Examples:
\(\frac(9x^2-1)(3x)\) \(\leq0\)
\(\frac(1)(2x)\) \(+\) \(\frac(x)(x+1)\) \(<\)\(\frac{1}{2}\)
\(\frac(6)(x+1)\) \(>\) \(\frac(x^2-5x)(x+1)\) .
When solving fractional rational inequalities, the method of intervals is used. Therefore, if the algorithm below causes you difficulties, see the article on .
How to solve fractional rational inequalities:
Algorithm for solving fractional rational inequalities.
- If $D \gt 0$, we get two different roots;
- If $D=0$, then the root will be one, but of the second multiplicity (what kind of multiplicity it is and how to take it into account - more on that later). Or we can say that the equation has two identical roots;
- For $D \lt 0$ there are no roots at all, and the sign of the polynomial $a((x)^(2))+bx+c$ for any $x$ coincides with the sign of the coefficient $a$. By the way, this is very useful fact, which for some reason they forget to talk about in algebra lessons.
- Factorization of the polynomial $P\left(x \right)$;
- Actually, bringing fractions to a common denominator.
- Factorize both denominators;
- Consider the first denominator and add to it the factors present in the second denominator, but not in the first. The resulting product will be the common denominator;
- Find out what factors each of the original fractions lacks so that the denominators become equal to the common one.
- Strict: $f\left(x \right) \gt 0$ or $f\left(x \right) \lt 0$;
- Nonstrict: $f\left(x \right)\ge 0$ or $f\left(x \right)\le 0$.
- Collect all non-zero elements on one side of the inequality sign. For example, on the left;
- Bring all fractions to a common denominator (if there are several such fractions), bring similar ones. Then, if possible, factorize into the numerator and denominator. One way or another, we will get an inequality of the form $\frac(P\left(x \right))(Q\left(x \right))\vee 0$, where the tick is the inequality sign.
- Equate the numerator to zero: $P\left(x \right)=0$. We solve this equation and get the roots $((x)_(1))$, $((x)_(2))$, $((x)_(3))$, ... Then we require that the denominator was not equal to zero: $Q\left(x \right)\ne 0$. Of course, in essence, we have to solve the equation $Q\left(x \right)=0$, and we get the roots $x_(1)^(*)$, $x_(2)^(*)$, $x_(3 )^(*)$, ... (in real problems there will hardly be more than three such roots).
- We mark all these roots (both with and without asterisks) on a single number line, and the roots without stars are painted over, and those with stars are punched out.
- We place the plus and minus signs, select the intervals that we need. If the inequality has the form $f\left(x \right) \gt 0$, then the answer will be the intervals marked with a "plus". If $f\left(x \right) \lt 0$, then we look at the intervals with "minuses".
- If $x=a$ is a root of even multiplicity, then the sign of the function does not change when passing through it;
- And vice versa, if $x=a$ is a root of odd multiplicity, then the sign of the function will change.
- The point $x=1$ has an even multiplicity, but is itself punctured. Therefore, it will have to be isolated in the answer: you need to write $x\in \left(-\infty ;1 \right)\bigcup \left(1;2 \right)$, and not $x\in \left(-\ infty ;2\right)$.
- The point $x=3$ also has an even multiplicity and is shaded. The arrangement of signs indicates that the point itself suits us, but a step to the left and right - and we find ourselves in an area that definitely does not suit us. Such points are called isolated and are written as $x\in \left\( 3 \right\)$.
- The interval $\left(-\infty ;1 \right)$, which literally means "all numbers less than one, but not one itself";
- The interval is $\left(1;2 \right)$, i.e. "all numbers between 1 and 2, but not the numbers 1 and 2 themselves";
- The set $\left\( 3 \right\)$, consisting of a single number - three;
- The interval $\left[ 4;5 \right)$ containing all numbers between 4 and 5, plus 4 itself, but not 5.
Examples:
Place signs on the intervals of the number axis. Let me remind you the rules for arranging signs:
We determine the sign in the rightmost interval - we take a number from this interval and substitute it into the inequality instead of x. After that, we determine the signs in brackets and the result of multiplying these signs;
Examples:
Highlight the spaces you want. If there is a separate root, then mark it with a flag so that you do not forget to include it in the answer (see example below).
Examples:
Write down in response the highlighted gaps and the roots marked with a flag (if any).
Examples:
Answer: \((-∞;-1)∪(-1;1,2]∪
Now let's complicate the task a little and consider not just polynomials, but the so-called rational fractions of the form:
where $P\left(x \right)$ and $Q\left(x \right)$ are the same polynomials of the form $((a)_(n))((x)^(n))+(( a)_(n-1))((x)^(n-1))+...+((a)_(0))$, or the product of such polynomials.
This will be a rational inequality. The fundamental point is the presence of the variable $x$ in the denominator. For example, here are rational inequalities:
\[\begin(align) & \frac(x-3)(x+7) \lt 0; \\ & \frac(\left(7x+1 \right)\left(11x+2 \right))(13x-4)\ge 0; \\ & \frac(3((x)^(2))+10x+3)(((\left(3-x \right))^(2))\left(4-((x)^( 2)) \right))\ge 0. \\ \end(align)\]
And this is not a rational, but the most common inequality, which is solved by the interval method:
\[\frac(((x)^(2))+6x+9)(5)\ge 0\]
Looking ahead, I’ll say right away: there are at least two ways to solve rational inequalities, but all of them in one way or another are reduced to the method of intervals already known to us. Therefore, before analyzing these methods, let's recall the old facts, otherwise there will be no sense from the new material.
What you already need to know
There are not many important facts. We really only need four.
