Let's work with quadratic equations. These are very popular equations! In its most general form, a quadratic equation looks like this:

For example:

Here A =1; b = 3; c = -4

Here A =2; b = -0,5; c = 2,2

Here A =-3; b = 6; c = -18

Well, you understand...

How to solve quadratic equations? If you have a quadratic equation in front of you in this form, then everything is simple. Let's remember Magic word discriminant . Rarely a high school student has not heard this word! The phrase “we solve through a discriminant” inspires confidence and reassurance. Because there is no need to expect tricks from the discriminant! It is simple and trouble-free to use. So, the formula for finding the roots of a quadratic equation looks like this:

The expression under the sign of the root is the one discriminant. As you can see, to find X, we use only a, b and c. Those. coefficients from a quadratic equation. Just carefully substitute the values a, b and c This is the formula we calculate. Let's substitute with your own signs! For example, for the first equation A =1; b = 3; c= -4. Here we write it down:

The example is almost solved:

That's all.

What cases are possible when using this formula? There are only three cases.

1. The discriminant is positive. This means the root can be extracted from it. Whether the root is extracted well or poorly is another question. What is important is what is extracted in principle. Then your quadratic equation has two roots. Two different solutions.

2. The discriminant is zero. Then you have one solution. Strictly speaking, this is not one root, but two identical. But this plays a role in inequalities, where we will study the issue in more detail.

3. The discriminant is negative. The square root of a negative number cannot be taken. Well, okay. This means there are no solutions.

Everything is very simple. And what, you think it’s impossible to make a mistake? Well, yes, how...
The most common mistakes are confusion with sign values a, b and c. Or rather, not with their signs (where to get confused?), but with the substitution of negative values ​​into the formula for calculating the roots. What helps here is a detailed recording of the formula with specific numbers. If there are problems with calculations, do that!

Suppose we need to solve the following example:

Here a = -6; b = -5; c = -1

Let's say you know that you rarely get answers the first time.

Well, don't be lazy. It will take about 30 seconds to write an extra line. And the number of errors will decrease sharply. So we write in detail, with all the brackets and signs:

It seems incredibly difficult to write out so carefully. But it only seems so. Give it a try. Well, or choose. What's better, fast or right? Besides, I will make you happy. After a while, there will be no need to write everything down so carefully. It will work out right on its own. Especially if you use practical techniques that are described below. This evil example with a bunch of minuses can be solved easily and without errors!

So, how to solve quadratic equations through the discriminant we remembered. Or they learned, which is also good. You know how to correctly determine a, b and c. Do you know how? attentively substitute them into the root formula and attentively count the result. Did you understand that keyword Here - attentively?

However, quadratic equations often look slightly different. For example, like this:

This incomplete quadratic equations . They can also be solved through a discriminant. You just need to understand correctly what they are equal to here. a, b and c.

Have you figured it out? In the first example a = 1; b = -4; A c? It's not there at all! Well yes, that's right. In mathematics this means that c = 0 ! That's all. Substitute zero into the formula instead c, and we will succeed. Same with the second example. Only we don’t have zero here With, A b !

But incomplete quadratic equations can be solved much more simply. Without any discrimination. Let's consider the first incomplete equation. What can you do on the left side? You can take X out of brackets! Let's take it out.

And what from this? And the fact that the product equals zero if and only if any of the factors equals zero! Don't believe me? Okay, then come up with two non-zero numbers that, when multiplied, will give zero!
Does not work? That's it...
Therefore, we can confidently write: x = 0, or x = 4

All. These will be the roots of our equation. Both are suitable. When substituting any of them into the original equation, we get the correct identity 0 = 0. As you can see, the solution is much simpler than using a discriminant.

The second equation can also be solved simply. Move 9 to the right side. We get:

All that remains is to extract the root from 9, and that’s it. It will turn out:

Also two roots . x = +3 and x = -3.

This is how all incomplete quadratic equations are solved. Either by placing X out of brackets, or by simply moving the number to the right and then extracting the root.
It is extremely difficult to confuse these techniques. Simply because in the first case you will have to extract the root of X, which is somehow incomprehensible, and in the second case there is nothing to take out of brackets...

