We will analyze two types of solving systems of equations:
1. Solution of the system by the substitution method.
2. Solution of the system by term-by-term addition (subtraction) of the equations of the system.
In order to solve the system of equations substitution method you need to follow a simple algorithm:
1. We express. From any equation, we express one variable.
2. Substitute. We substitute in another equation instead of the expressed variable, the resulting value.
3. We solve the resulting equation with one variable. We find a solution to the system.
To solve system by term-by-term addition (subtraction) need:
1. Select a variable for which we will make the same coefficients.
2. We add or subtract the equations, as a result we get an equation with one variable.
3. We solve the resulting linear equation. We find a solution to the system.
The solution of the system is the intersection points of the graphs of the function.
Let us consider in detail the solution of systems using examples.
Example #1:
Let's solve by the substitution method
Solving the system of equations by 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, hence it turns out that it is easiest to express the variable x from the second equation.
x=3+10y
2. After expressing, we substitute 3 + 10y in the first equation instead of the variable x.
2(3+10y)+5y=1
3. We solve the resulting equation with one variable.
2(3+10y)+5y=1 (open brackets)
6+20y+5y=1
25y=1-6
25y=-5 |: (25)
y=-5:25
y=-0.2
The solution of the system of equations 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 paragraph where we expressed we substitute y there.
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 by term-by-term addition (subtraction).
Solving a system of equations by the addition method3x-2y=1 (1 equation)
2x-3y=-10 (2nd equation)
1. Select a variable, let's say we select 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. Multiply the first equation by 2 and the second by 3 to get overall ratio 6.
3x-2y=1 |*2
6x-4y=2
2x-3y=-10 |*3
6x-9y=-30
2. From the first equation, subtract the second to get rid of the variable x. We solve the linear equation.
__6x-4y=2
5y=32 | :5
y=6.4
3. Find x. We substitute the found y in any of the equations, let's say in 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 point of intersection will be x=4.6; y=6.4
Answer: (4.6; 6.4)
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The free calculator offered to your attention has a rich arsenal of possibilities for mathematical calculations. It allows you to use the online calculator in various fields activities: educational, professional and commercial. Of course, the use of an online calculator is especially popular with students and schoolchildren, it makes it much easier for them to perform a variety of calculations.
However, the calculator can be a useful tool in some areas of business and for people of different professions. Of course, the need to use a calculator in business or work is determined primarily by the type of activity itself. If business and profession are associated with constant calculations and calculations, then it is worth trying out an electronic calculator and assessing the degree of its usefulness for a particular business.
This online calculator can
- Correctly execute standard mathematical functions written in one line like - 12*3-(7/2) and can handle numbers larger than we count huge numbers in an online calculator. We don’t even know how to call such a number correctly ( there are 34 characters and this is not the limit at all).
- Except tangent, cosine, sinus and other standard functions - the calculator supports calculation operations arc tangent, arc tangent and others.
- Available in the arsenal logarithms, factorials and other cool features
- This online calculator can make charts!!!
To plot graphs, the service uses a special button (gray graph is drawn) or a literal representation of this function (Plot). To build a graph in an online calculator, just write a function: plot(tan(x)),x=-360..360.
We took the simplest plot for the tangent, and after the decimal point, we indicated the range of the X variable from -360 to 360.
You can build absolutely any function, with any number of variables, for example: plot(cos(x)/3z, x=-180..360,z=4) Or even more complex than you can think of. We pay attention to the behavior of the variable X - the interval from and to is indicated using two points.
The only negative (although it is difficult to call it a negative) of this online calculator this is that he does not know how to build spheres and other three-dimensional figures - only a plane.
How to work with the Math Calculator
1. The display (calculator screen) displays the entered expression and the result of its calculation in ordinary characters, as we write on paper. This field is simply for viewing the current operation. The entry is displayed on the display as you type a mathematical expression in the input line.