Abbreviated multiplication formulas
Yes, yes: they will haunt us throughout the school math curriculum. And at the university too. There are quite a few of these formulas, but we only need the following:
\[\begin(align) & ((a)^(2))\pm 2ab+((b)^(2))=((\left(a\pm b \right))^(2)); \\ & ((a)^(2))-((b)^(2))=\left(a-b \right)\left(a+b \right); \\ & ((a)^(3))+((b)^(3))=\left(a+b \right)\left(((a)^(2))-ab+((b) ^(2))\right); \\ & ((a)^(3))-((b)^(3))=\left(a-b \right)\left(((a)^(2))+ab+((b)^( 2))\right). \\ \end(align)\]
Pay attention to the last two formulas - this is the sum and difference of cubes (and not the cube of the sum or difference!). They are easy to remember if you notice that the sign in the first bracket is the same as the sign in the original expression, and in the second bracket it is opposite to the sign in the original expression.
Linear equations
These are the most simple equations of the form $ax+b=0$, where $a$ and $b$ are regular numbers, and $a\ne 0$. This equation is easy to solve:
\[\begin(align) & ax+b=0; \\ &ax=-b; \\ & x=-\frac(b)(a). \\ \end(align)\]
I note that we have the right to divide by the coefficient $a$, because $a\ne 0$. This requirement is quite logical, since with $a=0$ we get this:
First, there is no $x$ variable in this equation. This, generally speaking, should not confuse us (this happens, say, in geometry, and quite often), but still we are no longer a linear equation.
Secondly, the solution of this equation depends solely on the coefficient $b$. If $b$ is also zero, then our equation is $0=0$. This equality is always true; hence $x$ is any number (usually written as $x\in \mathbb(R)$). If the coefficient $b$ is not zero, then the equality $b=0$ is never satisfied, i.e. no answers (written $x\in \varnothing $ and read "solution set is empty").
To avoid all these complexities, we simply assume $a\ne 0$, which does not in any way restrict us from further reflections.
Quadratic equations
Let me remind you that this is called a quadratic equation:
Here on the left is a polynomial of the second degree, and again $a\ne 0$ (otherwise, instead of quadratic equation we get linear). The following equations are solved through the discriminant:
The roots themselves are calculated according to the well-known formula:
\[((x)_(1,2))=\frac(-b\pm \sqrt(D))(2a)\]
Hence, by the way, the restrictions on the discriminant. Because the square root of negative number does not exist. Regarding the roots, many students have a terrible mess in their heads, so I specifically wrote down whole lesson: what is a root in algebra and how to calculate it - I highly recommend reading it. :)
Operations with rational fractions
Everything that was written above, you already know if you studied the method of intervals. But what we will analyze now has no analogues in the past - this is a completely new fact.
Definition. A rational fraction is an expression of the form
\[\frac(P\left(x \right))(Q\left(x \right))\]
where $P\left(x \right)$ and $Q\left(x \right)$ are polynomials.
It is obvious that it is easy to obtain an inequality from such a fraction - it is enough just to attribute the sign “greater than” or “less than” to the right. And a little further we will find that solving such problems is a pleasure, everything is very simple there.
Problems begin when there are several such fractions in one expression. They have to be brought to common denominator- and it is at this moment that a large number of offensive mistakes are made.
Therefore, for a successful solution rational equations Two skills must be firmly mastered:
How to factorize a polynomial? Very simple. Let we have a polynomial of the form
Let's equate it to zero. We get the $n$-th degree equation:
\[((a)_(n))((x)^(n))+((a)_(n-1))((x)^(n-1))+...+(( a)_(1))x+((a)_(0))=0\]
Let's say we solved this equation and got the roots $((x)_(1)),\ ...,\ ((x)_(n))$ (don't worry: in most cases there will be no more than two of these roots) . In this case, our original polynomial can be rewritten like this:
\[\begin(align) & P\left(x \right)=((a)_(n))((x)^(n))+((a)_(n-1))((x )^(n-1))+...+((a)_(1))x+((a)_(0))= \\ & =((a)_(n))\left(x -((x)_(1)) \right)\cdot \left(x-((x)_(2)) \right)\cdot ...\cdot \left(x-((x)_( n)) \right) \end(align)\]
That's all! Please note: the leading coefficient $((a)_(n))$ has not disappeared anywhere - it will be a separate factor in front of the brackets, and if necessary, it can be inserted into any of these brackets (practice shows that with $((a)_ (n))\ne \pm 1$ there are almost always fractions among the roots).
Task. Simplify the expression:
\[\frac(((x)^(2))+x-20)(x-4)-\frac(2((x)^(2))-5x+3)(2x-3)-\ frac(4-8x-5((x)^(2)))(x+2)\]
Decision. First, let's look at the denominators: they are all linear binomials, and there is nothing to factorize here. So let's factorize the numerators:
\[\begin(align) & ((x)^(2))+x-20=\left(x+5 \right)\left(x-4 \right); \\ & 2((x)^(2))-5x+3=2\left(x-\frac(3)(2) \right)\left(x-1 \right)=\left(2x- 3\right)\left(x-1\right); \\ & 4-8x-5((x)^(2))=-5\left(x+2 \right)\left(x-\frac(2)(5) \right)=\left(x +2 \right)\left(2-5x \right). \\\end(align)\]
Please note: in the second polynomial, the senior coefficient "2", in full accordance with our scheme, first appeared in front of the bracket, and then was included in the first bracket, since a fraction got out there.
The same thing happened in the third polynomial, only there the order of the terms is also confused. However, the coefficient “−5” ended up being included in the second bracket (remember: you can enter a factor in one and only one bracket!), which saved us from the inconvenience associated with fractional roots.
As for the first polynomial, everything is simple there: its roots are sought either in the standard way through the discriminant, or using the Vieta theorem.