Now take note of practical techniques that dramatically reduce the number of errors. The same ones that are due to inattention... For which it later becomes painful and offensive...

First appointment. Don’t be lazy before solving a quadratic equation and bring it to standard form. What does this mean?
Let's say that after all the transformations you get the following equation:

Don't rush to write the root formula! You'll almost certainly get the odds mixed up a, b and c. Construct the example correctly. First, X squared, then without square, then the free term. Like this:

And again, don’t rush! A minus in front of an X squared can really upset you. It's easy to forget... Get rid of the minus. How? Yes, as taught in the previous topic! We need to multiply the entire equation by -1. We get:

But now you can safely write down the formula for the roots, calculate the discriminant and finish solving the example. Decide for yourself. You should now have roots 2 and -1.

Reception second. Check the roots! According to Vieta's theorem. Don't be afraid, I'll explain everything! Checking last thing the equation. Those. the one we used to write down the root formula. If (as in this example) the coefficient a = 1, checking the roots is easy. It is enough to multiply them. The result should be a free member, i.e. in our case -2. Please note, not 2, but -2! Free member with your sign . If it doesn’t work out, it means they’ve already screwed up somewhere. Look for the error. If it works, you need to add the roots. Last and final check. The coefficient should be b With opposite familiar. In our case -1+2 = +1. A coefficient b, which is before the X, is equal to -1. So, everything is correct!
It’s a pity that this is so simple only for examples where x squared is pure, with a coefficient a = 1. But at least check in such equations! All less mistakes will.

Reception third. If your equation has fractional coefficients, get rid of the fractions! Multiply the equation by a common denominator as described in the previous section. When working with fractions, errors keep creeping in for some reason...

By the way, I promised to simplify the evil example with a bunch of minuses. Please! Here he is.

In order not to get confused by the minuses, we multiply the equation by -1. We get:

That's all! Solving is a pleasure!

So, let's summarize the topic.

Practical advice:

1. Before solving, we bring the quadratic equation to standard form and build it Right.

2. If there is a negative coefficient in front of the X squared, we eliminate it by multiplying the entire equation by -1.

3. If the coefficients are fractional, we eliminate the fractions by multiplying the entire equation by the corresponding factor.

4. If x squared is pure, its coefficient equal to one, the solution can be easily verified using Vieta's theorem. Do it!

Fractional equations. ODZ.

We continue to master the equations. We already know how to work with linear and quadratic equations. The last view left - fractional equations. Or they are also called much more respectably - fractional rational equations. It is the same.

Fractional equations.

As the name implies, these equations necessarily contain fractions. But not just fractions, but fractions that have unknown in denominator. At least in one. For example:

Let me remind you that if the denominators are only numbers, these are linear equations.

How to decide fractional equations? First of all, get rid of fractions! After this, the equation most often turns into linear or quadratic. And then we know what to do... In some cases it can turn into an identity, such as 5=5 or an incorrect expression, such as 7=2. But this rarely happens. I will mention this below.

But how to get rid of fractions!? Very simple. Applying the same identical transformations.

We need to multiply the entire equation by the same expression. So that all denominators are reduced! Everything will immediately become easier. Let me explain with an example. Let us need to solve the equation:

How were you taught in elementary school? We move everything to one side, bring it to a common denominator, etc. Forget it like a bad dream! This is what you need to do when you add or subtract fractions. Or you work with inequalities. And in equations, we immediately multiply both sides by an expression that will give us the opportunity to reduce all the denominators (i.e., in essence, by a common denominator). And what is this expression?

On the left side, reducing the denominator requires multiplying by x+2. And on the right, multiplication by 2 is required. This means that the equation must be multiplied by 2(x+2). Multiply:

This is a common multiplication of fractions, but I’ll describe it in detail:

Please note that I am not opening the bracket yet (x + 2)! So, in its entirety, I write it:

On the left side it contracts entirely (x+2), and on the right 2. Which is what was required! After reduction we get linear the equation:

And everyone can solve this equation! x = 2.

Let's solve another example, a little more complicated:

If we remember that 3 = 3/1, and 2x = 2x/ 1, we can write:

And again we get rid of what we don’t really like - fractions.