2. The expression input field is intended for writing the expression to be calculated. It should be noted here that the mathematical symbols used in computer programs do not always match those that we usually use on paper. In the overview of each function of the calculator, you will find the correct designation for a particular operation and examples of calculations in the calculator. On this page below is a list of all possible operations in the calculator, also indicating their correct spelling.
3. Toolbar - These are calculator buttons that replace the manual input of mathematical symbols that indicate the corresponding operation. Some calculator buttons (additional functions, unit converter, solution of matrices and equations, graphs) supplement the taskbar with new fields where data for a specific calculation is entered. The "History" field contains examples of writing mathematical expressions, as well as your six most recent entries.
Please note that when you press the buttons for calling additional functions, the converter of values, solving matrices and equations, plotting graphs, the entire calculator panel shifts up, covering part of the display. Fill in the required fields and press the "I" key (highlighted in red in the figure) to see the display in full size.
4. The numeric keypad contains numbers and arithmetic signs. The "C" button deletes the entire entry in the expression input field. To delete characters one by one, you need to use the arrow to the right of the input line.
Try to always close brackets at the end of an expression. For most operations, this is not critical, the online calculator will calculate everything correctly. However, in some cases errors are possible. For example, when raising to a fractional power, unclosed brackets will cause the denominator of the fraction in the exponent to go to the denominator of the base. On the display, the closing bracket is indicated in pale gray, it must be closed when the recording is completed.
Key | Symbol | Operation |
---|---|---|
pi | pi | constant pi |
e | e | Euler number |
% | % | Percent |
() | () | Open/Close Brackets |
, | , | Comma |
sin | sin(?) | Sine of an angle |
cos | cos(?) | Cosine |
tan | tan(y) | Tangent |
sinh | sinh() | Hyperbolic sine |
cash | cosh() | Hyperbolic cosine |
tanh | tanh() | Hyperbolic tangent |
sin-1 | asin() | Inverse sine |
cos-1 | acos() | inverse cosine |
tan-1 | atan() | inverse tangent |
sinh-1 | asinh() | Inverse hyperbolic sine |
cosh-1 | acosh() | Inverse hyperbolic cosine |
tanh-1 | atanh() | Inverse hyperbolic tangent |
x2 | ^2 | Squaring |
x 3 | ^3 | Cube |
x y | ^ | Exponentiation |
10 x | 10^() | Exponentiation in base 10 |
e x | exp() | Exponentiation of the Euler number |
vx | sqrt(x) | Square root |
3vx | sqrt3(x) | 3rd degree root |
yvx | square(x,y) | root extraction |
log 2 x | log2(x) | binary logarithm |
log | log(x) | Decimal logarithm |
ln | log(x) | natural logarithm |
log y x | log(x,y) | Logarithm |
I / II | Minimize/Call additional functions | |
unit | Unit converter | |
matrix | matrices | |
solve | Equations and systems of equations | |
Plotting | ||
Additional functions (call with II key) | ||
mod | mod | Division with remainder |
! | ! | Factorial |
i/j | i/j | imaginary unit |
Re | Re() | Selection of the whole real part |
Im | Im() | Exclusion of the real part |
|x| | abs() | The absolute value of a number |
Arg | arg() | Function argument |
nCr | ncr() | Binomial coefficient |
gcd | gcd() | GCD |
lcm | lcm() | NOC |
sum | sum() | The sum value of all solutions |
fac | factorize() | Prime factorization |
diff | diff() | Differentiation |
Deg | degrees | |
Rad | radians |
I. ax 2 \u003d 0 – incomplete quadratic equation (b=0, c=0 ). Solution: x=0. Answer: 0.
Solve equations.
2x·(x+3)=6x-x 2 .
Solution. Expand the brackets by multiplying 2x for each term in brackets:
2x2 +6x=6x-x2 ; moving the terms from the right side to the left side:
2x2 +6x-6x+x2=0; Here are similar terms:
3x 2 =0, hence x=0.