Let's go back to the original expression and rewrite it with the numerators decomposed into factors:
\[\begin(matrix) \frac(\left(x+5 \right)\left(x-4 \right))(x-4)-\frac(\left(2x-3 \right)\left( x-1 \right))(2x-3)-\frac(\left(x+2 \right)\left(2-5x \right))(x+2)= \\ =\left(x+5 \right)-\left(x-1 \right)-\left(2-5x \right)= \\ =x+5-x+1-2+5x= \\ =5x+4. \\ \end(matrix)\]
Answer: $5x+4$.
As you can see, nothing complicated. A bit of 7th-8th grade math and that's it. The point of all transformations is to turn a complex and scary expression into something simple and easy to work with.
However, this will not always be the case. So now we will consider a more serious problem.
But first, let's figure out how to bring two fractions to a common denominator. The algorithm is extremely simple:
Perhaps this algorithm will seem to you just a text in which there are “a lot of letters”. So let's take a look at a specific example.
Task. Simplify the expression:
\[\left(\frac(x)(((x)^(2))+2x+4)+\frac(((x)^(2))+8)(((x)^(3) )-8)-\frac(1)(x-2) \right)\cdot \left(\frac(((x)^(2)))(((x)^(2))-4)- \frac(2)(2-x) \right)\]
Decision. Such voluminous tasks are best solved in parts. Let's write out what is in the first bracket:
\[\frac(x)(((x)^(2))+2x+4)+\frac(((x)^(2))+8)(((x)^(3))-8 )-\frac(1)(x-2)\]
Unlike the previous problem, here the denominators are not so simple. Let's factorize each of them.
The square trinomial $((x)^(2))+2x+4$ cannot be factorized because the equation $((x)^(2))+2x+4=0$ has no roots (the discriminant is negative). We leave it unchanged.
The second denominator, the cubic polynomial $((x)^(3))-8$, upon closer examination is the difference of cubes and can be easily decomposed using the abbreviated multiplication formulas:
\[((x)^(3))-8=((x)^(3))-((2)^(3))=\left(x-2 \right)\left(((x) ^(2))+2x+4 \right)\]
Nothing else can be factored, since the first bracket contains a linear binomial, and the second one is a construction already familiar to us, which has no real roots.
Finally, the third denominator is a linear binomial that cannot be decomposed. Thus, our equation will take the form:
\[\frac(x)(((x)^(2))+2x+4)+\frac(((x)^(2))+8)(\left(x-2 \right)\left (((x)^(2))+2x+4 \right))-\frac(1)(x-2)\]
It is quite obvious that $\left(x-2 \right)\left(((x)^(2))+2x+4 \right)$ will be the common denominator, and to reduce all fractions to it, you need to multiply the first fraction to $\left(x-2 \right)$, and the last one to $\left(((x)^(2))+2x+4 \right)$. Then it remains only to bring the following:
\[\begin(matrix) \frac(x\cdot \left(x-2 \right))(\left(x-2 \right)\left(((x)^(2))+2x+4 \ right))+\frac(((x)^(2))+8)(\left(x-2 \right)\left(((x)^(2))+2x+4 \right))- \frac(1\cdot \left(((x)^(2))+2x+4 \right))(\left(x-2 \right)\left(((x)^(2))+2x +4 \right))= \\ =\frac(x\cdot \left(x-2 \right)+\left(((x)^(2))+8 \right)-\left(((x )^(2))+2x+4 \right))(\left(x-2 \right)\left(((x)^(2))+2x+4 \right))= \\ =\frac (((x)^(2))-2x+((x)^(2))+8-((x)^(2))-2x-4)(\left(x-2 \right)\left (((x)^(2))+2x+4 \right))= \\ =\frac(((x)^(2))-4x+4)(\left(x-2 \right)\ left(((x)^(2))+2x+4 \right)). \\ \end(matrix)\]
Pay attention to the second line: when the denominator is already common, i.e. instead of three separate fractions, we wrote one large one, you should not immediately get rid of the brackets. It is better to write an extra line and note that, say, there was a minus before the third fraction - and it will not go anywhere, but will “hang” in the numerator in front of the bracket. This will save you a lot of mistakes.
Well, in the last line it is useful to factorize the numerator. Moreover, this is an exact square, and the abbreviated multiplication formulas again come to our aid. We have:
\[\frac(((x)^(2))-4x+4)(\left(x-2 \right)\left(((x)^(2))+2x+4 \right))= \frac(((\left(x-2 \right))^(2)))(\left(x-2 \right)\left(((x)^(2))+2x+4 \right) )=\frac(x-2)(((x)^(2))+2x+4)\]
Now let's deal with the second bracket in the same way. Here I will simply write a chain of equalities:
\[\begin(matrix) \frac(((x)^(2)))(((x)^(2))-4)-\frac(2)(2-x)=\frac((( x)^(2)))(\left(x-2 \right)\left(x+2 \right))-\frac(2)(2-x)= \\ =\frac(((x) ^(2)))(\left(x-2 \right)\left(x+2 \right))+\frac(2)(x-2)= \\ =\frac(((x)^( 2)))(\left(x-2 \right)\left(x+2 \right))+\frac(2\cdot \left(x+2 \right))(\left(x-2 \right )\cdot \left(x+2 \right))= \\ =\frac(((x)^(2))+2\cdot \left(x+2 \right))(\left(x-2 \right)\left(x+2 \right))=\frac(((x)^(2))+2x+4)(\left(x-2 \right)\left(x+2 \right) ). \\ \end(matrix)\]
We return to the original problem and look at the product:
\[\frac(x-2)(((x)^(2))+2x+4)\cdot \frac(((x)^(2))+2x+4)(\left(x-2 \right)\left(x+2 \right))=\frac(1)(x+2)\]
Answer: \[\frac(1)(x+2)\].
