Machinery's Handbook, 31st Edition
34 Equation Solving Solving by Completing the Square.— An equation of the form x 2 + bx + c = 0 can be turned into the square of a sum. The steps are: • Move constant to the right side of the equal sign: x 2 + bx = – c
• Add ( b /2) 2 to both sides: x 2 + bx + ( b /2) 2 = – c + ( b /2) 2 • Note the left side is the square of a sum, that is: ( x + b /2) 2 • The right is a new constant, call it d : – c + ( b /2) 2 = d • The equation is thus converted to: ( x + b /2) 2 = d • Take the positive and negative square root of both sides: x = Example: Complete the square to solve x 2 + 4 x – 3 = 0. • x 2 + 4 x = 3 • x 2 + 4 x + (4/2) 2 = 3 + (4/2) 2 , that is, x 2 + 4 x + 4 = 3 + 4 • Left side: x 2 + 4 x + 4 = ( x + 2) 2 • Right side: 3 + 4 = 7 • So, ( x + 2) 2 = 7 • x + 2 = 7 ± • x = –2 7 ± • Solve for x : x b 2 – -- d ±
b 2 + --
d ± =
Completing the square is more involved if the leading coefficient a is not 1. Then it is preferable to use the quadratic formula , which can be used for any value of a in a quadratic equation. Using the Quadratic Formula.— The method of completing the square leads to the qua- dratic formula for finding the roots of equations with the form ax 2 + bx + c = 0: x b 2 4 ac – ± 2 a = -------------------- Example: In the equation, x 2 + 6 x + 5 = 0, a = 1, b = 6, and c = 5. x – 6 6 2 4 ( ) – 1 ( ) 5 ( ) ± 2 ( ) 1 ( ) --------------------------- –6 ( ) +4 2 ---------- –1 = = b – = or –6 ( ) – 4 2 ---------- = –5 Example: A right triangle has a hypotenuse of 5 cm and one leg that is 1 cm longer than the other; find the lengths of the two legs. Let x be the length of one leg and x + 1 be the length of the other; then by Pythagorean theorem (see page 62), x 2 + ( x + 1) 2 = 5 2 . Expanding this and setting all terms equal to zero gives x 2 + x – 12 = 0. Now referring to the basic formula, ax 2 + bx + c = 0, here, a = 1, b = 1, and c = –12. Substituting these values into the quadratic formula: x – 1 1 4 ( ) – 1 ( ) –12 ( ) ± ----------------------------- –1 ( ) +7 2 ---------- 3 or = = = x –1 ( ) – 7 2 ---------- –4 = = 2 ( ) 1 ( ) Since only the positive value, 3, makes sense in this case, the lengths of the two sides are x = 3 cm and x + 1 = 4 cm. Solving a Cubic Equation.— Just as quadratic equations may be simple to solve if they are of the form x 2 = c , a cubic equation of the form x 3 = c also is simple to solve. But, the given equation has the form: x 3 + ax + b = 0 then one of the real roots is:
1/3
1/3
a 3 27 --- b 2 4 + + ---
a 3 27 --- b 2 4 – + ---
b 2 – --
b 2 – --
x = The equation x 3 + px 2 + qx + r = 0 may be rewritten in the form x 1 3 + ax +
1 + b = 0 by substi
p 3 – -- for x in the given equation.
tuting x 1
Copyright 2020, Industrial Press, Inc.
ebooks.industrialpress.com
Made with FlippingBook - Share PDF online