Answer: 0.
II. ax2+bx=0 –incomplete quadratic equation (s=0 ). Solution: x (ax+b)=0 → x 1 =0 or ax+b=0 → x 2 =-b/a. Answer: 0; -b/a.
5x2 -26x=0.
Solution. Take out the common factor X for brackets:
x(5x-26)=0; each factor can be zero:
x=0 or 5x-26=0→ 5x=26, divide both sides of the equality by 5 and we get: x \u003d 5.2.
Answer: 0; 5,2.
Example 3 64x+4x2=0.
Solution. Take out the common factor 4x for brackets:
4x(16+x)=0. We have three factors, 4≠0, therefore, or x=0 or 16+x=0. From the last equality we get x=-16.
Answer: -16; 0.
Example 4(x-3) 2 +5x=9.
Solution. Applying the formula for the square of the difference of two expressions, open the brackets:
x 2 -6x+9+5x=9; transform to the form: x 2 -6x+9+5x-9=0; Here are similar terms:
x2-x=0; endure X outside the brackets, we get: x (x-1)=0. From here or x=0 or x-1=0→ x=1.
Answer: 0; 1.
III. ax2+c=0 –incomplete quadratic equation (b=0 ); Solution: ax 2 \u003d -c → x 2 \u003d -c / a.
If a (-c/a)<0 , then there are no real roots. If a (-s/a)>0
Example 5 x 2 -49=0.
Solution.
x 2 \u003d 49, from here x=±7. Answer:-7; 7.
Example 6 9x2-4=0.
Solution.
It is often required to find the sum of squares (x 1 2 + x 2 2) or the sum of cubes (x 1 3 + x 2 3) of the roots of a quadratic equation, less often - the sum of the reciprocals of the squares of the roots or the sum of arithmetic square roots from the roots of the quadratic equation:
Vieta's theorem can help with this:
x 2 +px+q=0
x 1 + x 2 \u003d-p; x 1 ∙ x 2 \u003d q.
Express through p and q:
1) the sum of the squares of the roots of the equation x2+px+q=0;
2) the sum of the cubes of the roots of the equation x2+px+q=0.
Solution.
1) Expression x 1 2 + x 2 2 obtained by squaring both sides of the equation x 1 + x 2 \u003d-p;
(x 1 +x 2) 2 \u003d (-p) 2; open the brackets: x 1 2 +2x 1 x 2 + x 2 2 =p 2; we express the desired amount: x 1 2 +x 2 2 \u003d p 2 -2x 1 x 2 \u003d p 2 -2q. We have a useful equation: x 1 2 +x 2 2 \u003d p 2 -2q.
2) Expression x 1 3 + x 2 3 represent by the formula of the sum of cubes in the form:
(x 1 3 +x 2 3)=(x 1 +x 2)(x 1 2 -x 1 x 2 +x 2 2)=-p (p 2 -2q-q)=-p (p 2 -3q).
Another useful equation: x 1 3 + x 2 3 \u003d-p (p 2 -3q).
Examples.
3) x 2 -3x-4=0. Without solving the equation, calculate the value of the expression x 1 2 + x 2 2.
Solution.
x 1 + x 2 \u003d-p \u003d 3, and the work x 1 ∙x 2 \u003d q \u003din example 1) equality:
x 1 2 +x 2 2 \u003d p 2 -2q. We have -p=x 1 +x 2 = 3 → p 2 =3 2 =9; q= x 1 x 2 = -4. Then x 1 2 + x 2 2 =9-2 (-4)=9+8=17.
Answer: x 1 2 + x 2 2 =17.
4) x 2 -2x-4=0. Calculate: x 1 3 +x 2 3 .
Solution.
By Vieta's theorem, the sum of the roots of this reduced quadratic equation x 1 + x 2 \u003d-p \u003d 2, and the work x 1 ∙x 2 \u003d q \u003d-four. Let us apply what we have obtained ( in example 2) equality: x 1 3 +x 2 3 \u003d-p (p 2 -3q) \u003d 2 (2 2 -3 (-4))=2 (4+12)=2 16=32.