The meaning of this problem is the same as the previous one: to show how much rational expressions can be simplified if you approach their transformation wisely.
And now, when you know all this, let's move on to the main topic of today's lesson - solving fractional rational inequalities. Moreover, after such preparation, the inequalities themselves will click like nuts. :)
The main way to solve rational inequalities
There are at least two approaches to solving rational inequalities. Now we will consider one of them - the one that is generally accepted in the school mathematics course.
But first, let's note important detail. All inequalities are divided into two types:
Inequalities of the second type are easily reduced to the first, as well as the equation:
This small "addition" $f\left(x \right)=0$ leads to such an unpleasant thing as filled points - we met them back in the interval method. Otherwise, there are no differences between strict and non-strict inequalities, so let's analyze the universal algorithm:
Practice shows that points 2 and 4 cause the greatest difficulties - competent transformations and the correct arrangement of numbers in ascending order. Well, at the last step, be extremely careful: we always place signs based on the last inequality written before moving on to the equations. This is a universal rule inherited from the interval method.
So, there is a scheme. Let's practice.
Task. Solve the inequality:
\[\frac(x-3)(x+7) \lt 0\]
Decision. We have a strict inequality of the form $f\left(x \right) \lt 0$. Obviously, points 1 and 2 from our scheme have already been completed: all the elements of the inequality are collected on the left, nothing needs to be brought to a common denominator. So let's move on to the third point.
Set the numerator to zero:
\[\begin(align) & x-3=0; \\ &x=3. \end(align)\]
And the denominator:
\[\begin(align) & x+7=0; \\ & ((x)^(*))=-7. \\ \end(align)\]
In this place, many people get stuck, because in theory you need to write down $x+7\ne 0$, as required by the ODZ (you can’t divide by zero, that’s all). But after all, in the future we will poke out the points that came from the denominator, so you should not complicate your calculations once again - write an equal sign everywhere and don’t worry. No one will deduct points for this. :)
Fourth point. We mark the obtained roots on the number line:
All points are punctured because the inequality is strict
Note: all points are punctured because the original inequality is strict. And here it doesn’t matter anymore: these points came from the numerator or from the denominator.
Well, look at the signs. Take any number $((x)_(0)) \gt 3$. For example, $((x)_(0))=100$ (but you could have just as well taken $((x)_(0))=3.1$ or $((x)_(0)) =1\000\000$). We get:
So, to the right of all the roots we have a positive area. And when passing through each root, the sign changes (this will not always be the case, but more on that later). Therefore, we proceed to the fifth point: we place the signs and choose the right one:
We return to the last inequality, which was before solving the equations. Actually, it coincides with the original one, because we did not perform any transformations in this task.
Since it is necessary to solve an inequality of the form $f\left(x \right) \lt 0$, I shaded the interval $x\in \left(-7;3 \right)$ - it is the only one marked with a minus sign. This is the answer.
Answer: $x\in \left(-7;3 \right)$
That's all! Is it difficult? No, it's not difficult. Indeed, it was an easy task. Now let's complicate the mission a little and consider a more "fancy" inequality. When solving it, I will no longer give such detailed calculations - I will simply indicate key points. In general, we will arrange it the way we would have done it on an independent work or exam. :)
Task. Solve the inequality:
\[\frac(\left(7x+1 \right)\left(11x+2 \right))(13x-4)\ge 0\]
Decision. This is a non-strict inequality of the form $f\left(x \right)\ge 0$. All non-zero elements are collected on the left, different denominators no. Let's move on to equations.
Numerator:
\[\begin(align) & \left(7x+1 \right)\left(11x+2 \right)=0 \\ & 7x+1=0\Rightarrow ((x)_(1))=-\ frac(1)(7); \\ & 11x+2=0\Rightarrow ((x)_(2))=-\frac(2)(11). \\ \end(align)\]
Denominator:
\[\begin(align) & 13x-4=0; \\ & 13x=4; \\ & ((x)^(*))=\frac(4)(13). \\ \end(align)\]
I don’t know what kind of pervert made up this problem, but the roots didn’t turn out very well: it will be difficult to arrange them on a number line. And if everything is more or less clear with the root $((x)^(*))=(4)/(13)\;$ (this is the only positive number - it will be on the right), then $((x)_(1 ))=-(1)/(7)\;$ and $((x)_(2))=-(2)/(11)\;$ require further study: which one is larger?
You can find this out, for example:
\[((x)_(1))=-\frac(1)(7)=-\frac(2)(14) \gt -\frac(2)(11)=((x)_(2 ))\]
I hope there is no need to explain why the numeric fraction $-(2)/(14)\; \gt -(2)/(11)\;$? If necessary, I recommend remembering how to perform actions with fractions.
And we mark all three roots on the number line:
The points from the numerator are shaded, from the denominator they are cut out
We put up signs. For example, you can take $((x)_(0))=1$ and find out the sign at this point:
\[\begin(align) & f\left(x \right)=\frac(\left(7x+1 \right)\left(11x+2 \right))(13x-4); \\ & f\left(1 \right)=\frac(\left(7\cdot 1+1 \right)\left(11\cdot 1+2 \right))(13\cdot 1-4)=\ frac(8\cdot 13)(9) \gt 0. \\\end(align)\]
The last inequality before the equations was $f\left(x \right)\ge 0$, so we are interested in the plus sign.
We got two sets: one is an ordinary segment, and the other is an open ray on the number line.
Answer: $x\in \left[ -\frac(2)(11);-\frac(1)(7) \right]\bigcup \left(\frac(4)(13);+\infty \right )$
An important note about the numbers that we substitute to find out the sign on the rightmost interval. It is not necessary to substitute a number close to the rightmost root. You can take billions or even "plus-infinity" - in this case, the sign of the polynomial in the bracket, numerator or denominator is determined solely by the sign of the leading coefficient.
