(A new question of the week)
Extraneous roots can not only confuse the final solution to a problem; they can also make it harder to solve in the first place if you don’t deal with them early. Here is a relatively complicated rational equation, two questions about its solutions, and several ways to make it easier to solve. We’ll solve it half a dozen ways!
Four fractions in one equation
The question came from Akanksha in mid-August:
See this image:
The equation given in the question (initial equation) and the equation got after simplifying the initial equation which is the final equation (underlined as cubic equation) are one and the same. But if I calculate the value of x using the initial equation then I get x=(25-√17)/4; x=(25+√17)/4. On the other hand, I get three different values of x i.e. x=6; x=(25-√17)/4 and x=(25+√17)/4 for the final equation.
Question: why I got 1 extra value of x in final equation despite the two equations being exactly same because they should have same no. of values of x?? And if I plug in the value of x=6 (got in final equation) in the initial equation then I get the value of equation as undefined!!?? Please explain.
My second question is that the equation given in the question is a linear equation because the highest degree of x is 1 but then also I have 2 values of x?? how??
Please clarify my two questions.
Thank you
Akanksha followed a very routine procedure, adding the fractions on each side (using common denominators in a method she called “cross-multiplication”), simplifying the equation (by a similar method also commonly called cross-multiplying), and ending up with a cubic equation to solve. She doesn’t say how she solved that, or how she solved the original equation (it turns out she used technology rather than completing the work herself); her initial question is not about how to solve, but where the extra solution came from. She also has a misunderstanding of the meaning of “linear”. We’ll answer those questions and more.
Simplifying it by substitution
Doctor Rick started with questions about the work:
Hi, Akanksha, thanks for writing to the Math Doctors.
I have some answers for you, but I also have some questions:
(1) You say you “calculate the value of x using the initial equation”. How did you do that? I assume you didn’t get the answer directly from the original equation; you manipulated it somehow, just not the same way as the method you showed, right?
(I was able to solve it by a method of my own, which began by substituting y = x – 6 just to make the numbers smaller — but that’s not necessary. My method did give just the two correct solutions.)
(2) Your solution just skips from the cubic equation to the solutions. How did you solve the cubic? (I have an idea there too, but you may have done something different.)
The substitution is not something I had thought of; let’s try it:
Letting \(y=x-6\), we replace \(x\) with \(y+6\), changing $$\frac{x-3}{x-6}+\frac{x-8}{x-7}=\frac{x-7}{x-6}+\frac{x-6}{x-5}$$ to $$\frac{y+3}{y}+\frac{y-2}{y-1}=\frac{y-1}{y}+\frac{y}{y+1}.$$
Adding fractions on each side, as before, we get $$\frac{(y+3)(y-1)+(y-2)y}{(y-1)y}=\frac{(y-1)(y+1)+y^2}{y(y+1)}$$ $$\frac{2y^2-3}{y^2-y}=\frac{2y^2-1}{y^2+y}$$
Cross-multiplying as Akanksha did (though that isn’t what I would normally do), we get $$(2y^2-3)(y^2+y)=(2y^2-1)(y^2-y)$$ $$2y^4+2y^3-3y^2-3y=2y^4-2y^3-y^2+y$$ $$4y^3-2y^2-4y=0$$
One benefit of the substitution, perhaps intended, is that because one factor became simply y, we can easily see that we can divide by y (which is legal because \(y=0\) can’t be a solution); in fact, we can divide by \(2y\): $$2y^2-y-2=0$$ This can be solved by the quadratic formula, yielding $$y=\frac{1\pm\sqrt{(-1)^2-4(2)(-2)}}{2(2)}=\frac{1\pm\sqrt{17}}{4}$$ To find x, we add 6 to this and get $$x=\frac{1\pm\sqrt{17}}{4}+6=\frac{25\pm\sqrt{17}}{4}$$ Akanksha had the right answer. And we have a preview of ways to eliminate that extra solution.
(Alternatively, we could have factored out \(2y\), obtaining \(y=0\) (i.e. \(x=6\)) as an extraneous solution; Doctor Rick evidently didn’t do that, since he got only the valid solutions.
We’ll see ways to solve the cubic later.
Extraneous roots
Now for some answers!
(1) Why did you get an extra solution when you solved the equation the way you showed?
We call the 6 you got an extraneous root. One common way that extraneous roots appear is when, in the course of the solution, we multiply by a number that could be zero.
Look at the denominators in the last equation you got before “cross-multiplying”: each has (x – 6) as a factor. (That’s easy to see by looking up two lines to before you expanded the denominators.) Now, when you “cross-multiply”, what you are really doing is multiplying each side of the equation by the product of the two denominators. If x = 6, then you are multiplying by zero! That is an invalid step in solving equations. Whatever the values of the numerators, the equation you got after the multiplication is 0*(numerator1) = 0*(numerator2), or 0 = 0, if x = 6. Thus x = 6 is a solution of the new equation.