We see that to reduce the denominator with X, we need to multiply the fraction by (x – 2). And a few are not a hindrance to us. Well, let's multiply. All left side and all right side:

Parentheses again (x – 2) I'm not revealing. I work with the bracket as a whole as if it were one number! This must always be done, otherwise nothing will be reduced.

With a feeling of deep satisfaction we reduce (x – 2) and we get an equation without any fractions, with a ruler!

Now let’s open the brackets:

We bring similar ones, move everything to the left side and get:

Classic quadratic equation. But the minus ahead is not good. You can always get rid of it by multiplying or dividing by -1. But if you look closely at the example, you will notice that it is best to divide this equation by -2! In one fell swoop, the minus will disappear, and the odds will become more attractive! Divide by -2. On the left side - term by term, and on the right - simply divide zero by -2, zero and we get:

We solve through the discriminant and check using Vieta’s theorem. We get x = 1 and x = 3. Two roots.

As you can see, in the first case the equation after the transformation became linear, but here it becomes quadratic. It happens that after getting rid of fractions, all the X's are reduced. Something remains, like 5=5. It means that x can be anything. Whatever it is, it will still be reduced. And it turns out to be pure truth, 5=5. But, after getting rid of fractions, it may turn out to be completely untrue, like 2=7. And this means that no solutions! Any X turns out to be untrue.

Realized the main solution fractional equations? It is simple and logical. We change the original expression so that everything we don’t like disappears. Or it interferes. In this case these are fractions. We will do the same with all kinds of complex examples with logarithms, sines and other horrors. We Always Let's get rid of all this.

However, we need to change the original expression in the direction we need according to the rules, yes... The mastery of which is preparation for the Unified State Exam in mathematics. So we are mastering it.

Now we will learn how to bypass one of main ambushes on the Unified State Exam! But first, let's see whether you fall into it or not?

Let's look at a simple example:

The matter is already familiar, we multiply both sides by (x – 2), we get:

I remind you, with brackets (x – 2) We work as if with one, integral expression!

Here I no longer wrote one in the denominators, it’s undignified... And I didn’t draw brackets in the denominators, except for x – 2 there is nothing, you don’t have to draw. Let's shorten:

Open the parentheses, move everything to the left, and give similar ones:

We solve, check, we get two roots. x = 2 And x = 3. Great.

Suppose the assignment says to write down the root, or their sum if there is more than one root. What are we going to write?

If you decide the answer is 5, you were ambushed. And the task will not be credited to you. They worked in vain... Correct answer is 3.

What's the matter?! And you try to do a check. Substitute the values ​​of the unknown into original example. And if at x = 3 everything will grow together wonderfully, we get 9 = 9, then when x = 2 It will be division by zero! What you absolutely cannot do. Means x = 2 is not a solution, and is not taken into account in the answer. This is the so-called extraneous or extra root. We simply discard it. The final root is one. x = 3.

How so?! – I hear indignant exclamations. We were taught that an equation can be multiplied by an expression! This is an identical transformation!

Yes, identical. Under a small condition - the expression by which we multiply (divide) - different from zero. A x – 2 at x = 2 equals zero! So everything is fair.

And now what i can do?! Don't multiply by expression? Should I check every time? Again it’s unclear!

Calmly! Don't panic!

In this difficult situation, three magic letters will save us. I know what you're thinking. Right! This ODZ . Area of ​​Acceptable Values.

Quadratic equations are studied in 8th grade, so there is nothing complicated here. The ability to solve them is absolutely necessary.

A quadratic equation is an equation of the form ax 2 + bx + c = 0, where the coefficients a, b and c are arbitrary numbers, and a ≠ 0.

Before studying specific solution methods, note that all quadratic equations can be divided into three classes:

1. Have no roots;
2. Have exactly one root;
3. They have two different roots.

This is an important difference between quadratic equations and linear ones, where the root always exists and is unique. How to determine how many roots an equation has? There is a wonderful thing for this - discriminant.

Discriminant

Let the quadratic equation ax 2 + bx + c = 0 be given. Then the discriminant is simply the number D = b 2 − 4ac.