Let's take another look at the $f\left(x \right)$ function from the last inequality:
It contains three polynomials:
\[\begin(align) & ((P)_(1))\left(x \right)=7x+1; \\ & ((P)_(2))\left(x \right)=11x+2; \\ & Q\left(x\right)=13x-4. \end(align)\]
All of them are linear binomials, and all of them have positive coefficients (numbers 7, 11 and 13). Therefore, when substituting very big numbers the polynomials themselves will also be positive. :)
This rule may seem overly complicated, but only at first, when we analyze very easy tasks. In serious inequalities, the "plus-infinity" substitution will allow us to figure out the signs much faster than the standard $((x)_(0))=100$.
We will face such challenges very soon. But first, let's look at an alternative way to solve fractional rational inequalities.
Alternative way
This technique was suggested to me by one of my students. I myself have never used it, but practice has shown that it is really more convenient for many students to solve inequalities in this way.
So, the original data is the same. Need to decide fractional rational inequality:
\[\frac(P\left(x \right))(Q\left(x \right)) \gt 0\]
Let's think: why is the polynomial $Q\left(x \right)$ "worse" than the polynomial $P\left(x \right)$? Why do we have to consider individual groups roots (with and without an asterisk), think about punched points, etc.? It's simple: a fraction has a domain of definition, according to which the fraction makes sense only when its denominator is different from zero.
Otherwise, there are no differences between the numerator and the denominator: we also equate it to zero, look for the roots, then mark them on the number line. So why not replace the fractional bar (in fact, the division sign) with the usual multiplication, and write all the requirements of the DHS as a separate inequality? For example, like this:
\[\frac(P\left(x \right))(Q\left(x \right)) \gt 0\Rightarrow \left\( \begin(align) & P\left(x \right)\cdot Q \left(x \right) \gt 0, \\ & Q\left(x \right)\ne 0. \\ \end(align) \right.\]
Please note: this approach will allow you to reduce the problem to the method of intervals, but it will not complicate the solution at all. After all, anyway, we will equate the polynomial $Q\left(x \right)$ to zero.
Let's see how it works on real tasks.
Task. Solve the inequality:
\[\frac(x+8)(x-11) \gt 0\]
Decision. So, let's move on to the interval method:
\[\frac(x+8)(x-11) \gt 0\Rightarrow \left\( \begin(align) & \left(x+8 \right)\left(x-11 \right) \gt 0 , \\ & x-11\ne 0. \\ \end(align) \right.\]
The first inequality is solved elementarily. Just set each parenthesis to zero:
\[\begin(align) & x+8=0\Rightarrow ((x)_(1))=-8; \\ & x-11=0\Rightarrow ((x)_(2))=11. \\ \end(align)\]
With the second inequality, everything is also simple:
We mark the points $((x)_(1))$ and $((x)_(2))$ on the real line. All of them are punctured because the inequality is strict:
The right point turned out to be punctured twice. This is fine.Pay attention to the point $x=11$. It turns out that it is “twice gouged out”: on the one hand, we gouge it out because of the severity of inequality, on the other hand, because of the additional requirement of ODZ.
In any case, it will be just a punctured point. Therefore, we put signs for the inequality $\left(x+8 \right)\left(x-11 \right) \gt 0$ - the last one we saw before we started solving the equations:
We are interested in positive regions, since we are solving an inequality of the form $f\left(x \right) \gt 0$, and we will color them. It remains only to write down the answer.
Answer. $x\in \left(-\infty ;-8 \right)\bigcup \left(11;+\infty \right)$
Using this solution as an example, I would like to warn you against a common mistake among novice students. Namely: never open parentheses in inequalities! On the contrary, try to factor everything - this will simplify the solution and save you a lot of problems.
Now let's try something more difficult.
Task. Solve the inequality:
\[\frac(\left(2x-13 \right)\left(12x-9 \right))(15x+33)\le 0\]
Decision. This is a non-strict inequality of the form $f\left(x \right)\le 0$, so here you need to carefully monitor the filled points.
Let's move on to the interval method:
\[\left\( \begin(align) & \left(2x-13 \right)\left(12x-9 \right)\left(15x+33 \right)\le 0, \\ & 15x+33\ ne 0. \\ \end(align) \right.\]
Let's move on to the equation:
\[\begin(align) & \left(2x-13 \right)\left(12x-9 \right)\left(15x+33 \right)=0 \\ & 2x-13=0\Rightarrow ((x )_(1))=6.5; \\ & 12x-9=0\Rightarrow ((x)_(2))=0.75; \\ & 15x+33=0\Rightarrow ((x)_(3))=-2,2. \\ \end(align)\]
We take into account the additional requirement:
We mark all the obtained roots on the number line:
If a point is both punched out and filled in at the same time, it is considered punched out.Again, two points "overlap" each other - this is normal, it will always be so. It is only important to understand that a point marked both as punched out and filled in is actually a punched out point. Those. "poking out" - more strong action than "painting".
This is absolutely logical, because by puncturing we mark points that affect the sign of the function, but do not themselves participate in the answer. And if at some point the number ceases to suit us (for example, it does not fall into the ODZ), we delete it from consideration until the very end of the task.
In general, stop philosophizing. We arrange the signs and paint over those intervals that are marked with a minus sign:
Answer. $x\in \left(-\infty ;-2,2 \right)\bigcup \left[ 0,75;6,5 \right]$.
And again I wanted to draw your attention to this equation:
\[\left(2x-13 \right)\left(12x-9 \right)\left(15x+33 \right)=0\]
Once again: never open parentheses in such equations! You're only making it harder for yourself. Remember: the product is zero when at least one of the factors is zero. Consequently, this equation simply “falls apart” into several smaller ones, which we solved in the previous problem.