But looking at the original equation, when we set x = 6, two of the fractions have 0 in the denominator! That’s not allowed; we can’t divide by zero. So when x = 6, the original equation becomes, not 0 = 0, but (no number) = (no number). The (original) equation is not true when x = 6, so 6 is not one of its solutions.
This is closely related to what happened when we did the substitution above: We had y in the denominators on both sides, and after cross-multiplying, we had y on both sides as a factor. If \(y=0\) (that is, \(x=6\)), then that equation, but not the original, would be true. What I did above was to divide both sides by y, eliminating this problem! We’ll see below that we could have just avoided multiplying by this y (or \(x-6\)) in the first place.
When you claim that the original equation is “exactly the same” as the cubic equation, do you mean that they are equivalent? When we manipulate an equation step by step, generally each step produces an equation equivalent to the previous equation — meaning that it has exactly the same roots. That’s the whole idea of solving an equation: we want to end up with an equation that is equivalent to the original, but whose solution is obvious — like x = 3.
But there are certain things we might do that do not produce an equivalent equation; multiplying by a quantity that might be 0 is one of those things. Take a look at our blog on this topic:
That page is very much worth reading at this point. Also, for alternative ways to solve this sort of equation, see
Many Ways to Solve a Proportion
A linear equation with two solutions?
(2) Why does the original equation have two solutions if it is a linear equation?
The answer is that the equation is not a linear equation. You seem to have too loose a definition of a linear equation. A linear equation is an equation between polynomials (a sum of terms, each of the form axn — a constant times a power of the variable), in which the greatest power of x is 1. Your equation is not between polynomials, because there are variables in the denominator, which a polynomial cannot have.
What it is, is a rational equation whose numerators and denominators, in the original form, are linear; the equation as a whole is equivalent to a cubic equation; that better explains its behavior. (We’ll see that a quadratic equation works even better.)
Simplifying by division
Doctor Fenton had a different suggestion for simplifying the problem:
Another thing you can do with terms like these is to write
x-3 (x-6+6) – 3 (x-6)+(6-3) (x-6) + 3 x-6 3 3 --- = ----------- = ----------- = --------- = --- + --- = 1 + --- x-6 x-6 x-6 x-6 x-6 x-6 x-6If you do this for each term and simplify, you get a much simpler equation
3 6 1 1 --- - --- = --- - --- x-6 x-2 x-6 x-5 .
Essentially what he has done is to carry out each division, producing “mixed numbers” where all the fractions are proper. We’ve reduced the degree of the numerators. Let’s continue the work:
$$\frac{x-3}{x-6}+\frac{x-8}{x-7}=\frac{x-7}{x-6}+\frac{x-6}{x-5}$$
$$\left(1+\frac{3}{x-6}\right)+\left(1-\frac{1}{x-7}\right)=\left(1-\frac{1}{x-6}\right)+\left(1-\frac{1}{x-5}\right)$$
$$2+\frac{3}{x-6}-\frac{1}{x-7}=2-\frac{1}{x-6}-\frac{1}{x-5}$$
$$\frac{3}{x-6}-\frac{1}{x-7}=-\frac{1}{x-6}-\frac{1}{x-5}$$
(It looks like Doctor Fenton misread a 7 as a 2, and also made a little sign error.)
There are many routes from here; for now I’ll use something like Akanksha’s approach and combine fractions first, starting with the two with the same denominator:
$$\frac{3}{x-6}+\frac{1}{x-6}=\frac{1}{x-7}-\frac{1}{x-5}$$
$$\frac{4}{x-6}=\frac{1}{x-7}-\frac{1}{x-5}$$
$$\frac{4}{x-6}=\frac{(x-5)-(x-7)}{(x-7)(x-5)}$$
$$\frac{4}{x-6}=\frac{2}{(x-7)(x-5)}$$
$$4(x-7)(x-5)=2(x-6)$$
$$2(x^2-12x+35)=1(x-6)$$
$$2x^2-25x+76=0$$
$$x=\frac{25\pm\sqrt{25^2-4(2)(76)}}{2(2)}=\frac{25\pm\sqrt{17}}{4}$$
as we saw before. There was no extraneous solution this time, at least in part because I combined the fractions with the same denominator.
Solving the cubic
Akanksha replied, first acknowledging that “solve” meant using technology:
In reply To Rick Peterson Sir
1) I used calculator on Google (equation solver calculator) to find out the value of x. I input the first equation as the original one and got 2 values of x. Next, I entered the final equation in the input field and got 3 values of x.
