test the above result, let f = 256 and n = 8. Then we have by the last expression for the value of x -
Now this value of x may be tested in the original expression -
6 nx + 4 x2 = f,
for which we have -
6x8x4 + 4x42=f,
192 + 64 = f
the correct value as above.