Taking into account the multiplicity of roots
From the previous problems, it is easy to see that it is the non-strict inequalities that are the most difficult, because in them you have to keep track of the filled points.
But there is an even greater evil in the world - these are multiple roots in inequalities. Here it is already necessary to follow not some filled points there - here the inequality sign may not suddenly change when passing through these same points.
We have not yet considered anything like this in this lesson (although a similar problem was often encountered in the interval method). So let's introduce a new definition:
Definition. The root of the equation $((\left(x-a \right))^(n))=0$ is equal to $x=a$ and is called the root of the $n$th multiplicity.
Actually, we are not particularly interested in the exact value of the multiplicity. The only important thing is whether this very number $n$ is even or odd. Because:
A special case of a root of odd multiplicity are all the previous problems considered in this lesson: there the multiplicity is equal to one everywhere.
And further. Before we start solving problems, I would like to draw your attention to one subtlety that seems obvious to an experienced student, but drives many beginners into a stupor. Namely:
The multiplicity root $n$ occurs only when the entire expression is raised to this power: $((\left(x-a \right))^(n))$, and not $\left(((x)^( n))-a\right)$.
Once again: the bracket $((\left(x-a \right))^(n))$ gives us the root $x=a$ of multiplicity $n$, but the bracket $\left(((x)^(n)) -a \right)$ or, as often happens, $(a-((x)^(n)))$ gives us a root (or two roots, if $n$ is even) of the first multiplicity, no matter what is equal to $n$.
Compare:
\[((\left(x-3 \right))^(5))=0\Rightarrow x=3\left(5k \right)\]
Everything is clear here: the whole bracket was raised to the fifth power, so at the output we got the root of the fifth degree. And now:
\[\left(((x)^(2))-4 \right)=0\Rightarrow ((x)^(2))=4\Rightarrow x=\pm 2\]
We got two roots, but both of them have the first multiplicity. Or here's another one:
\[\left(((x)^(10))-1024 \right)=0\Rightarrow ((x)^(10))=1024\Rightarrow x=\pm 2\]
And do not be confused by the tenth degree. The main thing is that 10 is an even number, so we have two roots at the output, and both of them again have the first multiplicity.
In general, be careful: multiplicity occurs only when the degree applies to the entire bracket, not just the variable.
Task. Solve the inequality:
\[\frac(((x)^(2))((\left(6-x \right))^(3))\left(x+4 \right))(((\left(x+7 \right))^(5)))\ge 0\]
Decision. Let's try to solve it alternative way- through the transition from the particular to the product:
\[\left\( \begin(align) & ((x)^(2))((\left(6-x \right))^(3))\left(x+4 \right)\cdot ( (\left(x+7 \right))^(5))\ge 0, \\ & ((\left(x+7 \right))^(5))\ne 0. \\ \end(align )\right.\]
We deal with the first inequality using the interval method:
\[\begin(align) & ((x)^(2))((\left(6-x \right))^(3))\left(x+4 \right)\cdot ((\left( x+7 \right))^(5))=0; \\ & ((x)^(2))=0\Rightarrow x=0\left(2k \right); \\ & ((\left(6-x \right))^(3))=0\Rightarrow x=6\left(3k \right); \\ & x+4=0\Rightarrow x=-4; \\ & ((\left(x+7 \right))^(5))=0\Rightarrow x=-7\left(5k \right). \\ \end(align)\]
Additionally, we solve the second inequality. In fact, we have already solved it, but so that the reviewers do not find fault with the solution, it is better to solve it again:
\[((\left(x+7 \right))^(5))\ne 0\Rightarrow x\ne -7\]
Note that there are no multiplicities in the last inequality. Indeed: what difference does it make how many times to cross out the point $x=-7$ on the number line? At least once, at least five times - the result will be the same: a punctured point.
Let's note everything that we got on the number line:
As I said, the $x=-7$ point will eventually be punched out. The multiplicities are arranged based on the solution of the inequality by the interval method.
It remains to place the signs:
Since the point $x=0$ is a root of even multiplicity, the sign does not change when passing through it. The remaining points have an odd multiplicity, and everything is simple with them.
Answer. $x\in \left(-\infty ;-7 \right)\bigcup \left[ -4;6 \right]$
Pay attention to $x=0$ again. Because of the even multiplicity, an interesting effect arises: everything to the left of it is painted over, to the right - too, and the point itself is completely painted over.
As a consequence, it does not need to be isolated when recording a response. Those. you don't have to write something like $x\in \left[ -4;0 \right]\bigcup \left[ 0;6 \right]$ (although formally such an answer would also be correct). Instead, we immediately write $x\in \left[ -4;6 \right]$.
Such effects are possible only for roots of even multiplicity. And in the next task, we will encounter the reverse "manifestation" of this effect. Ready?
Task. Solve the inequality:
\[\frac(((\left(x-3 \right))^(4))\left(x-4 \right))(((\left(x-1 \right))^(2)) \left(7x-10-((x)^(2)) \right))\ge 0\]
Decision. This time we will follow the standard scheme. Set the numerator to zero:
\[\begin(align) & ((\left(x-3 \right))^(4))\left(x-4 \right)=0; \\ & ((\left(x-3 \right))^(4))=0\Rightarrow ((x)_(1))=3\left(4k \right); \\ & x-4=0\Rightarrow ((x)_(2))=4. \\ \end(align)\]
And the denominator:
\[\begin(align) & ((\left(x-1 \right))^(2))\left(7x-10-((x)^(2)) \right)=0; \\ & ((\left(x-1 \right))^(2))=0\Rightarrow x_(1)^(*)=1\left(2k \right); \\ & 7x-10-((x)^(2))=0\Rightarrow x_(2)^(*)=5;\ x_(3)^(*)=2. \\ \end(align)\]
Since we are solving a non-strict inequality of the form $f\left(x \right)\ge 0$, the roots from the denominator (which have asterisks) will be cut out, and those from the numerator will be painted over.