2) I used the equation solver calculator to calculate x in cubic equation.
I am satisfied with your answer that x=6 is an extraneous solution.
In case I don’t want to use any other method (like factorise), can I use cross multiplication?? because I don’t know that the equation by which I am multiplying has x as 0 and that is an invalid step.
Can you help me in concluding whether the original equation = final equation??
Did the steps I use are correct to find out the value of x as x=(25-√17)/4 and x=(25+√17)/4?? (x=6 is extraneous solution)
There is just a small misunderstanding of the validity of cross-multiplication to be clarified.
Doctor Rick responded:
Thanks, now I see that you didn’t really solve either equation yourself; you let technology do it for you. I wondered about that.
Clearly you’ve learned some techniques for solving rational equations, but this one led you to a difficult equation, a cubic.
There are ways to solve the cubic “by hand”, as I mentioned; the Rational Root Theorem provides a way to list every rational number that could possibly be a root of a polynomial, and then there is a nice way (“synthetic division”) to test each possible rational root. Unfortunately, your cubic equation isn’t very nice; it has something like 32 possible rational roots!
So it’s good to know a few other techniques or tricks, and try different things. We have suggested a few.
The Rational Root Theorem tells us that if our cubic equation, which we can first simplify, by dividing by 2, to $$2x^3-37x^2+226x-456=0,$$ has any rational roots, then (in lowest terms) the numerator must be a factor of 456 (namely, 1, 2, 3, 4, 6, 8, 12, 19, 24, 38, 57, 76, 114, 152, 228, or 456), and the denominator must be a factor of 2 (namely 1 or 2). Counting signs, this gives the following possible roots: $$1, 2, 3, 4, 6, 8, 12, 19, 24, 38, \\57, 76, 114, 152, 228, 456, \\\frac{1}{2}, \frac{3}{2}, \frac{19}{2}, \frac{57}{2}, \frac{1}{4}, \frac{3}{4}, \frac{19}{4}, \frac{57}{4}$$ and their negatives (48 in all).
We can then divide the polynomial by each of these in turn, and when we get to the divisor \((x-6)\), we find that the quotient is \(2x^2-25x+76\), and we’ve already found its solutions. Throwing away the extraneous root 6, we have the answer.
And we could in fact have known to divide first by \((x-6)\), by noting that it was a factor of both denominators. It would be even better if we had just eliminated it in the first place!
How to recognize non-equivalent equations
But on to the questions you asked:
Can you help me in concluding whether the original equation = final equation?
You already know that your original and final equations are not equivalent (that’s the proper term, not “equal” or “the same equation”), because they do not have identical solution sets. But probably what you are asking is how to tell whether they are equivalent without solving both!
As the blog I showed you says, we need to be able to recognize what sort of steps might produce an extraneous root (or, on the other hand, miss a root). The first “risky step” listed in the blog is the one you did:
multiplying by an expression containing the variable, which you do in solving rational equations, and which might result in unintentionally multiplying by zero;
Then you asked,
In case I don’t want to use any other method (like factorise), can I use cross multiplication?? because I don’t know that the equation by which I am multiplying has x as 0 and that is an invalid step.
The problem with your solution was multiplying by a quantity that could be zero. (It isn’t that x is zero, but that there is some value of x for which the multiplier is zero. That value of x turns out to be 6.) We don’t need to know this will happen, just to realize that it might happen. As the quote above says, just multiplying by an expression that contains the variable raises this possibility!
So the main issue with extraneous roots is not to avoid making them (though we sometimes can), but to recognize when they may exist, so we can check them. (Other methods we’ve shown can avoid it in this case, but that is not always possible.)
You ask, what if you want to use this risky method? Sometimes we really can’t find another method! And sometimes the risky method is the quickest, so we choose to do it anyway. There is nothing wrong with this. However, when we’re done, we need to check all solutions in the original equation to see whether they really make the equation true. That’s the main point of the blog. We aren’t done until we do that.
In your problem, as I already pointed out, if you plug x = 6 into the original equation, you’ll get two fractions with 0 in the denominator, and that means the solution is extraneous. The other two solutions will work fine.
But in one sense, you don’t even need to recognize the possibility:
Now, what if you don’t notice that what you are doing could introduce an extraneous root (or lose a true root)? The fact is, we should check all the solutions anyway. That’s a good practice all the time, because extraneous or missing roots are not the only way we can go wrong! We might have made some careless error, like getting a sign wrong, or even multiplying 2 by 3 and getting 5. It happens! We all make mistakes, so we all need to check.
As I tell students, sometimes you have to check because it’s really part of the work; other times you should check merely because you’re human. In either case, check! Especially when it counts.
