## Sunday, November 7, 2021

### Solving Equations with Radicals and Extraneous Solutions

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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