Solving Equations with Radicals and Extraneous Solutions
Introduction
For equations such as:
x^2 = a
Solving for x requires us to take the root of both sides with the solutions:
x = +a and x = -a.
However when have something in the form of:
√(f(x)) = a
we can take the square of both sides and solve for x.
f(x) = a^2
Sometimes our solutions of x may not work out. This is where we run into the case of extraneous solutions.
What is an extraneous solution? This is a solution that may appear to be a valid solution by solving an equation through valid methods, but the catch is that this solution does not solve the equation. Think of an extraneous solution as a "false solution". We are going to see some examples in the next section.
Principal Square Root
When we refer to the square root of a number, most of the time we are referring to the principal square root. That is, the principal square root only refers to the non-negative square root.
Example: The principal square root of 256 is:
√256 = 16
Calculators use the principal square root for their square root function.
Example Problems where Extraneous Solutions May Exist
In all the examples, √ is used for the principal square root. In fact, unless when specified, the √ symbol will always refer to the principal square root. Other root operators, such as cube root and quartic root, will operating the same way.
Example 1:
x - 3 = √(3x - 9)
Start by squaring both sides:
x^2 - 6x + 9 = 3x - 9
Subtract (3x - 9) from both sides:
x^2 - 9x + 18 = 0
The left side can be factored into:
(x - 6)(x - 3) = 0
Giving potential solutions of:
x = 6 and x = 3.
Checking both solutions:
x = 6: 6 - 3 = 3; √(6*3 - 9) = 3
x = 3: 3 - 3 = 0; √(3*3 - 9) = 0
In this first case, both solutions, x = 3 and x = 6, are valid.
Example 2:
√(5x - 1) + 7 = 3
An attempt to solve for x:
√(5x - 1) = -4
5x - 1 = 16
5x = 17
x = 17/5
But checking for when x = 17/5:
√(5 * 17/5 - 1) + 7 = √(16) + 7 = 11 ≠ 3
In this case, x = 17/5 is an extraneous solution, and does not solve the equation.
Example 3:
x + √(4x + 1) = 5
An attempt to solve for x:
√(4x + 1) = 5 - x
4x + 1 = 25 - 10x + x^2
0 = 24 -14x + x^2
0 = (x - 2)(x - 12)
Possible solutions: x = 2, x = 12
Checking:
x = 2: 2 + √(4*2 + 1) = 2 + √(9) = 5
x = 12: 12 + √(4*12 + 1) = 12 + √(49) = 19 ≠ 5
Hence only x = 2 is a true solution, as x = 12 is extraneous.
The bottom line: when solving equations involving radicals and roots, it is best to check your answers. This is not a good practice in school, but in life.
Eddie
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