Extraneous Solutions Of Radical Equations

A radical equation is an equation that contains a square root, cube root, or other higher root of the variable in the original problem. "Radical" is the term used for the ${\sqrt {}}$ symbol, so the problem is called a "radical equation."^[1] To solve a radical equation, you have to eliminate the root by isolating it, squaring or cubing the equation, and and then simplifying to find your answer. Withal, this process tin create answers that appear to exist right, simply are not, considering of the squaring process. These are called extraneous solutions. Y'all must learn to identify and discard the extraneous solutions.

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Isolate the radical term. The first step to solving a radical equation is to move the radical term to stand alone on one side of the equation. Move all other terms to the reverse side. In this step, if possible, combine any other similar terms that may exist.^[2]
- Consider the sample problem ${\sqrt {ten-1}}+iv=x-3$ . Your start stride is to isolate the radical on the left side of the equation, as follows:
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Echo the previous steps if necessary. If your original problem contained 2 or more radical terms, then the outset circular of isolating and squaring may not accept removed all the radicals. If that is the case, then yous should, once again, manipulate your equation to isolate the radical that remains and square each side again.^[4]
- An example of such a problem would be something like ${\sqrt {2x-v}}-{\sqrt {x-i}}=1$ . Because of the two radicals, you will need to do this procedure twice.

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Consolidate and combine like terms. Later on you have eliminated all the radicals from the problem, move all the terms to one side of the equation and combine like terms.^[five]
- Returning to the working sample problem, this looks as follows:
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Solve the equation. In most cases, this step will create a quadratic polynomial. This is an equation that contains an $x^{2}$ term as its highest variable. If the original radical was something other than a square root (such equally a cube root or fourth root, for example), then you may accept a more hard trouble. We will focus on the quadratic for this article. You may be able to solve the quadratic equation by factoring, or you can go direct to the quadratic formula.^[6]
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Make up one's mind your solutions. Factoring the quadratic equation in this instance suggests ii possible solutions. Because the quadratic equation is equal to 0, you detect the solutions by setting each cistron equal to 0 and then solve.^[7]

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Recognize the potential for an extraneous solution. Call back that after isolating the radical on ane side of the equation, you then squared both sides to remove the radical sign. This is a necessary footstep to solving the trouble. Still, the squaring operation is what creates the inapplicable solutions.^[eight]
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Test each of your solutions in the original trouble. After you lot accept found the solutions to your problem, y'all may have found one, two or more unlike possible values for the variable. Yous demand to check each of these in the original problem to see which work. Recall that the original problem here was ${\sqrt {x-i}}+4=x-3$ .^[9]
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Discard the extraneous solution and report your issue. The inapplicable solution is incorrect and can be discarded. Whatever remains is the answer to your problem. In this example, you would written report that $x=10$ .^[10]

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Question

What is an instance of an extraneous solution?

Sergeantpro

Community Answer

For example, say that you desire to find the length of a foursquare with the given area of 25 cm squared. You do that by taking the square root of 25. If you take the square root of 25, yous will take ii answers, which are 5 and -5. However, -5 cm would not work since negative lengths do not really make much sense in geometry. -5 cm would exist, in this instance, the extraneous solution, since it is a valid answer but doesn't solve the problem.