# extraneous roots

by bacon
Tags: extraneous, roots
 P: 67 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: $$\sqrt{x}-\sqrt[4]{x} -2=0$$ . . . $$(x-16)(x-1)=0$$ 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. $$(x-16)(x-1)=0$$ $$x^{2}-17x+16=0$$ $$x^{2}-17x+16+25x=25x$$ $$x^{2}+8x+16=25x$$ $$(x+4)^{2}=25x$$ $$x+4=5\sqrt{x}$$ $$x+4-4\sqrt{x}=5\sqrt{x}-4\sqrt{x}$$ $$x+4-4\sqrt{x}=\sqrt{x}$$ $$(\sqrt{x}-2)^{2}=\sqrt{x}$$ $$(\sqrt{x}-2)^{2}=\sqrt{x}$$ equation A $$\sqrt{x}-2=\sqrt[4]{x}$$ equation B $$\sqrt{x}-\sqrt[4]{x}-2=0$$ 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 $$((\sqrt{x}-2)^{2})^{\frac{1}{2}}$$ 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.
 P: 258 because equation B is not your original equation, you changed the power, and because of that you changed the roots
 HW Helper P: 3,353 $$x = 5, x^2= 25, x=5, -5$$. 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.

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