Online calculator. Solving exponential equations. Horner's scheme. Examples 2x 0 solve equation
to solve mathematics. Find quickly solving a mathematical equation in mode online. The website www.site allows solve the equation almost any given algebraic, trigonometric or transcendental equation online. When studying almost any branch of mathematics at different stages you have to decide equations online. To get an answer immediately, and most importantly an accurate answer, you need a resource that allows you to do this. Thanks to the site www.site solve equations online will take a few minutes. The main advantage of www.site when solving mathematical equations online- this is the speed and accuracy of the response provided. The site is able to solve any algebraic equations online, trigonometric equations online, transcendental equations online, and also equations with unknown parameters in mode online. Equations serve as a powerful mathematical apparatus solutions practical problems. With the help mathematical equations it is possible to express facts and relationships that may seem confusing and complex at first glance. Unknown quantities equations can be found by formulating the problem in mathematical language in the form equations And decide received task in mode online on the website www.site. Any algebraic equation, trigonometric equation or equations containing transcendental features you can easily decide online and get the exact answer. When studying natural sciences, you inevitably encounter the need solving equations. In this case, the answer must be accurate and must be obtained immediately in the mode online. Therefore for solving mathematical equations online we recommend the site www.site, which will become your indispensable calculator for solve algebraic equations online, trigonometric equations online, and also transcendental equations online or equations with unknown parameters. For practical problems of finding the roots of various mathematical equations resource www.. Solving equations online yourself, it is useful to check the received answer using online equation solving on the website www.site. You need to write the equation correctly and instantly get online solution, after which all that remains is to compare the answer with your solution to the equation. Checking the answer will take no more than a minute, it’s enough solve equation online and compare the answers. This will help you avoid mistakes in decision and correct the answer in time when solving equations online be it algebraic, trigonometric, transcendental or equation with unknown parameters.
Let us analyze two types of solutions to systems of equations:
1. Solving the system using the substitution method.
2. Solving the system by term-by-term addition (subtraction) of the system equations.
In order to solve the system of equations by substitution method you need to follow a simple algorithm:
1. Express. From any equation we express one variable.
2. Substitute. We substitute the resulting value into another equation instead of the expressed variable.
3. Solve the resulting equation with one variable. We find a solution to the system.
To decide system by term-by-term addition (subtraction) method need to:
1. Select a variable for which we will make identical coefficients.
2. We add or subtract equations, resulting in an equation with one variable.
3. Solve the resulting linear equation. We find a solution to the system.
The solution to the system is the intersection points of the function graphs.
Let us consider in detail the solution of systems using examples.
Example #1:
Let's solve by substitution method
Solving a system of equations using the substitution method2x+5y=1 (1 equation)
x-10y=3 (2nd equation)
1. Express
It can be seen that in the second equation there is a variable x with a coefficient of 1, which means that it is easiest to express the variable x from the second equation.
x=3+10y
2.After we have expressed it, we substitute 3+10y into the first equation instead of the variable x.
2(3+10y)+5y=1
3. Solve the resulting equation with one variable.
2(3+10y)+5y=1 (open the brackets)
6+20y+5y=1
25y=1-6
25y=-5 |: (25)
y=-5:25
y=-0.2
The solution to the equation system is the intersection points of the graphs, therefore we need to find x and y, because the intersection point consists of x and y. Let's find x, in the first point where we expressed it we substitute y.
x=3+10y
x=3+10*(-0.2)=1
It is customary to write points in the first place we write the variable x, and in the second place the variable y.
Answer: (1; -0.2)
Example #2:
Let's solve using the term-by-term addition (subtraction) method.
Solving a system of equations using the addition method3x-2y=1 (1 equation)
2x-3y=-10 (2nd equation)
1. We choose a variable, let’s say we choose x. In the first equation, the variable x has a coefficient of 3, in the second - 2. We need to make the coefficients the same, for this we have the right to multiply the equations or divide by any number. We multiply the first equation by 2, and the second by 3 and get a total coefficient of 6.
3x-2y=1 |*2
6x-4y=2
2x-3y=-10 |*3
6x-9y=-30
2. Subtract the second from the first equation to get rid of the variable x. Solve the linear equation.
__6x-4y=2
5y=32 | :5
y=6.4
3. Find x. We substitute the found y into any of the equations, let’s say into the first equation.
3x-2y=1
3x-2*6.4=1
3x-12.8=1
3x=1+12.8
3x=13.8 |:3
x=4.6
The intersection point will be x=4.6; y=6.4
Answer: (4.6; 6.4)
Do you want to prepare for exams for free? Tutor online for free. No joke.
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:
- Have no roots;
- Have exactly one root;
- 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:
- If D< 0, корней нет;
- If D = 0, there is exactly one root;
- 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:
- x 2 − 8x + 12 = 0;
- 5x 2 + 3x + 7 = 0;
- 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 the 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.
- x 2 − 2x − 3 = 0;
- 15 − 2x − x 2 = 0;
- 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:
- x 2 + 9x = 0;
- 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:
- 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;
- 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 bracketsThe 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:
- x 2 − 7x = 0;
- 5x 2 + 30 = 0;
- 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.
We offer you a convenient free online calculator for solving quadratic equations. You can quickly get and understand how they are solved using clear examples.
To produce solve quadratic equation online, first bring the equation to its general form:
ax 2 + bx + c = 0
Fill in the form fields accordingly:
How to solve a quadratic equation
How to solve a quadratic equation: | Types of roots: |
1.
Reduce the quadratic equation to its general form: General view Аx 2 +Bx+C=0 Example: 3x - 2x 2 +1=-1 Reduce to -2x 2 +3x+2=0 2.
Find the discriminant D. 3.
Finding the roots of the equation. |
1.
Real roots. Moreover. x1 is not equal to x2 The situation occurs when D>0 and A is not equal to 0. 2.
The real roots are the same. x1 equals x2 3.
Two complex roots. x1=d+ei, x2=d-ei, where i=-(1) 1/2 5.
The equation has countless solutions. 6.
The equation has no solutions. |
To consolidate the algorithm, here are a few more illustrative examples of solutions to quadratic equations.
Example 1. Solving an ordinary quadratic equation with different real roots.
x 2 + 3x -10 = 0
In this equation
A=1, B=3, C=-10
D=B 2 -4*A*C = 9-4*1*(-10) = 9+40 = 49
We will denote the square root as the number 1/2!
x1=(-B+D 1/2)/2A = (-3+7)/2 = 2
x2=(-B-D 1/2)/2A = (-3-7)/2 = -5
To check, let's substitute:
(x-2)*(x+5) = x2 -2x +5x – 10 = x2 + 3x -10
Example 2. Solving a quadratic equation with matching real roots.
x 2 – 8x + 16 = 0
A=1, B = -8, C=16
D = k 2 – AC = 16 – 16 = 0
X = -k/A = 4
Let's substitute
(x-4)*(x-4) = (x-4)2 = X 2 – 8x + 16
Example 3. Solving a quadratic equation with complex roots.
13x 2 – 4x + 1 = 0
A=1, B = -4, C=9
D = b 2 – 4AC = 16 – 4*13*1 = 16 - 52 = -36
The discriminant is negative – the roots are complex.
X1=(-B+D 1/2)/2A = (4+6i)/(2*13) = 2/13+3i/13
x2=(-B-D 1/2)/2A = (4-6i)/(2*13) = 2/13-3i/13
, where I is the square root of -1
Here are actually all the possible cases of solving quadratic equations.
We hope that our online calculator will be very useful for you.
If the material was useful, you can