Answer: x 1 3 + x 2 3 =32.
Question: what if we are given a non-reduced quadratic equation? Answer: it can always be “reduced” by dividing term by term by the first coefficient.
5) 2x2 -5x-7=0. Without solving, calculate: x 1 2 + x 2 2.
Solution. We are given a complete quadratic equation. Divide both sides of the equation by 2 (the first coefficient) and get the following quadratic equation: x 2 -2.5x-3.5 \u003d 0.
By Vieta's theorem, the sum of the roots is 2,5 ; the product of the roots is -3,5 .
We solve in the same way as an example 3) using the equality: x 1 2 +x 2 2 \u003d p 2 -2q.
x 1 2 +x 2 2 =p 2 -2q= 2,5 2 -2∙(-3,5)=6,25+7=13,25.
Answer: x 1 2 + x 2 2 = 13,25.
6) x2 -5x-2=0. Find:
Let us transform this equality and, by replacing the sum of the roots in terms of the Vieta theorem, -p, and the product of the roots through q, we get another useful formula. When deriving the formula, we used equality 1): x 1 2 +x 2 2 \u003d p 2 -2q.
In our example x 1 + x 2 \u003d -p \u003d 5; x 1 ∙x 2 \u003d q \u003d-2. Substitute these values into the resulting formula:
7) x 2 -13x+36=0. Find:
Let's transform this sum and get a formula by which it will be possible to find the sum of arithmetic square roots from the roots of a quadratic equation.
We have x 1 + x 2 \u003d -p \u003d 13; x 1 ∙x 2 \u003d q \u003d 36. Substitute these values into the derived formula:
Advice : always check the possibility of finding the roots of the quadratic equation by suitable way, after all 4 reviewed useful formulas allow you to quickly complete the task, first of all, in cases where the discriminant is an “inconvenient” number. In all simple cases, find the roots and operate on them. For example, in the last example, we select the roots using the Vieta theorem: the sum of the roots should be equal to 13 , and the product of the roots 36 . What are these numbers? Of course, 4 and 9. Now calculate the sum of the square roots of these numbers: 2+3=5. That's it!
I. Vieta's theorem for the reduced quadratic equation.
The sum of the roots of the reduced quadratic equation x 2 +px+q=0 is equal to the second coefficient, taken with the opposite sign, and the product of the roots is equal to the free term:
x 1 + x 2 \u003d-p; x 1 ∙ x 2 \u003d q.
Find the roots of the given quadratic equation using Vieta's theorem.
Example 1) x 2 -x-30=0. This is the reduced quadratic equation ( x 2 +px+q=0), the second coefficient p=-1, and the free term q=-30. First, make sure that the given equation has roots, and that the roots (if any) will be expressed as integers. For this it is sufficient that the discriminant be full square whole number.
Finding the discriminant D=b 2 - 4ac=(-1) 2 -4∙1∙(-30)=1+120=121= 11 2 .
Now, according to the Vieta theorem, the sum of the roots must be equal to the second coefficient, taken with the opposite sign, i.e. ( -p), and the product is equal to the free term, i.e. ( q). Then:
x 1 + x 2 =1; x 1 ∙ x 2 \u003d -30. We need to choose such two numbers so that their product is equal to -30 , and the sum is unit. These are the numbers -5 and 6 . Answer: -5; 6.
Example 2) x 2 +6x+8=0. We have the reduced quadratic equation with the second coefficient p=6 and free member q=8. Make sure that there are integer roots. Let's find the discriminant D1 D1=3 2 -1∙8=9-8=1=1 2 . The discriminant D 1 is the perfect square of the number 1 , so the roots of this equation are integers. We choose the roots according to the Vieta theorem: the sum of the roots is equal to –p=-6, and the product of the roots is q=8. These are the numbers -4 and -2 .
