The slide rule finds an interesting application in the solution of
equations of the second and third degree; and although the process is
essentially one of trial and error, it may often serve as an efficient
substitute for the more laborious algebraic methods, particularly when
the conditions of the problem or the operator’s knowledge of the theory
of equations enables some idea to be obtained as to the character of the
result sought. The principle may be thus briefly explained:—If 1 on C is
set to _x_ on D (Fig. 38), we find _x_(_x_) = _x_^2 on D under _x_ on C.
If, however, with the slide set as before, instead of reading under _x_,
we read under _x_ + _m_ on C, the result on D will now be _x_(_x_ + _m_)
= _x_^2 + _mx_ = _q_. Hence to solve the equation _x_^2 + _mx_ − _q_ =
0, we reverse the above process, and setting the cursor to _q_ on D, we
move the slide until the number on C under the cursor, and that on D
under 1 on C, _differ by m_. It is obvious from the setting that the
_product_ of these numbers = _q_, and as their difference = _m_, they
are seen to be the roots of the equation as required. For the equation
_x_^2 − _mx_ + _q_ = 0, we require _m_ to equal the _sum_ of the roots.
Hence, setting the cursor as before to _q_ on D, we move the slide until
the number on C under the cursor, and that on D under 1 on C, are
_together equal to_ _m_, these numbers being the roots sought. The
alternative equations _x_^2 − _mx_ − _q_ = 0, and _x_^2 + _mx_ + _q_ = 0
are deducible from the others by changing the signs of the roots, and
need not be further considered.
[Illustration: FIG. 38.]
EX.—Find the roots of _x_^2 − 8_x_ + 9 = 0.
Set the cursor to 9 on D, and move the slide to the right until when
6·64 is found under the cursor, 1·355 on D is under 1 on C. These
numbers are the roots required.
The upper scales can of course be used; indeed, in general they are to
be preferred.
EX.—Find the roots of _x_^2 + 12·8_x_ + 39·4 = 0.
Set the cursor to 39·4 on A, and move the slide to the right until we
read 7·65 on B under the cursor, and 5·15 on A over 1 on B. The roots
are therefore −7·65 and −5.15.
With a little consideration of the relative value of the upper and lower
scales, the student interested will readily perceive how equations of
the third degree may be similarly resolved. The subject is not of
sufficient general importance to warrant a detailed examination being
made of the several expressions which can be dealt with in the manner
suggested; but the author gives the following example as affording some
indication of the adaptability of the method to practical calculations.
EX.—A hollow copper ball, 7·5 in. in diameter and 2 lb. in weight,
floats in water. To what depth will it sink?
The water displaced = 27·7 × 2 = 55·4 cub. in. The cubic contents of
the immersed segment will be (π)/(3)(3_r_ _x_^2 − _x_^3), _r_ being
the radius and _x_ the depth of immersion. Hence (π)/(3)(3_r_ _x_^2 −
_x_^3) = 55·4, and 11·25_x_^2 − _x_^3 = 52·9.
To solve this equation we place the cursor to 52·9 on A, and move the
slide until the reading on D under 1 and that on B under the cursor
together amount to 11·25. In this way find 2·45 on D under 1, with 8·8
on B under the cursor _c_, _c_, as a pair of values of which the sum
is 11·25. Hence we conclude that _x_ = 2·45 in. is the result sought.
With the rule thus set (Fig. 39) the student will note that the slide
is displaced to the right by an amount which represents _x_ on D, and
therefore _x_^2 on A; while the length on B from 1 to the cursor line
represents 11·25 − _x_. Hence the upper scale setting gives
_x_^2(11·25 − _x_) = 11·25_x_^2 − _x_^3 = 52·9 as required.
[Illustration: FIG. 39.]
When in doubt as to the method to be pursued in any given case, the
student should work synthetically, building up a simple example of an
analogous character to that under consideration, and so deducing the
plan to be followed in the reverse process.