Did the steps I use are correct to find out the value of x as x=(25-√17)/4 and x=(25+√17)/4?? (x=6 is extraneous solution)
Well, they are correct as far as they go, except that they produce an extraneous solution! Therefore your work isn’t complete until you do the checks I just mentioned.
And your solution is not the best, because there are ways to solve the problem without getting a cubic equation. But we can’t really know that we’re not going down the best path until we get in trouble. Finding a better way takes experience, so I won’t fault you for not finding one on the first try.
Simplifying from the start
So, how can we solve it without getting the cubic and its three solutions? We’ve seen a couple; I saw another and jumped into the discussion to show it:
If I may add yet another suggestion, here is what I would have done (because I would not have thought of Doctor Fenton’s nice idea!): When I got to
I would see that (x-6) is in both denominators, and that 6 is not a valid solution, so I would immediately multiply both sides by (x-6), leaving
After cross-multiplying and simplifying, this leaves a quadratic (whose solution is easy, and correct).
Here is what happens if we cross-multiply now:
$$\frac{(x-3)(x-7)+(x-6)(x-8)}{x-7}=\frac{(x-7)(x-5)+(x-6)^2}{x-5}$$
$$\frac{(x^2-10x+21)+(x^2-14x+48)}{x-7}=\frac{(x^2-12x+35)+(x^2-12x+36)}{x-5}$$
$$\frac{2x^2-24x+69}{x-7}=\frac{2x^2-24x+71)}{x-5}$$
$$(2x^2-24x+69)(x-5)=(2x^2-24x+71)(x-7)$$
$$2x^3-34x^2+189x-345=2x^3-38x^2+239x-497$$
$$4x^2-50x+152=0$$
$$2x^2-25x+76=0$$
$$x=\frac{25\pm\sqrt{17}}{4}$$
We never saw a cubic, or an extraneous root.
Another similar way is to use the trick I included in my version of Doctor Fenton’s approach: We can see that two of the denominators are the same, and combine those fractions at the start. Here is a version of that approach:
$$\frac{x-3}{x-6}+\frac{x-8}{x-7}=\frac{x-7}{x-6}+\frac{x-6}{x-5}$$
Subtract one fraction from each side:
$$\frac{x-3}{x-6}-\frac{x-7}{x-6}=\frac{x-6}{x-5}-\frac{x-8}{x-7}$$
Combine each pair of fractions, using a common denominator (which is easy on the left this time):
$$\frac{(x-3)-(x-7)}{x-6}=\frac{(x-6)(x-7)-(x-8)(x-5)}{(x-5)(x-7)}$$
$$\frac{4}{x-6}=\frac{(x^2-13x+42)-(x^2-13x+40)}{(x-5)(x-7)}$$
$$\frac{4}{x-6}=\frac{2}{(x-5)(x-7)}$$
This simplified far more than we were expecting, due to the particular numbers! Now we can divide both sides by 2 to simplify the equation, and then cross-multiply:
$$\frac{2}{x-6}=\frac{1}{(x-5)(x-7)}$$
$$2(x-5)(x-7)=1(x-6)$$
$$2x^2-24x+70=x-6$$
$$2x^2-25x+76=0$$
As before, we get two solutions from the quadratic equation, and nothing extraneous.
Multiplying by the LCD to clear fractions
In general, looking for such simplifications may both prevent extraneous solutions, and make the work easier.
Another way, incidentally, is rather than cross-multiplying, to multiply both sides by the LCD. This will have similar results. The extraneous solution you got is a common effect of cross-multiplication, and therefore a reason to prefer the LCD.
Here is the work using the LCD:
$$\frac{x-3}{x-6}+\frac{x-8}{x-7}=\frac{x-7}{x-6}+\frac{x-6}{x-5}$$
The LCD is \((x-6)(x-7)(x-5)\); we don’t need two of the same factor \((x-6)\). (That’s what’s going to help!) Multiplying every term by this cancels each denominator and leaves
$$(x-3)(x-7)(x-5)+(x-8)(x-6)(x-5)=(x-7)(x-7)(x-5)+(x-6)(x-6)(x-7)$$
$$(x^2-10x+21)(x-5)+(x^2-14x+48)(x-5)=(x^2-14x+49)(x-5)+(x^2-12x+36)(x-7)$$
$$(x^3-15x^2+71x-105)+(x^3-19x^2+118x-240)=(x^3-19x^2+119x-245)+(x^3-19x^2+120x-252)$$
$$2x^3-34x^2+189x-345=2x^3-38x^2+239x-497$$
$$4x^2-50x+152=0$$
$$x=\frac{25\pm\sqrt{17}}{4}$$
We’ve seen several tricks that are specific to this problem, and other more generic ideas that may help with many rational equations. Have fun applying them!