We arrange the signs and stroke the areas marked with a "plus":
The point $x=3$ is isolated. This is part of the answer
Before writing down the final answer, take a close look at the picture:
We combine all the obtained pieces into a common set and write down the answer.
Answer: $x\in \left(-\infty ;1 \right)\bigcup \left(1;2 \right)\bigcup \left\( 3 \right\)\bigcup \left[ 4;5 \right) $
Definition. Solving the inequality means find the set of all its solutions, or prove that this set is empty.
It would seem: what can be incomprehensible here? Yes, the fact of the matter is that sets can be specified in different ways. Let's rewrite the answer to the last problem:
We literally read what is written. The variable "x" belongs to a certain set, which is obtained by the union (symbol "U") of four separate sets:
The third point is of interest here. Unlike intervals, which define infinite sets of numbers and only denote only the boundaries of these sets, the set $\left\( 3 \right\)$ defines exactly one number by enumeration.
To understand that we are listing the specific numbers included in the set (and not setting boundaries or anything else), curly braces are used. For example, the notation $\left\( 1;2 \right\)$ means exactly "a set consisting of two numbers: 1 and 2", but not a segment from 1 to 2. In no case do not confuse these concepts.
Multiplicity addition rule
Well, at the end of today's lesson, a little tin from Pavel Berdov. :)
Attentive students have probably already wondered: what will happen if the numerator and denominator contain identical roots? So the following rule works:
Multiplicities of identical roots are added. Always. Even if this root occurs in both the numerator and the denominator.
Sometimes it's better to decide than to talk. Therefore, we solve the following problem:
Task. Solve the inequality:
\[\frac(((x)^(2))+6x+8)(\left(((x)^(2))-16 \right)\left(((x)^(2))+ 9x+14 \right))\ge 0\]
\[\begin(align) & ((x)^(2))+6x+8=0 \\ & ((x)_(1))=-2;\ ((x)_(2))= -4. \\ \end(align)\]
So far, nothing special. Set the denominator to zero:
\[\begin(align) & \left(((x)^(2))-16 \right)\left(((x)^(2))+9x+14 \right)=0 \\ & ( (x)^(2))-16=0\Rightarrow x_(1)^(*)=4;\ x_(2)^(*)=-4; \\ & ((x)^(2))+9x+14=0\Rightarrow x_(3)^(*)=-7;\ x_(4)^(*)=-2. \\ \end(align)\]
Two identical roots are found: $((x)_(1))=-2$ and $x_(4)^(*)=-2$. Both have the first multiplicity. Therefore, we replace them with one root $x_(4)^(*)=-2$, but with a multiplicity of 1+1=2.
In addition, there are also identical roots: $((x)_(2))=-4$ and $x_(2)^(*)=-4$. They are also of the first multiplicity, so only $x_(2)^(*)=-4$ of multiplicity 1+1=2 remains.
Please note: in both cases, we left exactly the “cut out” root, and threw out the “painted over” one from consideration. Because even at the beginning of the lesson, we agreed: if a point is both punched out and painted over at the same time, then we still consider it punched out.
As a result, we have four roots, and all of them turned out to be gouged out:
\[\begin(align) & x_(1)^(*)=4; \\ & x_(2)^(*)=-4\left(2k \right); \\ & x_(3)^(*)=-7; \\ & x_(4)^(*)=-2\left(2k \right). \\ \end(align)\]
We mark them on the number line, taking into account the multiplicity:
We place the signs and paint over the areas of interest to us:
Everything. No isolated points and other perversions. You can write down the answer.
Answer. $x\in \left(-\infty ;-7 \right)\bigcup \left(4;+\infty \right)$.
multiplication rule
Sometimes an even more unpleasant situation occurs: an equation that has multiple roots is itself raised to a certain power. This changes the multiplicities of all the original roots.
This is rare, so most students do not have experience in solving such problems. And the rule here is:
When an equation is raised to a power $n$, the multiplicity of all its roots also increases by a factor of $n$.
In other words, raising to a power results in multiplying multiplicities by the same power. Let's take this rule as an example:
Task. Solve the inequality:
\[\frac(x((\left(((x)^(2))-6x+9 \right))^(2))((\left(x-4 \right))^(5)) )(((\left(2-x \right))^(3))((\left(x-1 \right))^(2)))\le 0\]
Decision. Set the numerator to zero:
The product is equal to zero when at least one of the factors is equal to zero. Everything is clear with the first multiplier: $x=0$. And here's where the problems start:
\[\begin(align) & ((\left(((x)^(2))-6x+9 \right))^(2))=0; \\ & ((x)^(2))-6x+9=0\left(2k \right); \\ & D=((6)^(3))-4\cdot 9=0 \\ & ((x)_(2))=3\left(2k \right)\left(2k \right) \ \ & ((x)_(2))=3\left(4k \right) \\ \end(align)\]
As you can see, the equation $((x)^(2))-6x+9=0$ has a unique root of the second multiplicity: $x=3$. The whole equation is then squared. Therefore, the multiplicity of the root will be $2\cdot 2=4$, which we finally wrote down.
\[((\left(x-4 \right))^(5))=0\Rightarrow x=4\left(5k \right)\]
No problem with the denominator either:
\[\begin(align) & ((\left(2-x \right))^(3))((\left(x-1 \right))^(2))=0; \\ & ((\left(2-x \right))^(3))=0\Rightarrow x_(1)^(*)=2\left(3k \right); \\ & ((\left(x-1 \right))^(2))=0\Rightarrow x_(2)^(*)=1\left(2k \right). \\ \end(align)\]
In total, we got five points: two punched out and three filled in. There are no coinciding roots in the numerator and denominator, so we just mark them on the number line:
We arrange the signs taking into account the multiplicities and paint over the intervals of interest to us:
Again one isolated point and one punctured
Because of the roots of even multiplicity, we again received a couple of “non-standard” elements. This is $x\in \left[ 0;1 \right)\bigcup \left(1;2 \right)$, not $x\in \left[ 0;2 \right)$, and also an isolated point $ x\in \left\( 3 \right\)$.
