Solution of special equations in one unknown

Equations of the form x2n + axn + b = 0

Let xn = z, so that you arrive at a quadratic equation in z; n can then even be a fraction!

Task: Solve x4 - 5x2 + 4 = 0.

With z = x2, the equation becomes z2 - 5z + 4 = 0 and z1=4, z2 = 1. Since z = x2, you must still solve the equations x² = 4 and x² = 1. when you find the roots of the given equation:

x1 = +2, x2 = -2, x3 = +1, x4 = -1.

You can expand this method:

Task: Find the roots of .

Let to obtain z² + 8 = 6z, so that z1 = +4, z2= 2.

Now you must still solve the equations

They have the roots x1 = +8, x2 = -5, x3 = +7, x4 = -4.

Equations with roots

Task: Solve the equation .

You take powers, in order to remove the root. If you were to simply square the equation, a root would remain on the left hand side. You must first reorganize the equation, so that the root is alone on one side.

  x² - 3x - 3 = -4x + 4   x = 7

Task: Solve

This time you must square twice:

     
85 + 7x - 2x² = 49 + 7x + x²/4   9x²=144, x1= +4, x2=-4   +4

Sometimes the manipulations become simpler by shifting in the initial equation only one root to the other side.

You can also solve this task by introduction of a new variable. Let or x = z² - 5, so that the given equation becomes

.

After taking the square, you find 27 - 2z² = 36 - 12z + z² or z1=3, z2=1, whence x = z² - 5 yields again x1= +4, x2 = -4.

Tasks: Test

1. Square the equation: or

2. Square the equation again: whence x1= 12, x2 = 4.

When you substitute these results into the original equation, you find that they do not satisfy it. Why?

In order for the two roots to be principal values, x must be at least 3. If

Hence the difference cannot be equal to -2. The given equation is impossible and contains a contradiction.

However, you found roots! The quadratic equation could arise out of as well as , since the different leading signs cancel during squaring. Let us see whether the roots x1=12, x2=4 satisfy the new equation. You will discover that they also do not satisfy the equation. However, due to the second squaring, the quadratic equation could arise out of and the last equation out of Finally, you find the equation

with the roots x1= 12, x2 = 4.

Whenever you employ squaring, test your results!

Exponential equations

Task: Solve the equation 2x = 1/8.

You can take the logarithm:

xlog 2 = log 1 - log 8   x = -log 8/log 2 = -0.903/0.301 = -3,

or also, if it does not cause difficulties as in the present task, take the power on both sides with the same base:

2x = 2-3   x = -3.

If you assume that the base a > 0, you can obtain from ax = ay the equation
x = y, provided a 1! 13 = 14 does not yield 3 = 4!

Example: Find x from the equation xx = 10000!

log x·log x = log 10000   [log x]² = 4   log x = ±2   x1= 100   x2= 1/100

last next