The other day a student ns was helping challenge to face asked just how she can know when to check for extraneous services of an equation. I gave her a quick version the my conventional answer, and the light went on! this day I desire to share this thoughts here, because they are really important in numerous ways. In looking for references, I found that I have actually been saying essentially the exact same thing because that a long time. Few of it, ns don’t think I’ve ever seen a textbook state, though I’m sure I didn’t design it myself.
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What reasons extraneous solutions?
Here is a question that will serve together a great overview. Buy it in 2013 asked,
Extraneous RoutesI understand that radical, log, absolute value, and fractional to work sometimes introduce extraneous roots. Once else perform I need to check for them? Is it just by an equation?In instance you are not acquainted with the concept, an extraneous solution (also called an extraneous root) is a systems you gain in the procedure of solving, that transforms out no to be a systems of the initial equation. That is not actually a solution, and is not inherent in the problem itself, however is introduced by what friend do.
I sometimes highlight the idea by imagining that we work-related in a lab examining blood samples. Expect we include some reagent come the sample in order come determine, say, whether there is any kind of arsenic in the blood, and also then (still making use of the same sample) we do a check for some other class of prisoner – however the reagent we supplied for the first test is one of them. Then us will discover that chemistry in our sample, not because it was originally there (though it may have actually been), but since we put it there! We need to recognize that we presented it into the sample, and also either neglect it or execute some various other test to watch if it have to be had on ours report. In the exact same way, we need to pay fist to what we have done the may introduce an extraneous solution, so we can examine for it in the end.
Sarah’s inquiry was a vast one, for this reason I provided a rapid survey that the concept, emphasizing the it is not mainly the type of equation that calls for the examine (though that is a an excellent clue), yet what friend do in addressing it:
You have to check for extraneous roots whenever -- in the procedure of addressing -- you have done something the is not guaranteed to create an equivalent equation.I wouldn"t desire to insurance claim to offer a finish list; any time you usage a an approach you haven"t provided before, girlfriend should identify for yourself whether it drops in this category. However the most familiar of these space multiplying by an expression containing the variable, i m sorry you perform in addressing rational equations, and also which might result in unintentionally multiplying by zero; squaring or raising to one more even power, which you execute in solving radical equations or equations v fractional exponents, and also which loser information about signs; and also simplifying logarithmic expressions, i beg your pardon can readjust the domain of an expression. Details things you execute in fixing absolute values can likewise do this, though in many cases this have the right to be handled by paying close attention to conditions rather than simply checking at the end.What matters most is no the type of equation you are addressing so much as the points you do in addressing it. Because that example, sometimes people solve absolute worth equations by squaring -- that drops under one classification I listed. Other world will do something various that does no carry any kind of risk.Let’s look at every of these key types, to to fill in part details.
Multiplying by the variable:
Suppose we start with the equation (x = 3), and then (just because that fun) main point both political parties by x. What perform we get? (x^2 = 3x). If you fix this brand-new equation (be careful!), you find that it has two solutions, 3 and 0. The initial equation had only one solution, 3 (obviously). For this reason the new equation is not tantamount to the original; the new one has an extra solution, 0. Why? due to the fact that when you main point by 0, both sides end up being 0, and also what may have been a false equation is now a true one. So any type of value of x because that which the multiplier is zero will certainly look prefer a solution, whether or no it was.
Of course, we wouldn’t multiply that equation by x to settle it; yet we would do this to resolve a rational equation favor (frac1x + 2 = fracx + 1x); after multiply by x, we have (1 + 2x = x + 1), and we find that the systems is (x = 0). However if x is zero, then us multiplied through zero, for this reason of course the will show up to it is in a solution! In fact, if we check (x = 0) in the original equation, we find that both sides are undefined, which means that 0 is not a solution.
Note that there is other extra happening here: as soon as we multiplied by x, we additionally changed the domain the the equation, by removed the denominator. In fact, this is what we really require to check for: If our declared solution is no in the domain of the initial equation, it is extraneous, and must be ignored.
Squaring each side:
Suppose we start with the same equation (x = 3), and then square both sides. This time we gain (x^2 = 9), which has two solutions, 3 and -3. Again, the new equation is not equivalent to the original; we presented the extra “solution” -3. Why? because when girlfriend square, you shed information around signs. If we had actually started with (x = -3), we would certainly have ended up v the same equation. So once we deal with the squared equation, we room actually fixing both original equations in ~ once; our “solutions” might be solutions of one, or the other, or both, and we don’t recognize which until we check.
This commonly happens as soon as we are addressing a radical equation. As an example, we can solve (sqrtx + 6 = x) through squaring, which yields (x + 6 = x^2). The remedies of this brand-new equation room 3 and -2. The very first of these, 3, is in reality a solution of the initial equation: (sqrt3 + 6 = 3). But -2 is not a solution: (sqrt-2 + 6 = 2), not -2. What happened? us actually resolved the negate equation, (-sqrtx + 6 = x), of i m sorry -2 is the solution!
There is much more going top top here, too. Occasionally (though no usually), the equation is undefined due to the fact that the radicand i do not care negative; an example of this is (sqrt2x + 1 = sqrtx). Squaring yields (2x + 1 = x), whose equipment is (x = -1). When we check this, we uncover that -1 is no in the domain that the equation, therefore that rather than gift a solution of the negated equation, it is not a systems of either. The usual problem, together in the first example above, can likewise be explained as a range issue: since the radical prize represents only the positive square root, the is this limit on the range that brought about our -2 not to it is in a solution.