You need to know this formula by heart. Where it comes from is not important now. Another thing is important: by the sign of the discriminant you can determine how many roots a quadratic equation has. Namely:

1. If D< 0, корней нет;
2. If D = 0, there is exactly one root;
3. If D > 0, there will be two roots.

Please note: the discriminant indicates the number of roots, and not at all their signs, as for some reason many people believe. Take a look at the examples and you will understand everything yourself:

Task. How many roots do quadratic equations have:

1. x 2 − 8x + 12 = 0;
2. 5x 2 + 3x + 7 = 0;
3. x 2 − 6x + 9 = 0.

Let's write out the coefficients for the first equation and find the discriminant:
a = 1, b = −8, c = 12;
D = (−8) 2 − 4 1 12 = 64 − 48 = 16

So the discriminant is positive, so the equation has two different roots. We analyze the second equation in a similar way:
a = 5; b = 3; c = 7;
D = 3 2 − 4 5 7 = 9 − 140 = −131.

The discriminant is negative, there are no roots. The last equation left is:
a = 1; b = −6; c = 9;
D = (−6) 2 − 4 1 9 = 36 − 36 = 0.

The discriminant is zero - the root will be one.

Please note that coefficients have been written down for each equation. Yes, it’s long, yes, it’s tedious, but you won’t mix up the odds and make stupid mistakes. Choose for yourself: speed or quality.

By the way, if you get the hang of it, after a while you won’t need to write down all the coefficients. You will perform such operations in your head. Most people start doing this somewhere after 50-70 solved equations - in general, not that much.

Roots of a quadratic equation

Now let's move on to the solution itself. If the discriminant D > 0, the roots can be found using the formulas:

Basic formula for the roots of a quadratic equation

When D = 0, you can use any of these formulas - you will get the same number, which will be the answer. Finally, if D< 0, корней нет — ничего считать не надо.

1. x 2 − 2x − 3 = 0;
2. 15 − 2x − x 2 = 0;
3. x 2 + 12x + 36 = 0.

First equation:
x 2 − 2x − 3 = 0 ⇒ a = 1; b = −2; c = −3;
D = (−2) 2 − 4 1 (−3) = 16.

D > 0 ⇒ the equation has two roots. Let's find them:

Second equation:
15 − 2x − x 2 = 0 ⇒ a = −1; b = −2; c = 15;
D = (−2) 2 − 4 · (−1) · 15 = 64.

D > 0 ⇒ the equation again has two roots. Let's find them

\[\begin(align) & ((x)_(1))=\frac(2+\sqrt(64))(2\cdot \left(-1 \right))=-5; \\ & ((x)_(2))=\frac(2-\sqrt(64))(2\cdot \left(-1 \right))=3. \\ \end(align)\]

Finally, the third equation:
x 2 + 12x + 36 = 0 ⇒ a = 1; b = 12; c = 36;
D = 12 2 − 4 1 36 = 0.

D = 0 ⇒ the equation has one root. Any formula can be used. For example, the first one:

As you can see from the examples, everything is very simple. If you know the formulas and can count, there will be no problems. Most often, errors occur when substituting negative coefficients into the formula. Here again, the technique described above will help: look at the formula literally, write down each step - and very soon you will get rid of mistakes.

Incomplete quadratic equations

It happens that a quadratic equation is slightly different from what is given in the definition. For example:

1. x 2 + 9x = 0;
2. x 2 − 16 = 0.

It is easy to notice that these equations are missing one of the terms. Such quadratic equations are even easier to solve than standard ones: they don’t even require calculating the discriminant. So, let's introduce a new concept:

The equation ax 2 + bx + c = 0 is called an incomplete quadratic equation if b = 0 or c = 0, i.e. the coefficient of the variable x or the free element is equal to zero.

Of course, a very difficult case is possible when both of these coefficients are equal to zero: b = c = 0. In this case, the equation takes the form ax 2 = 0. Obviously, such an equation has a single root: x = 0.

Let's consider the remaining cases. Let b = 0, then we obtain an incomplete quadratic equation of the form ax 2 + c = 0. Let us transform it a little:

Since the arithmetic square root exists only of a non-negative number, the last equality makes sense only for (−c /a) ≥ 0. Conclusion:

1. If in an incomplete quadratic equation of the form ax 2 + c = 0 the inequality (−c /a) ≥ 0 is satisfied, there will be two roots. The formula is given above;
2. If (−c /a)< 0, корней нет.

As you can see, a discriminant was not required—there are no complex calculations at all in incomplete quadratic equations. In fact, it is not even necessary to remember the inequality (−c /a) ≥ 0. It is enough to express the value x 2 and see what is on the other side of the equal sign. If there is a positive number, there will be two roots. If it is negative, there will be no roots at all.

Now let's look at equations of the form ax 2 + bx = 0, in which the free element is equal to zero. Everything is simple here: there will always be two roots. It is enough to factor the polynomial:

Taking the common factor out of brackets

The product is zero when at least one of the factors is zero. This is where the roots come from. In conclusion, let’s look at a few of these equations:

Task. Solve quadratic equations:

1. x 2 − 7x = 0;
2. 5x 2 + 30 = 0;
3. 4x 2 − 9 = 0.

x 2 − 7x = 0 ⇒ x · (x − 7) = 0 ⇒ x 1 = 0; x 2 = −(−7)/1 = 7.

5x 2 + 30 = 0 ⇒ 5x 2 = −30 ⇒ x 2 = −6. There are no roots, because a square cannot be equal to a negative number.

4x 2 − 9 = 0 ⇒ 4x 2 = 9 ⇒ x 2 = 9/4 ⇒ x 1 = 3/2 = 1.5; x 2 = −1.5.

First, what is a quadratic equation? A quadratic equation is an equation of the form ax^2+bx+c=0, where x is a variable, a, b and c are some numbers, and a is not equal to zero.

Step 2

To solve a quadratic equation, we need to know the formula of its roots, that is, for starters, the discriminant formula of the quadratic equation. It looks like this: D=b^2-4ac. You can derive it yourself, but usually this is not required, just remember the formula (!) You will really need it in the future. There is also a formula for the quarter discriminant, more about it a little later.

Step 3

Let's take as an example the equation 3x^2-24x+21=0. I'll solve it in two ways.

Step 4

Method 1. Discriminant.
3x^2-24x+21=0
a=3, b=-24, c=21
D=b^2-4ac
D=576-4*63=576-252=324=18^2
D>
x1.2= (-b 18)/6=42/6=7
x2=(-(-24)-18)/6=6/6=1

Step 5

It's time to remember the quarter discriminant formula, which can greatly facilitate the solution of our equation =) so, here's what it looks like: D1=k^2-ac (k=1/2b)
Method 2. Quarter Discriminant.
3x^2-24x+21=0
a=3, b=-24, c=21
k=-12
D1=k^2 – ac
D1=144-63=81=9^2
D1>0, which means the equation has 2 roots
x1,2= k +/ square root of D1)/a
x1= (-(-12) +9)/3=21/3=7
x2= (-(-12) -9)/3=3/3=1

Did you rate how much easier the solution is? ;)
Thank you for your attention, I wish you success in your studies =)

• In our case, in the equations D and D1 were >0 and we got 2 roots each. If there were D=0 and D1=0, then we would get one root each, and if there were D<0 и D1<0 соответственно, то у уравнений корней бы не было вовсе.
• Through the root of the discriminant (D1) it is possible to solve only those equations in which the term b is even(!)

With this math program you can solve quadratic equation.

The program not only gives the answer to the problem, but also displays the solution process in two ways:
- using a discriminant
- using Vieta's theorem (if possible).

Moreover, the answer is displayed as exact, not approximate.
For example, for the equation \(81x^2-16x-1=0\) the answer is displayed in the following form:

$$ x_1 = \frac(8+\sqrt(145))(81), \quad x_2 = \frac(8-\sqrt(145))(81) $$ and not like this: \(x_1 = 0.247; \quad x_2 = -0.05\)

This program can be useful for high school students in general education schools when preparing for tests and exams, when testing knowledge before the Unified State Exam, and for parents to control the solution of many problems in mathematics and algebra.

Or maybe it’s too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get your math or algebra homework done as quickly as possible? In this case, you can also use our programs with detailed solutions.

In this way, you can conduct your own training and/or training of your younger brothers or sisters, while the level of education in the field of solving problems increases.

If you are not familiar with the rules for entering a quadratic polynomial, we recommend that you familiarize yourself with them.

Rules for entering a quadratic polynomial
Any Latin letter can act as a variable.

For example: \(x, y, z, a, b, c, o, p, q\), etc.
Numbers can be entered as whole or fractional numbers.

Moreover, fractional numbers can be entered not only in the form of a decimal, but also in the form of an ordinary fraction.
In decimal fractions, the fractional part can be separated from the whole part by either a period or a comma.
For example, you can enter decimal fractions like this: 2.5x - 3.5x^2

Rules for entering ordinary fractions.
Only a whole number can act as the numerator, denominator and integer part of a fraction.

The denominator cannot be negative.

When entering a numerical fraction, the numerator is separated from the denominator by a division sign: /
The whole part is separated from the fraction by the ampersand sign: &
Input: 3&1/3 - 5&6/5z +1/7z^2
Result: \(3\frac(1)(3) - 5\frac(6)(5) z + \frac(1)(7)z^2\)

When entering an expression you can use parentheses. In this case, when solving a quadratic equation, the introduced expression is first simplified.
For example: 1/2(y-1)(y+1)-(5y-10&1/2)

Decide

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A little theory.

Quadratic equation and its roots. Incomplete quadratic equations

Each of the equations
\(-x^2+6x+1.4=0, \quad 8x^2-7x=0, \quad x^2-\frac(4)(9)=0 \)
looks like
\(ax^2+bx+c=0, \)
where x is a variable, a, b and c are numbers.
In the first equation a = -1, b = 6 and c = 1.4, in the second a = 8, b = -7 and c = 0, in the third a = 1, b = 0 and c = 4/9. Such equations are called quadratic equations.

Definition.
Quadratic equation is called an equation of the form ax 2 +bx+c=0, where x is a variable, a, b and c are some numbers, and \(a \neq 0 \).

The numbers a, b and c are the coefficients of the quadratic equation. The number a is called the first coefficient, the number b is the second coefficient, and the number c is the free term.

In each of the equations of the form ax 2 +bx+c=0, where \(a\neq 0\), the largest power of the variable x is a square. Hence the name: quadratic equation.

Note that a quadratic equation is also called an equation of the second degree, since its left side is a polynomial of the second degree.

A quadratic equation in which the coefficient of x 2 is equal to 1 is called given quadratic equation. For example, the quadratic equations given are the equations
\(x^2-11x+30=0, \quad x^2-6x=0, \quad x^2-8=0 \)

If in a quadratic equation ax 2 +bx+c=0 at least one of the coefficients b or c is equal to zero, then such an equation is called incomplete quadratic equation. Thus, the equations -2x 2 +7=0, 3x 2 -10x=0, -4x 2 =0 are incomplete quadratic equations. In the first of them b=0, in the second c=0, in the third b=0 and c=0.

There are three types of incomplete quadratic equations:
1) ax 2 +c=0, where \(c \neq 0 \);
2) ax 2 +bx=0, where \(b \neq 0 \);
3) ax 2 =0.

Let's consider solving equations of each of these types.

To solve an incomplete quadratic equation of the form ax 2 +c=0 for \(c \neq 0 \), move its free term to the right side and divide both sides of the equation by a:
\(x^2 = -\frac(c)(a) \Rightarrow x_(1,2) = \pm \sqrt( -\frac(c)(a)) \)

Since \(c \neq 0 \), then \(-\frac(c)(a) \neq 0 \)

If \(-\frac(c)(a)>0\), then the equation has two roots.

If \(-\frac(c)(a) To solve an incomplete quadratic equation of the form ax 2 +bx=0 with \(b \neq 0 \) factor its left side and obtain the equation
\(x(ax+b)=0 \Rightarrow \left\( \begin(array)(l) x=0 \\ ax+b=0 \end(array) \right. \Rightarrow \left\( \begin (array)(l) x=0 \\ x=-\frac(b)(a) \end(array) \right.

This means that an incomplete quadratic equation of the form ax 2 +bx=0 for \(b \neq 0 \) always has two roots.

An incomplete quadratic equation of the form ax 2 =0 is equivalent to the equation x 2 =0 and therefore has a single root 0.

Formula for the roots of a quadratic equation

Let us now consider how to solve quadratic equations in which both the coefficients of the unknowns and the free term are nonzero.

Let us solve the quadratic equation in general form and as a result we obtain the formula for the roots. This formula can then be used to solve any quadratic equation.

Solve the quadratic equation ax 2 +bx+c=0

Dividing both sides by a, we obtain the equivalent reduced quadratic equation
\(x^2+\frac(b)(a)x +\frac(c)(a)=0 \)

Let's transform this equation by selecting the square of the binomial:
\(x^2+2x \cdot \frac(b)(2a)+\left(\frac(b)(2a)\right)^2- \left(\frac(b)(2a)\right)^ 2 + \frac(c)(a) = 0 \Rightarrow \)

\(x^2+2x \cdot \frac(b)(2a)+\left(\frac(b)(2a)\right)^2 = \left(\frac(b)(2a)\right)^ 2 - \frac(c)(a) \Rightarrow \) \(\left(x+\frac(b)(2a)\right)^2 = \frac(b^2)(4a^2) - \frac( c)(a) \Rightarrow \left(x+\frac(b)(2a)\right)^2 = \frac(b^2-4ac)(4a^2) \Rightarrow \) \(x+\frac(b )(2a) = \pm \sqrt( \frac(b^2-4ac)(4a^2) ) \Rightarrow x = -\frac(b)(2a) + \frac( \pm \sqrt(b^2 -4ac) )(2a) \Rightarrow \) \(x = \frac( -b \pm \sqrt(b^2-4ac) )(2a) \)

The radical expression is called discriminant of a quadratic equation ax 2 +bx+c=0 (“discriminant” in Latin - discriminator). It is designated by the letter D, i.e.
\(D = b^2-4ac\)

Now, using the discriminant notation, we rewrite the formula for the roots of the quadratic equation:
\(x_(1,2) = \frac( -b \pm \sqrt(D) )(2a) \), where \(D= b^2-4ac \)

It's obvious that:
1) If D>0, then the quadratic equation has two roots.
2) If D=0, then the quadratic equation has one root \(x=-\frac(b)(2a)\).
3) If D Thus, depending on the value of the discriminant, a quadratic equation can have two roots (for D > 0), one root (for D = 0) or have no roots (for D When solving a quadratic equation using this formula, it is advisable to do the following way:
1) calculate the discriminant and compare it with zero;
2) if the discriminant is positive or equal to zero, then use the root formula; if the discriminant is negative, then write down that there are no roots.

Vieta's theorem

The given quadratic equation ax 2 -7x+10=0 has roots 2 and 5. The sum of the roots is 7, and the product is 10. We see that the sum of the roots is equal to the second coefficient taken with the opposite sign, and the product of the roots is equal to the free term. Any reduced quadratic equation that has roots has this property.

The sum of the roots of the above quadratic equation is equal to the second coefficient taken with the opposite sign, and the product of the roots is equal to the free term.

Those. Vieta's theorem states that the roots x 1 and x 2 of the reduced quadratic equation x 2 +px+q=0 have the property:
\(\left\( \begin(array)(l) x_1+x_2=-p \\ x_1 \cdot x_2=q \end(array) \right. \)

I hope that after studying this article you will learn how to find the roots of a complete quadratic equation.

Using the discriminant, only complete quadratic equations are solved; to solve incomplete quadratic equations, other methods are used, which you will find in the article “Solving incomplete quadratic equations.”

What quadratic equations are called complete? This equations of the form ax 2 + b x + c = 0, where coefficients a, b and c are not equal to zero. So, to solve a complete quadratic equation, we need to calculate the discriminant D.

D = b 2 – 4ac.

Depending on the value of the discriminant, we will write down the answer.

If the discriminant is a negative number (D< 0),то корней нет.

If the discriminant is zero, then x = (-b)/2a. When the discriminant is a positive number (D > 0),

then x 1 = (-b - √D)/2a, and x 2 = (-b + √D)/2a.

For example. Solve the equation x 2– 4x + 4= 0.

D = 4 2 – 4 4 = 0

x = (- (-4))/2 = 2

Answer: 2.

Solve Equation 2 x 2 + x + 3 = 0.

D = 1 2 – 4 2 3 = – 23

Answer: no roots.

Solve Equation 2 x 2 + 5x – 7 = 0.

D = 5 2 – 4 2 (–7) = 81

x 1 = (-5 - √81)/(2 2)= (-5 - 9)/4= – 3.5

x 2 = (-5 + √81)/(2 2) = (-5 + 9)/4=1

Answer: – 3.5; 1.

So let’s imagine the solution of complete quadratic equations using the diagram in Figure 1.

Using these formulas you can solve any complete quadratic equation. You just need to be careful to the equation was written as a polynomial of the standard form

A x 2 + bx + c, otherwise you may make a mistake. For example, in writing the equation x + 3 + 2x 2 = 0, you can mistakenly decide that

a = 1, b = 3 and c = 2. Then

D = 3 2 – 4 1 2 = 1 and then the equation has two roots. And this is not true. (See solution to example 2 above).

Therefore, if the equation is not written as a polynomial of the standard form, first the complete quadratic equation must be written as a polynomial of the standard form (the monomial with the largest exponent should come first, that is A x 2 , then with less – bx and then a free member With.

When solving the reduced quadratic equation and a quadratic equation with an even coefficient in the second term, you can use other formulas. Let's get acquainted with these formulas. If in a complete quadratic equation the second term has an even coefficient (b = 2k), then you can solve the equation using the formulas shown in the diagram in Figure 2.

A complete quadratic equation is called reduced if the coefficient at x 2 is equal to one and the equation takes the form x 2 + px + q = 0. Such an equation can be given for solution, or it can be obtained by dividing all coefficients of the equation by the coefficient A, standing at x 2 .

Figure 3 shows a diagram for solving the reduced square
equations. Let's look at an example of the application of the formulas discussed in this article.

Example. Solve the equation

3x 2 + 6x – 6 = 0.

Let's solve this equation using the formulas shown in the diagram in Figure 1.

D = 6 2 – 4 3 (– 6) = 36 + 72 = 108

√D = √108 = √(36 3) = 6√3

x 1 = (-6 - 6√3)/(2 3) = (6 (-1- √(3)))/6 = –1 – √3

x 2 = (-6 + 6√3)/(2 3) = (6 (-1+ √(3)))/6 = –1 + √3

Answer: –1 – √3; –1 + √3

You can notice that the coefficient of x in this equation is an even number, that is, b = 6 or b = 2k, whence k = 3. Then let’s try to solve the equation using the formulas shown in the diagram of the figure D 1 = 3 2 – 3 · (– 6 ) = 9 + 18 = 27

√(D 1) = √27 = √(9 3) = 3√3

x 1 = (-3 - 3√3)/3 = (3 (-1 - √(3)))/3 = – 1 – √3

x 2 = (-3 + 3√3)/3 = (3 (-1 + √(3)))/3 = – 1 + √3

Answer: –1 – √3; –1 + √3. Noticing that all the coefficients in this quadratic equation are divisible by 3 and performing the division, we get the reduced quadratic equation x 2 + 2x – 2 = 0 Solve this equation using the formulas for the reduced quadratic
equations figure 3.

D 2 = 2 2 – 4 (– 2) = 4 + 8 = 12

√(D 2) = √12 = √(4 3) = 2√3

x 1 = (-2 - 2√3)/2 = (2 (-1 - √(3)))/2 = – 1 – √3

x 2 = (-2 + 2√3)/2 = (2 (-1+ √(3)))/2 = – 1 + √3

Answer: –1 – √3; –1 + √3.

As you can see, when solving this equation using different formulas, we received the same answer. Therefore, having thoroughly mastered the formulas shown in the diagram in Figure 1, you will always be able to solve any complete quadratic equation.

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