Solve Extraneous Roots: Why Do They Arise?

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In summary, the conversation discusses the issue of extraneous roots and why they arise. The original equation \sqrt{x}-\sqrt[4]{x}-2=0 has roots of 16 and 1, but when put back into the equation, 1 is found to be extraneous. The conversation also includes a step-by-step explanation of how the extraneous root arises and why it only affects x=1, not x=16. The conclusion is that extraneous roots occur when a mathematical operation is violated for that root, as seen in the example of changing the power in equation B.
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bacon
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Sorry for the length of the post, the problem I've included is not difficult but I wanted to have an example to help illustrate my question.
solve:

[tex]\sqrt{x}-\sqrt[4]{x} -2=0 [/tex]
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[tex](x-16)(x-1)=0[/tex]

The roots are 16 and 1, however when one puts them back into the original equation, 1 is found to be extraneous leaving 16 as the only solution. My question is, why do extraneous roots arise?
I attempted to answer the question myself by reversing the above process and putting 1 in for x at each step to see when the equation becomes "invalid" for the extraneous root.

[tex](x-16)(x-1)=0[/tex]

[tex]x^{2}-17x+16=0[/tex]

[tex]x^{2}-17x+16+25x=25x[/tex]

[tex]x^{2}+8x+16=25x[/tex]

[tex](x+4)^{2}=25x[/tex]

[tex]x+4=5\sqrt{x}[/tex]

[tex]x+4-4\sqrt{x}=5\sqrt{x}-4\sqrt{x}[/tex]

[tex]x+4-4\sqrt{x}=\sqrt{x}[/tex]

[tex](\sqrt{x}-2)^{2}=\sqrt{x}[/tex]

[tex](\sqrt{x}-2)^{2}=\sqrt{x}[/tex] equation A

[tex]\sqrt{x}-2=\sqrt[4]{x}[/tex] equation B

[tex]\sqrt{x}-\sqrt[4]{x}-2=0[/tex]

Putting 1 in for x in equation A works but B does not. It seems that going from A to B creates the problem. When one takes the square root of equation A the left side becomes

[tex]((\sqrt{x}-2)^{2})^{\frac{1}{2}}[/tex]

If I understood CompuChip's answer correctly to one of my previous posts, the inner to outer priority is not followed. If 1 is in for x, then -1 is the value in the first set of parenthesis and then -1 squared is 1, and then the square root is also 1. However if 1 is not in for x , since the roots are not known when one first goes through the problem, the squared to the 1/2 power gives what's in the parenthesis to the first power, which is just what's in the parenthesis. Then when 1 is in for x, we have -1 to the first power which is -1. The order of operations makes a difference for x=1 but does not for x=16.
Is it true then, that extraneous roots arise because some mathematical operation is violated for that root?
Thanks for any answers.
 
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  • #2
because equation B is not your original equation, you changed the power, and because of that you changed the roots
 
  • #3
[tex]x = 5, x^2= 25, x=5, -5[/tex]. Exactly the same as that, but more disguised =] In this same one, when you squared it, you introduced the erroneous negative square root when only the positive root applies.
 

FAQ: Solve Extraneous Roots: Why Do They Arise?

What are extraneous roots?

Extraneous roots are solutions that appear to be valid solutions to an equation, but when substituted back into the original equation, they do not satisfy the equation.

How do extraneous roots arise?

Extraneous roots can arise when solving equations that involve both radicals and variables. This can occur when squaring both sides of an equation, which can introduce additional solutions that do not actually satisfy the original equation.

Why is it important to identify and eliminate extraneous roots?

It is important to identify and eliminate extraneous roots because they can lead to incorrect solutions and can cause confusion in mathematical equations and applications. It is also important to ensure that the solutions are accurate and valid.

How can we identify extraneous roots?

To identify extraneous roots, we can substitute the solutions back into the original equation and check if they satisfy the equation. If they do not, then they are extraneous roots.

How can we avoid extraneous roots?

To avoid extraneous roots, we can carefully examine the steps we take while solving an equation and avoid squaring both sides of an equation without considering the possibility of introducing extraneous roots. We can also check our solutions by substituting them back into the original equation before considering them as valid solutions.

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