Here is an instance from 2001 where a student discovered this for himself:
Simplifying a logarithmic or various other expression:Domain concerns are the normal culprit in logarithmic equations. Right here is one example: (logx + logx+3= 1). If we simplify (condense) the left side, us get (logx^2 + 3x = 1) and also then (x^2 + 3x = 10), whose solution is (x = 2, -5). But due to the fact that of the domain the the log, the negative solution is extraneous: (log2 + log2+3= 1) is true, but (log-5 + log-5+3= 1) is not. The problem here is the condensing a logarithmic expression (or some other species of expression) can readjust the domain. For more on this, see
Are properties of Logarithms lacking Something?
How can you recognize extraneous solutions?Hidden in the details over is critical fact: each kind of extraneous solution has its own particular test. Students too regularly miss the truth that as soon as they check a solution, it have the right to fail because that two an extremely different reasons: It might fail because it is an extraneous systems that we presented by our job-related (in which case the check is critical part the the work-related itself); or it may fail just since it is wrong, a an outcome of a failure in our work. If the is extraneous, we have the right to just ignore it (and say there is no solution, if we discovered no non-extraneous solutions); if the is erroneous, we need to go back and resolve our work.
So once I teach about this, I constantly show just how to differentiate an extraneous solution from one erroneous solution.
Here is a concern from 2012:
Extraneous root Checked less TediouslyGiven (x - 4)/x + x/3 = 6I solved this for x and also found 2 roots: x = <15 - SQRT(273)>/2 x = <15 + SQRT(273)>/2Can anybody help me examine whether one of these root is extraneous or not?Plugging in these values of x and also satisfying the left hand side and right hand side would certainly take as well long. For basic values that x, like 2 and also 3, it"s simpler to uncover the extraneous roots. However values of x that have two parts and also contain square roots, as in mine case, cause challenge in recognize the extraneous roots.Other than plugging in values, is there a simple method of checking for extraneous roots?Muhammad knew the need to check the solutions, yet the examine here appeared too hard, because the solutions were ugly expressions. I offered him two pieces that advice that deserve to simplify the check process. First, since any kind of extraneous systems to a rational equation will certainly fail by making terms undefined, he only demands to inspect that:
In this kind of equation, the source of extraneous roots is multiplication by x, which deserve to introduce one extraneous source if x = 0 turns out to be a root of the brand-new equation, since multiplication by 0 go not develop an identical equation. To put it one more way, one extraneous root will prove to it is in extraneous just by making a denominator the the initial equation zero.So all you need to do come test for an extraneous root is to make sure each solution is in the domain of the equation -- the is, none makes the denominator zero. In this case, that is clear that they don"t.Second, check is also needed in order come make certain you didn’t make a mistake; but there, the accuracy the a calculator is great enough; girlfriend don’t need specific verification uneven you room told to:
The first root, for example, is about -0.76135582. I would save that value in mine calculator to avoid having actually to retype it three times, and also use the in the equation: (x - 4)/x + x/3 = (-0.76135582-4)/-0.76135582 + -0.76135582/3 = 6.25378527 + -0.25378527 = 6If this had actually come out to 5.99999999, ns would take into consideration it verified!I likewise gave a warning that many students must hear: once they have actually learned around extraneous solutions, they often start composing “no solution” at any time a inspect fails, even if that is a linear equation the can’t probably be extraneous! Their knowledge of extraneous solutions seems to have displaced what they previously knew, that they themselves are the many common cause of failed checks. Only disregard failed solutions that girlfriend know room extraneous.
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Here is a concern from 2017 around recognizing extraneous solutions brought about by squaring:
One variable in two RadicalsSolve for x: sqrt(2x - 5) - sqrt(x - 2) = 2I make the efforts squaring each term individually and also then squaring the 2, yet my roots are not the root in the solution. ...Here Phinah had several different issues, so i went v the whole procedure of solving; but at the end (before gift asked) i pointed out just how checking works here:
I"ll point out one extra thing: when you are checking your answer, you have the right to recognize one extraneous root (as opposed to an error early out to, because that example, an arithmetic mistake) if it satisfies one equation obtained by transforming the sign of a radical.In this case, as soon as you inspect x = 3, you obtain sqrt(2x - 5) - sqrt(x - 2) = 2 sqrt(2*3 - 5) - sqrt(3 - 2) = 2 sqrt(1) - sqrt(1) = 2This is false. But it i do not care true if you readjust a sign to sqrt(1) + sqrt(1) = 2So this is in reality a systems of sqrt(2x - 5) + sqrt(x - 2) = 2This is identical from the provided equation after ~ squaring. The is why the extraneous equipment arises.This is typical: If the examine fails, check out if changing the authorize on one or both radicals would make the correct. If so, climate the equipment is extraneous and you have the right to quietly overcome it out; however if not (if friend got, say, (sqrt3 – sqrt2 = 2), which is just false), climate you have actually to find the error, since your job-related was wrong.
Some connected issues
There space some various other things worth discussing that are regarded extraneous solutions, yet I will just carry out links:
Here I offer an alternative (which I had never believed of previously) the simplifies the inspect in part cases:
Avoiding the final Step that Checking for Extraneous SolutionsHere I disputed the the contrary issue, whereby you can lose a valid solution quite than uncover an invalid one:
Root Propagation and LossAnd here I discussed comparable issues in trigonometric equations: