How to solve an equation with x squared. Square root: calculation formulas

I came up with such a cool theorem for them,
and they decide through the discriminant:-(((
(c) Francois Viet “Non-existent statements”

Root formula, or the long way

Everyone who attended even the slightest bit of mathematics lessons in the 8th grade knows the formula for the roots of a quadratic equation. The solution using the root formula is often called in common parlance “the solution through the discriminant.” Let us briefly recall the formula for roots.

[You can also view the contents of this article at video format ]

The quadratic equation has the form ax 2 +bx+c= 0, where a, b, c- some numbers. For example, in Eq. 2x 2 + 3x – 5 = 0 these numbers are equal: a = 2, b = 3. c= -5. Before solving any quadratic equation, you need to “see” these numbers and understand what they equal.

Next, the so-called discriminant is calculated using the formula D=b^2-4ac. In our case D = 3^2 – 4 \cdot 2 \cdot (-5) = 9 + 40 = 49. Then the root is extracted from the discriminant: \sqrt(D) = \sqrt(49) = 7 .

After the discriminant has been calculated, the root formula is used: x_1=\frac(-b-\sqrt(D))(2a); x_2=\frac(-b+\sqrt(D))(2a):

x_1=\frac(-3-7)(2 \cdot 2)=\frac(-10)(4)=-2.5
x_2= \frac(-3+7)(2 \cdot 2)=\frac(4)(4)=1

And thus, the equation is solved. It has two roots: 1 and -2.5.

But this equation, like many others proposed in school textbooks/problems, could be solved in a much faster way if you knew a couple of life hacks. And we’re not just talking about Vieta’s theorem, although it is a useful tool.

Life hack first. If a + b + c= 0, then x_1=1, x_2=\frac(c)(a) .

It only applies if all three coefficients in a quadratic equation are a, b, c when added they give 0. For example, we had the equation 2x 2 + 3x – 5 = 0 . Adding all three coefficients, we get 2 + 3 – 5, which is equal to 0. In this case, you can not count the discriminant and not apply the root formula. Instead, you can immediately write that

x_1=1,
x_2=\frac(c)(a)=\frac(-5)(2)=-2.5

(note that we got the same result in the roots formula).

People often ask whether x_1=1 will always work? Yes, whenever a + b + c = 0.

Life hack second. If a + c = b, then x_1=-1, x_2=-\frac(c)(a) .

Let the equation be given 5x 2 + 6x + 1 = 0 . In him a = 5, b = 6, c= 1. If we add up the “extreme” coefficients a And c, we get 5+1 = 6, which is exactly equal to the “average” coefficient b. This means we can do without a discriminant! We immediately write down:

x_1=-1,
x_2=-\frac(c)(a)=\frac(-1)(5)=-0.2

Life hack third(the theorem inverse to Vieta’s theorem). If a= 1, then

Consider the equation x 2 – 12x+ 35 = 0. It contains a = 1, b = -12, c = 35. It does not fit either the first or the second life hack - the conditions are not met. If it fit the first or second, then we would do without Vieta’s theorem.

The very use of Vieta's theorem implies an understanding of some useful techniques.

First appointment. Don’t be shy about writing down the view system itself \begin(cases) x_1+x_2 = -b \\ x_1 \cdot x_2 = c \end(cases), which is obtained by using Vieta’s theorem. There is no need to try at all costs to solve the equation absolutely orally, without written notes, as “advanced users” do.

For our equation x 2 – 12x+ 35 = 0 this system has the form

\begin(cases) x_1+x_2 = 12 \\ x_1 \cdot x_2 = 35 \end(cases)

Now we need to verbally select the numbers x_1 and x_2 that satisfy our system, i.e. the total is 12, and when multiplied it is 35.

So, second appointment is that you need to start the selection not with the sum, but with the product. Let's look at the second equation of the system and ask ourselves: what numbers, when multiplied, give 35? If everything is in order with the multiplication table, then the answer immediately comes to mind: 7 and 5. And only now let’s substitute these numbers into the first equation: we will have 7 + 5 = 12, which is a true equality. So, the numbers 7 and 5 satisfy both equations, so we immediately write:

x_1 = 7, x_2 = 5

Third reception is that if the numbers cannot be found quickly (within 15-20 seconds), then, regardless of the reason, you need to calculate the discriminant and use the root formula. Why? Because the roots may not be found if the equation does not have them at all (the discriminant is negative), or the roots are numbers that are not integers.

Training exercises for solving quadratic equations

Practice! Try solving the following equations. Look at each equation in the following order:

• if the equation fits the first life hack (when a + b + c = 0), then we solve it with its help;
• if the equation fits the second life hack (when a + c = b), then we solve it with its help;
• if the equation fits the third life hack (Viete’s theorem), we solve it with its help;
• and only in the most extreme case - if nothing fits and/or it was not possible to solve using Vieta’s theorem - we calculate the discriminant. Again: discriminant - last but not least!

1. Solve the equation x 2 + 3x + 2 = 0
   View solution and answer

   See lifehack two
   In this equation, a = 1, b = 3, c = 2. Thus, a + c = b, whence x_1=-1, x_2 = -\frac(c)(a) = -\frac(2)(1)=-2.
   Answer: -1, -2.

2. Solve the equation x 2 + 8x – 9 = 0
   View solution and answer

   See life hack first
   In this equation, a = 1, b = 8, c = -9. Thus, a + b + c = 0, whence x_1=1, x_2 = \frac(c)(a) = \frac(-9)(1)=-9.
   Answer: 1, -9.

3. Solve the equation 15x 2 – 11x + 2 = 0
   View solution and answer

   This equation (the only one from the entire list) does not fall under any of the life hacks, so we will solve it using the root formula:
   D=b^2-4ac = (-11)^2 – 4 \cdot 15 \cdot 2 = 121 – 120 = 1.x_1=\frac(11-1)(2 \cdot 15)=\frac(10)(30)=\frac(1)(3)x_2= \frac(11+1)(2 \cdot 15)=\frac(12)(30)=\frac(2)(5) Answer: \frac(1)(3), \frac(2)(5).
   

4. Solve the equation x 2 + 9x + 20 = 0
   View solution and answer

   
   \begin(cases) x_1+x_2 = -9 \\ x_1 \cdot x_2 = 20 \end(cases)
   By selection we establish that x_1 = -4, x_2 = -5.
   Answer: -4, -5.
   

5. Solve the equation x 2 – 7x – 30 = 0
   View solution and answer

   See life hack three (Vieta’s theorem)
   In this equation a = 1, so we can write that \begin(cases) x_1+x_2 = 7 \\ x_1 \cdot x_2 = -30 \end(cases)
   By selection we establish that x_1 = 10, x_2 = -3.
   Answer: 10, -3.

6. Solve the equation x 2 – 19x + 18 = 0
   View solution and answer

   See life hack first
   In this equation, a = 1, b = -19, c = 18. Thus, a + b + c = 0, whence x_1=1, x_2 = \frac(c)(a) = \frac(18)(1)=18.
   Answer: 1, 18.

7. Solve the equation x 2 + 7x + 6 = 0
   View solution and answer

   See lifehack two
   In this equation, a = 1, b = 7, c = 6. Thus, a + c = b, whence x_1=-1, x_2 = -\frac(c)(a) = -\frac(6)(1)=-6.
   Answer: -1, -6.

8. Solve the equation x 2 – 8x + 12 = 0
   View solution and answer

   See life hack three (Vieta’s theorem)
   In this equation a = 1, so we can write that \begin(cases) x_1+x_2 = 8 \\ x_1 \cdot x_2 = 12 \end(cases)
   By selection we establish that x_1 = 6, x_2 = 2.
   Answer: 6, 2.

9. Solve the equation x 2 – x – 6 = 0
   View solution and answer

   See life hack three (Vieta’s theorem)
   In this equation a = 1, so we can write that \begin(cases) x_1+x_2 = 1 \\ x_1 \cdot x_2 = -6 \end(cases)
   By selection we establish that x_1 = 3, x_2 = -2.
   Answer: 3, -2.

10. Solve the equation x 2 – 15x – 16 = 0
    View solution and answer

    See lifehack two
    In this equation, a = 1, b = -15, c = -16. Thus, a + c = b, whence x_1=-1, x_2 = -\frac(c)(a) = -\frac(-16)(1)=16.
    Answer: -1, 16.

11. Solve the equation x 2 + 11x – 12 = 0
    View solution and answer

    See life hack first
    In this equation, a = 1, b = 11, c = -12. Thus, a + b + c = 0, whence x_1=1, x_2 = \frac(c)(a) = \frac(-12)(1)=-12.
    Answer: 1, -12.

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. They 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 errors.

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.

", that is, equations of the first degree. In this lesson we will look at what is called a quadratic equation and how to solve it.

What is a quadratic equation?

Important!

The degree of an equation is determined by the highest degree to which the unknown stands.

If the maximum power in which the unknown is “2”, then you have a quadratic equation.

Examples of quadratic equations

• 5x 2 − 14x + 17 = 0
• −x 2 + x +

  1

  3

  = 0
• x 2 + 0.25x = 0
• x 2 − 8 = 0

Important! The general form of a quadratic equation looks like this:

A x 2 + b x + c = 0

“a”, “b” and “c” are given numbers.

• “a” is the first or highest coefficient;
• “b” is the second coefficient;
• “c” is a free member.

To find “a”, “b” and “c” you need to compare your equation with the general form of the quadratic equation “ax 2 + bx + c = 0”.

Let's practice determining the coefficients "a", "b" and "c" in quadratic equations.

5x 2 − 14x + 17 = 0 −7x 2 − 13x + 8 = 0 −x 2 + x +

The equation	Odds
a = 5 b = −14 c = 17
a = −7 b = −13 c = 8

1

3

= 0

• a = −1
• b = 1
• c =

  1

  3

x 2 + 0.25x = 0

• a = 1
• b = 0.25
• c = 0

x 2 − 8 = 0

• a = 1
• b = 0
• c = −8

How to Solve Quadratic Equations

Unlike linear equations, a special method is used to solve quadratic equations. formula for finding roots.