Answer. $x\in \left[ 0;1 \right)\bigcup \left(1;2 \right)\bigcup \left\( 3 \right\)\bigcup \left[ 4;+\infty \right)$
As you can see, everything is not so difficult. The main thing is attentiveness. The last section of this lesson is devoted to transformations - the very ones that we discussed at the very beginning.
Preconversions
The inequalities we will discuss in this section are not complex. However, unlike the previous tasks, here you will have to apply skills from the theory of rational fractions - factorization and reduction to a common denominator.
We discussed this issue in detail at the very beginning of today's lesson. If you are not sure that you understand what it is about, I strongly recommend that you go back and repeat. Because there is no point in cramming the methods for solving inequalities if you "swim" in the conversion of fractions.
AT homework By the way, there will also be many similar tasks. They are placed in a separate subsection. And there you will find very non-trivial examples. But this will be in the homework, but now let's analyze a couple of such inequalities.
Task. Solve the inequality:
\[\frac(x)(x-1)\le \frac(x-2)(x)\]
Decision. Moving everything to the left:
\[\frac(x)(x-1)-\frac(x-2)(x)\le 0\]
We reduce to a common denominator, open the brackets, give like terms in the numerator:
\[\begin(align) & \frac(x\cdot x)(\left(x-1 \right)\cdot x)-\frac(\left(x-2 \right)\left(x-1 \ right))(x\cdot \left(x-1 \right))\le 0; \\ & \frac(((x)^(2))-\left(((x)^(2))-2x-x+2 \right))(x\left(x-1 \right)) \le0; \\ & \frac(((x)^(2))-((x)^(2))+3x-2)(x\left(x-1 \right))\le 0; \\ & \frac(3x-2)(x\left(x-1 \right))\le 0. \\\end(align)\]
Now we have a classical fractional rational inequality, the solution of which is no longer difficult. I propose to solve it by an alternative method - through the method of intervals:
\[\begin(align) & \left(3x-2 \right)\cdot x\cdot \left(x-1 \right)=0; \\ & ((x)_(1))=\frac(2)(3);\ ((x)_(2))=0;\ ((x)_(3))=1. \\ \end(align)\]
Don't forget the constraint that comes from the denominator:
We mark all the numbers and restrictions on the number line:
All roots have first multiplicity. No problem. We just place the signs and paint over the areas we need:
It's all. You can write down the answer.
Answer. $x\in \left(-\infty ;0 \right)\bigcup \left[ (2)/(3)\;;1 \right)$.
Of course, this was a very simple example. So now let's take a closer look at the problem. And by the way, the level of this task is quite consistent with independent and control work on this topic in 8th grade.
Task. Solve the inequality:
\[\frac(1)(((x)^(2))+8x-9)\ge \frac(1)(3((x)^(2))-5x+2)\]
Decision. Moving everything to the left:
\[\frac(1)(((x)^(2))+8x-9)-\frac(1)(3((x)^(2))-5x+2)\ge 0\]
Before bringing both fractions to a common denominator, we decompose these denominators into factors. Suddenly the same brackets will come out? With the first denominator it's easy:
\[((x)^(2))+8x-9=\left(x-1 \right)\left(x+9 \right)\]
The second one is a little more difficult. Feel free to add a constant multiplier to the bracket where the fraction was found. Remember: the original polynomial had integer coefficients, so it is highly likely that the factorization will also have integer coefficients (in fact, it always will, except when the discriminant is irrational).
\[\begin(align) & 3((x)^(2))-5x+2=3\left(x-1 \right)\left(x-\frac(2)(3) \right)= \\ & =\left(x-1 \right)\left(3x-2 \right) \end(align)\]
As you can see, there is a common bracket: $\left(x-1 \right)$. We return to the inequality and bring both fractions to a common denominator:
\[\begin(align) & \frac(1)(\left(x-1 \right)\left(x+9 \right))-\frac(1)(\left(x-1 \right)\ left(3x-2\right))\ge 0; \\ & \frac(1\cdot \left(3x-2 \right)-1\cdot \left(x+9 \right))(\left(x-1 \right)\left(x+9 \right )\left(3x-2 \right))\ge 0; \\ & \frac(3x-2-x-9)(\left(x-1 \right)\left(x+9 \right)\left(3x-2 \right))\ge 0; \\ & \frac(2x-11)(\left(x-1 \right)\left(x+9 \right)\left(3x-2 \right))\ge 0; \\ \end(align)\]
Set the denominator to zero:
\[\begin(align) & \left(x-1 \right)\left(x+9 \right)\left(3x-2 \right)=0; \\ & x_(1)^(*)=1;\ x_(2)^(*)=-9;\ x_(3)^(*)=\frac(2)(3) \\ \end( align)\]
No multiplicities and no coinciding roots. We mark four numbers on a straight line:
We place the signs:
We write down the answer.
Answer: $x\in \left(-\infty ;-9 \right)\bigcup \left((2)/(3)\;;1 \right)\bigcup \left[ 5,5;+\infty \ right)$.
