Lesson 3 · from §101 of the 1910 text
Quadratic Equations with Literal Coefficients
Quadratic equations with literal coefficients
Equations whose coefficients are letters are solved as in the examples which follow. Treat the letters other than x as known numbers.
Example 1. Solve for x: 2a^2x^2 - ax - 1 = 0.
| Transposing, | 2a^2x^2 - ax = 1. |
| Dividing by 2a^2, | x^2 - \dfrac{x}{2a} = \dfrac{1}{2a^2}. |
| Completing the square, | x^2 - \dfrac{x}{2a} + \left(-\dfrac{1}{4a}\right)^2 = \dfrac{1}{2a^2} + \dfrac{1}{16a^2}. |
| Then | \left(x - \dfrac{1}{4a}\right)^2 = \dfrac{9}{16a^2}. |
| Extracting the square root, | x - \dfrac{1}{4a} = \pm\dfrac{3}{4a}. |
| Whence | x = \dfrac1a or -\dfrac{1}{2a}. |
Check: substituting \frac1a for x: 2a^2\left(\frac1a\right)^2 - a\left(\frac1a\right) - 1 = 2 - 1 - 1 = 0. Substituting -\frac{1}{2a}: 2a^2\left(-\frac{1}{2a}\right)^2 - a\left(-\frac{1}{2a}\right) - 1 = \frac12 + \frac12 - 1 = 0.
Example 2. Solve for x: x^2 + 2x = ax + 2a.
Collect the terms in x: x^2 + (2 - a)x = 2a. Half the coefficient of x is \frac{2 - a}{2}, so x^2 + (2 - a)x + \left(\frac{2 - a}{2}\right)^2 = 2a + \frac{4 - 4a + a^2}{4}, \qquad \left(x + \frac{2 - a}{2}\right)^2 = \frac{4 + 4a + a^2}{4}. Then x + \frac{2 - a}{2} = \pm\frac{2 + a}{2}, and x = a or -2.
Type literal roots the same way as numbers, for example 2a, -a or -a ± 2. Letters under a radical sign stand for positive numbers.
Try one:
Quick check
Solve for x: x^2 - 2ax - 3a^2 = 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.
Solve for x by completing the square, and check:
x^2 - ax - 2a^2 = 0
x^2 + 2ax + a^2 - 4 = 0
x^2 + 1 = a + 2x
3x^2 - ax = 10a^2
3mx + 2m^2 = 2x^2
a^2x^2 - 7ax + 10 = 0
4x^2 + 4ax - 3a^2 = 0
x^2 + 4\sqrt a\,x - 5a = 0
2x^2 + 9x\sqrt h = 5h
a^2x^2 + 2ab = a^2 + b^2
5x^2 + ax = x
x(x - b) = a(a + b)
1 - \dfrac4x + \dfrac4{x^2} = \dfrac{b}{4x^2}
x^2 + 2x = ax + 2a
x^2 - (a + 1)x + a = 0
x^2 + bx + cx + bc = 0
x^2 - ax + 4x - 4a = 0
x^2 + 2a^2b^2 = a^2x + 2b^2x
\dfrac1a - \dfrac1{5x} = 1 - \dfrac{5x}a
x^2 + bx + c = 0
ax^2 + bx + c = 0