Remember!

To solve a quadratic equation you need:

• bring the quadratic equation to the general form “ax 2 + bx + c = 0”. That is, only “0” should remain on the right side;
• use formula for roots:

Let's look at an example of how to use the formula to find the roots of a quadratic equation. Let's solve a quadratic equation.

X 2 − 3x − 4 = 0

The equation “x 2 − 3x − 4 = 0” has already been reduced to the general form “ax 2 + bx + c = 0” and does not require additional simplifications. To solve it, we just need to apply formula for finding the roots of a quadratic equation.

Let us determine the coefficients “a”, “b” and “c” for this equation.

x 1;2 =
x 1;2 =
x 1;2 =
x 1;2 =

It can be used to solve any quadratic equation.

In the formula “x 1;2 = ” the radical expression is often replaced
“b 2 − 4ac” for the letter “D” and is called discriminant. The concept of a discriminant is discussed in more detail in the lesson “What is a discriminant”.

Let's look at another example of a quadratic equation.

x 2 + 9 + x = 7x

In this form, it is quite difficult to determine the coefficients “a”, “b” and “c”. Let's first reduce the equation to the general form “ax 2 + bx + c = 0”.

X 2 + 9 + x = 7x
x 2 + 9 + x − 7x = 0
x 2 + 9 − 6x = 0
x 2 − 6x + 9 = 0

Now you can use the formula for the roots.

X 1;2 =
x 1;2 =
x 1;2 =
x 1;2 =
x =

6

2

x = 3
Answer: x = 3

There are times when quadratic equations have no roots. This situation occurs when the formula contains a negative number under the root.

A quadratic equation, or algebraic equation of the 2nd degree with one unknown, in general form is written as follows:

Ax 2 + bx + c = 0,

• a, b, c are known coefficients, and a ≠ 0.
• x is unknown.

3x 2 + 8x - 5 = 0.

2. Types of quadratic equations

Dividing both sides of the equation by a, we get reduced quadratic equation:

x 2 + px + q = 0,

• p = b/a
• q = c/a

If one of the coefficients b, c or both are equal to 0 at the same time, then a quadratic equation is called incomplete.

• x 2 +8x-5=0 is a complete reduced quadratic equation.
• 3x 2 -5=0 is not a complete unreduced quadratic equation.
• x 2 -8x=0 is not a complete reduced quadratic equation.

Incomplete quadratic equation of the form

X 2 = m

the simplest and most important, because the solution of any quadratic equation is reduced to it.

Three cases are possible:

• m = 0, x = 0
• m > 0, x = ±√‾m
• m< 0, x = ±i√‾m. Где i — мнимая единица, равная √‾-1.

3. Solving a quadratic equation

The roots of an unreduced complete quadratic equation are found by the formula

x = (-b ± √‾(b 2 - 4ac)) / 2a

x = (7 ± √‾(1)) / 6

4. Properties of the roots of a quadratic equation. Discriminant.

According to the formula for the roots of a quadratic equation, there can be three cases, determined by the radical expression (b 2 - 4ac). It's called discriminant(discriminating).

Denoting the discriminant with the letter D, we can write:

• D > 0, the equation has two different real roots.
• D = 0, the equation has two equal real roots.
• D< 0, уравнение имеет два различных мнимых корня.

x = (-b ± √‾(b 2 - 4ac)) / 2a

x = (7 ± √‾(7 2 - 4×3×4)) / (2×3)

x = (7 ± √‾(1)) / 6

5. Formulas useful in life

Often there are problems of converting volume into area or length and the inverse problem - converting area into volume. For example, boards are sold in cubes (cubic meters), and we need to calculate how much wall area can be covered with boards contained in a certain volume, see.

The use of equations is widespread in our lives. They are used in many calculations, construction of structures and even sports. Man used equations in ancient times, and since then their use has only increased. The discriminant allows you to solve any quadratic equation using a general formula, which has the following form:

The discriminant formula depends on the degree of the polynomial. The above formula is suitable for solving quadratic equations of the following form:

The discriminant has the following properties that you need to know:

* "D" is 0 when the polynomial has multiple roots (equal roots);

* "D" is a symmetric polynomial with respect to the roots of the polynomial and is therefore a polynomial in its coefficients; moreover, the coefficients of this polynomial are integers regardless of the extension in which the roots are taken.

Let's say we are given a quadratic equation of the following form:

1 equation

According to the formula we have:

Since \, the equation has 2 roots. Let's define them:

Where can I solve an equation using a discriminant online solver?

You can solve the equation on our website https://site. The free online solver will allow you to solve online equations of any complexity in a matter of seconds. All you need to do is simply enter your data into the solver. You can also watch the video instructions and find out how to solve the equation on our website. And if you have any questions, you can ask them in our VKontakte group http://vk.com/pocketteacher. Join our group, we are always happy to help you.