Actually: -4-2=-6=-p; -4∙(-2)=8=q. Answer: -4; -2.
Example 3) x 2 +2x-4=0. In this reduced quadratic equation, the second coefficient p=2, and the free term q=-4. Let's find the discriminant D1, since the second coefficient is an even number. D1=1 2 -1∙(-4)=1+4=5. The discriminant is not a perfect square of a number, so we do conclusion: the roots of this equation are not integers and cannot be found using Vieta's theorem. So, we solve this equation, as usual, according to the formulas (in this case formulas). We get:
Example 4). Write a quadratic equation using its roots if x 1 \u003d -7, x 2 \u003d 4.
Solution. The desired equation will be written in the form: x 2 +px+q=0, moreover, based on the Vieta theorem –p=x1 +x2=-7+4=-3 →p=3; q=x 1 ∙x 2=-7∙4=-28 . Then the equation will take the form: x2 +3x-28=0.
Example 5). Write a quadratic equation using its roots if:
II. Vieta's theorem for the complete quadratic equation ax2+bx+c=0.
The sum of the roots is minus b divided by a, the product of the roots is With divided by a:
x 1 + x 2 \u003d -b / a; x 1 ∙ x 2 \u003d c / a.
Example 6). Find the sum of the roots of a quadratic equation 2x2 -7x-11=0.
Solution.
We are convinced that this equation will have roots. To do this, it is enough to write an expression for the discriminant, and without calculating it, just make sure that the discriminant is greater than zero. D=7 2 -4∙2∙(-11)>0 . And now let's use theorem Vieta for complete quadratic equations.
x 1 + x 2 =-b:a=- (-7):2=3,5.
Example 7). Find the product of the roots of a quadratic equation 3x2 +8x-21=0.
Solution.
Let's find the discriminant D1, since the second coefficient ( 8 ) is an even number. D1=4 2 -3∙(-21)=16+63=79>0 . The quadratic equation has 2 root, according to the Vieta theorem, the product of the roots x 1 ∙ x 2 \u003d c: a=-21:3=-7.
I. ax 2 +bx+c=0 is a general quadratic equation
Discriminant D=b 2 - 4ac.
If a D>0, then we have two real roots:
If a D=0, then we have a single root (or two equal roots) x=-b/(2a).
If D<0, то действительных корней нет.
Example 1) 2x2 +5x-3=0.
Solution. a=2; b=5; c=-3.
D=b 2-4ac=5 2 -4∙2∙(-3)=25+24=49=7 2 >0; 2 real roots.
4x2 +21x+5=0.
Solution. a=4; b=21; c=5.
D=b 2-4ac=21 2 - 4∙4∙5=441-80=361=19 2 >0; 2 real roots.
II. ax2+bx+c=0 – special quadratic equation for an even second
coefficient b
Example 3) 3x2 -10x+3=0.
Solution. a=3; b\u003d -10 (even number); c=3.
Example 4) 5x2-14x-3=0.
Solution. a=5; b= -14 (even number); c=-3.
Example 5) 71x2 +144x+4=0.
Solution. a=71; b=144 (even number); c=4.
Example 6) 9x 2 -30x+25=0.
Solution. a=9; b\u003d -30 (even number); c=25.
III. ax2+bx+c=0 – quadratic equation private type, provided: a-b+c=0.
The first root is always minus one and the second root is minus With divided by a:
x 1 \u003d -1, x 2 \u003d - c / a.
Example 7) 2x2+9x+7=0.
Solution. a=2; b=9; c=7. Let's check the equality: a-b+c=0. We get: 2-9+7=0 .
Then x 1 \u003d -1, x 2 \u003d -c / a \u003d -7 / 2 \u003d -3.5. Answer: -1; -3,5.
IV. ax2+bx+c=0 – quadratic equation of a particular form under the condition : a+b+c=0.
The first root is always equal to one, and the second root is equal to With divided by a:
x 1 \u003d 1, x 2 \u003d c / a.
