. Graphical and mechanical computation . olve the equationwe get w3 — 210 , 9000az a6 w3 — 210 w — 9000 = o. If weNow let a = 10, and we have w3 — w — 9 = o. The index line joining p = — andq = —9 cuts the curve in w = ; hence w = (Fig. 606). We may similarly build an alignment chart for the quadratic equa-tion w2 + pw + q = o. The method of construction of the curved axisfor this equation differs from the construction of the curved axis for thecubic only in that w2 replaces w3, so that DD = z = —w2/(w +1).Again we note that the points A (q = o), D (w = w0), and E (p = —w0)must
. Graphical and mechanical computation . olve the equationwe get w3 — 210 , 9000az a6 w3 — 210 w — 9000 = o. If weNow let a = 10, and we have w3 — w — 9 = o. The index line joining p = — andq = —9 cuts the curve in w = ; hence w = (Fig. 606). We may similarly build an alignment chart for the quadratic equa-tion w2 + pw + q = o. The method of construction of the curved axisfor this equation differs from the construction of the curved axis for thecubic only in that w2 replaces w3, so that DD = z = —w2/(w +1).Again we note that the points A (q = o), D (w = w0), and E (p = —w0)must lie on a straight line, since these values of q, w, and p satisfy theequation w2 -f- pw + q = o. We may use this fact in constructing thepoints of the curve. The complete curve is drawn in Fig. 606 (curvemarked Q); by means of it we can find the positive roots of theequation. Example 3. Solve the equation w2 — w + = 0. The indexline joining p — — and q = cuts the curve in w = andw = ?10 112 ALIGNMENT CHART FOR SOLUTION OF QUADRATIC AND CUBIC EQUATIONS. Fig. 6o&. Art. 60 THE SOLUTION OF CUBIC AND QUADRATIC EQUATIONS 113 The complete cubic equation w3 + nw2 -f- pw + q = o may be trans-formed into the equation w3 + pw + g = o by the substitution w = w , or we may proceed to solve it directly by means of an alignment chart. In the construction of the curved axis we have AD = zx = low and DD = z = - w3 -\- nw2 •UP + n - w -\-1 ~ w -\- 1 \ w-\-\) \ wJr\l In Fig. 60c, let the curve AQ correspond to the quadratic w2 -\- pw + q = oand the curve A C0 to the cubic w3 + pw -f- q = o (where n = o). Then w3 DQ= - vf DC0 = DD = DC0 + n- DQ. w + 1 w -\- 1 Thus starting at C0, we simply lay off, along DD, n times the fixed dis-tance DQ to arrive at the point Cn of the curve corresponding to the com-plete cubic equation. We thus rapidly layoff from Co in either direction a uniformscale with interval equal to DQ, and getthe points d, C2, C
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