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


All original content copyright, © 2011-2021.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 


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