Example 8) 2x2 -9x+7=0.
Solution. a=2; b=-9; c=7. Let's check the equality: a+b+c=0. We get: 2-9+7=0 .
Then x 1 \u003d 1, x 2 \u003d c / a \u003d 7/2 \u003d 3.5. Answer: 1; 3,5.
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Quadratic equations are studied in grade 8, so there is nothing complicated here. The ability to solve them is essential.
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 methods of solving, we note that all quadratic equations can be divided into three classes:
- Have no roots;
- They have exactly one root;
- They have two different roots.
This is an important difference between quadratic and linear equations, 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 .
This formula must be known 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 think. Take a look at the examples and you will understand everything yourself:
A task. How many roots do quadratic equations have:
- x 2 - 8x + 12 = 0;
- 5x2 + 3x + 7 = 0;
- x 2 − 6x + 9 = 0.
We write 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 the same way:
a = 5; b = 3; c = 7;
D \u003d 3 2 - 4 5 7 \u003d 9 - 140 \u003d -131.
The discriminant is negative, there are no roots. The last equation remains:
a = 1; b = -6; c = 9;
D = (−6) 2 − 4 1 9 = 36 − 36 = 0.
Discriminant zero- the root will be one.
Note that coefficients have been written out for each equation. Yes, it's long, yes, it's tedious - but you won't mix up the odds and don't make stupid mistakes. Choose for yourself: speed or quality.
By the way, if you “fill your hand”, after a while you will no longer need to write out 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 so many.
The roots of a quadratic equation
Now let's move on to the solution. If the discriminant D > 0, the roots can be found using the formulas:
The basic formula for the roots of a quadratic equation
When D = 0, you can use any of these formulas - you 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 - x2 = 0;
- x2 + 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 be able to count, there will be no problems. Most often, errors occur when negative coefficients are substituted into the formula. Here, again, the technique described above will help: look at the formula literally, paint each step - and get rid of mistakes very soon.
Incomplete quadratic equations
It happens that the quadratic equation is somewhat different from what is given in the definition. For example:
- x2 + 9x = 0;
- x2 − 16 = 0.
It is easy to see that one of the terms is missing in these equations. Such quadratic equations are even easier to solve than standard ones: they do not even need to calculate 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 \u003d c \u003d 0. In this case, the equation takes the form ax 2 \u003d 0. Obviously, such an equation has a single root: x \u003d 0.
Let's consider other cases. Let b \u003d 0, then we get an incomplete quadratic equation of the form ax 2 + c \u003d 0. Let's slightly transform it:
Since the arithmetic square root exists only from a non-negative number, the last equality only makes sense when (−c / a ) ≥ 0. Conclusion:
- If an incomplete quadratic equation of the form ax 2 + c = 0 satisfies the inequality (−c / a ) ≥ 0, there will be two roots. The formula is given above;
- If (−c / a )< 0, корней нет.
As you can see, the discriminant was not required - in incomplete quadratic equations no complex calculations at all. In fact, it is not even necessary to remember the inequality (−c / a ) ≥ 0. It is enough to express the value of x 2 and see what is on the other side of the equal sign. If there positive number there will be two roots. If negative, there will be no roots at all.
Now let's deal with 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 factorize the polynomial:
Taking the common factor out of the bracketThe product is equal to zero when at least one of the factors is equal to zero. This is where the roots come from. In conclusion, we will analyze several of these equations:
A task. Solve quadratic equations:
- x2 − 7x = 0;
- 5x2 + 30 = 0;
- 4x2 − 9 = 0.
x 2 − 7x = 0 ⇒ x (x − 7) = 0 ⇒ x 1 = 0; x2 = −(−7)/1 = 7.
5x2 + 30 = 0 ⇒ 5x2 = -30 ⇒ x2 = -6. There are no roots, because the 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 \u003d -1.5.