Lesson 2 · from §100 of the 1910 text
Harder Equations; Equations in the Quadratic Form
Harder equations
The rule of completing the square works whatever the coefficients are.
- Radical coefficients. In x^2 + 4\sqrt5\,x = 25, half the coefficient of x is 2\sqrt5, and its square is 20: (x + 2\sqrt5)^2 = 45, so x + 2\sqrt5 = \pm 3\sqrt5, and x = \sqrt5 or -5\sqrt5.
- Fractional equations. Clear of fractions first, then complete the square. Then test each root in the original equation: a value that makes a denominator zero is not a root.
- Equations in other letters. Solve for the letter that is used, y, t, v, R, … exactly as for x.
Equations in the quadratic form
Some equations of the fourth degree can be solved by completing the square, when only x^4, x^2, and a constant appear. For example, solve x^4 - 5x^2 + 4 = 0.
| Transposing, | x^4 - 5x^2 = -4. |
| Completing the square, | x^4 - 5x^2 + \frac{25}{4} = -4 + \frac{25}{4} = \frac94. |
| Extracting the square root, | x^2 - \frac52 = \pm\frac32. |
| Hence | x^2 = 4 or 1, |
| whence | x = \pm 2 or \pm 1. |
Check as usual. Such an equation has four roots; type all four, for instance ±2, ±1.
Typing roots
Separate the roots with commas: 2, -8. A pair of roots with a double sign may be typed with ± or +-: 1 ± sqrt(5) means 1 + \sqrt5 and 1 - \sqrt5, and (3 +- sqrt(17))/4 works too. Give irrational roots in simplest radical form: \sqrt{20} is marked "equal, but not simplified"; write 2\sqrt5. Where decimal values are asked for, type them to three places.
Try one:
Quick check
Solve: x^4 - 10x^2 + 9 = 0
Exercises
How to type math
Use ^ for powers (a^2), / for fractions ((a+b)/2), sqrt(x) for square roots, and cbrt(x) for cube roots. You can leave out the multiplication sign: 3ab means 3 \cdot a \cdot b. Mixed numbers like 4 1/2 work too. The preview under each box shows how your answer was read.
x^2 + 4\sqrt5\,x = 25
x^2 - 3\sqrt2\,x + 4 = 0
1 - \dfrac{3\sqrt3}{2R} + \dfrac{3}{2R^2} = 0
\dfrac{2x}{9} + \dfrac23 = \dfrac{7}{18x}
\dfrac13 + \dfrac x9 - \dfrac2x = 0
\dfrac{2}{5v^2} + \dfrac14 - \dfrac{11}{10v} = 0
\dfrac{3a}{2} + \dfrac12 - \dfrac{1}{3a} = 0
\dfrac{25c}{2} - \dfrac{40}{c} - 40 = 0
(x - 4)^2 - 3(x - 9) = 15
(x - 2)(x + 3) = x(5x - 9) - 2
x - \dfrac{3}{x + 2} = 0
\dfrac{v^2}{v - 5} + \dfrac52 = 0
\dfrac{3}{t - 7} + \dfrac t4 = 0
\dfrac{s}{s - 2} + \dfrac{s - 2}{s} = \dfrac52
\dfrac{7}{2y - 3} - \dfrac{5}{1 - y} = 12
\dfrac{3 + x}{4 + x} - \dfrac{x - 5}{x - 6} = \dfrac{1}{12}
If y = 2, solve for x the equation x^2 - xy - 3y^2 = -12.
If x = -3, solve x^2 - 4xy + x^3 + y^2 + 5 = 0 for y.
Solve these equations in the quadratic form (each has four roots):
x^4 - 13x^2 + 36 = 0
4x^4 - 5x^2 + 1 = 0
9x^4 - 37x^2 + 4 = 0
4k^4 = 9k^2 - 2
9s^4 + 12 = 31s^2
4v^4 + 5 = 21v^2
Point out the error in the following:
| (1) | 9 - 30 = 49 - 70. |
| (2) | 9 - 30 + 25 = 49 - 70 + 25. |
| (3) | (3 - 5)^2 = (7 - 5)^2. |
| (4) | 3 - 5 = 7 - 5. |
| (5) | 3 = 7. |
Which line is the first one that does not follow from the line before it?