To solve an equation means to find such values of the unknown for which the equality will be true.
Equation solution
- Let's represent the equation in the following form:
2x * x - 3 * x = 0.
- We see that the terms of the equation on the left side have a common factor x. Let's take it out of brackets and write:
x * (2x - 3) = 0.
- The resulting expression is the product of the factors x and (2x - 3). Recall that the product is equal to 0 if at least one of the factors is equal to 0. So, we can write the equalities:
x = 0 or 2x - 3 = 0.
- So one of the roots of the original equation is x 1 = 0.
- Find the second root by solving the equation 2x - 3 = 0.
In this expression, 2x is the minuend, 3 is the subtrahend, and 0 is the difference. To find the minuend, you need to add the subtrahend to the difference:
In the last expression, 2 and x are factors, 3 is the product. To find the unknown factor, you need to divide the product by the known factor:
Thus, we have found the second root of the equation: x 2 \u003d 1.5.
Checking the correctness of the solution
In order to find out if the equation is solved correctly, it is necessary to substitute the numerical values of x into it and perform the necessary arithmetic operations. If as a result of calculations it turns out that the left and right parts of the expression have the same value, then the equation is solved correctly.
Let's check:
- Let's calculate the value of the original expression at x 1 = 0 and get:
2 * 0 2 - 3 * 0 = 0,
0 = 0, right.
- Let's calculate the value of the expression at x 2 = 0 and get:
2 * 1,5 2 - 3 * 1,5 = 0,
2 * 2,25 - 4,5 = 0,
0 = 0, right.
- So the equation is correct.
Answer: x 1 \u003d 0, x 2 \u003d 1.5.
Solve the Equation X 2 +(1-x) 2 =x
Prove that there are no integers that increase by a factor of 5 by rearranging the initial digit to the end.
In a certain kingdom, every two are either friends or enemies. Everyone can at some point quarrel with all friends and make peace with all enemies. It turned out that every three people can become friends in this way. Prove that then all people in this kingdom can become friends.
In a triangle, one of the medians is perpendicular to one of the bisectors. Prove that one of the sides of this triangle is twice the other.
Tasks for holding a district (city) Olympiad for schoolchildren in mathematics.
In shooting from a target, the athlete knocked out only 8.9 and 10 points each. In total, having made more than 11 shots, he knocked out exactly 100 points. How many shots did the athlete make, and what were the hits?
Prove the truth of the inequality:
3. Solve the equation:
Find a three-digit number that decreases by a factor of 7 after being crossed out in it average figure.
IN triangle ABC bisectors are drawn from vertices A and B. Then straight lines are drawn from vertex C parallel to these bisectors. The points D and E of the intersection of these lines with the bisectors are connected. It turned out that the lines DE and AB are parallel. Prove that triangle ABC is isosceles.
Tasks for holding a district (city) Olympiad for schoolchildren in mathematics.
Solve the system of equations:
On the sides AB and AD of the parallelogram ABCD, points E and K are taken, respectively, so that the segment EK is parallel to the diagonal BD. Prove that the areas of triangles ALL and SDO are equal.
They decided to seat a group of tourists in buses so that each bus had the same number of passengers. At first, 22 people were put on each bus, but it turned out that it was not possible to put one tourist in this case. When one bus left empty, then all the tourists boarded the remaining buses equally. How many buses were originally there and how many tourists were in the group, if it is known that no more than 32 people can fit in each bus?
Tasks for holding a district (city) Olympiad for schoolchildren in mathematics.
Solve the system of equations:
Prove that four distances from a point of a circle to a vertex of a square inscribed in it cannot simultaneously be rational numbers.
Possible solutions to problems
1. Answer: x=1, x=0.5
From the permutation of the initial digit to the end, the significance of the number will not change. In this case, according to the condition of the problem, they should get a number that is 5 times greater than the first number. Therefore, the first digit of the desired number should be equal to 1 and only 1. (because if the first digit is 2 or more, then the value will change, 2 * 5 = 10). When rearranging 1 to the end, the resulting number ends in 1, therefore it is not divisible by 5.
It follows from the condition that if A and B are friends, then C is either their common enemy or a common friend (otherwise the three of them cannot be reconciled). Let us take all the friends of person A. It follows from what has been said that they are all friendly with each other and are at enmity with the rest. Let A and his friends now take turns quarreling with friends and making peace with enemies. After that, everyone will be friends.
Indeed, let A be the first to quarrel with his friends and make peace with his enemies, but then each of his former friends will put up with him, and former enemies will remain friends. So, all people turn out to be friends of A, and, consequently, friends among themselves.
The number 111 is divisible by 37, so the sum is also divisible by 37.
By condition, the number is divisible by 37, so the sum
Divisible by 37.
Note that the indicated median and bisector cannot come out of the same vertex, since otherwise the angle at this vertex would be greater than 180 0 . Let now in the triangle ABC the bisector AD and the median CE intersect at point F. Then AF is the bisector and the height in the triangle ACE, which means that this triangle is isosceles (AC \u003d AE), and since CE is the median, then AB \u003d 2AE and, therefore, AB = 2AC.
Possible solutions to problems
1. Answer: 9 shots for 8 points,
2 shots for 9 points,
1 shot for 10 points.
Let x shots were made by an athlete, knocking out 8 points, y shots for 9 points, z shots for 10 points. Then you can create a system:
Using the first equation of the system, we write:
It follows from this system that x+ y+ z=12
Multiply the second equation by (-8) and add it to the first. We get that y+2 z=4 , where y=4-2 z, y=2(2- z) . Hence, at is an even number, i.e. y=2t, Where .
Hence,
3. Answer: x = -1/2, x = -4
After reducing the fractions to the same denominator, we get
4. Answer: 105
Denote by x, y, z respectively the first, second and third digit of the desired three digit number. Then it can be written as . Crossing out the middle digit will result in a two-digit number. According to the condition of the problem, i.e. unknown numbers x, y, z satisfy the equation
7(10 x+ z)=100 x+10 y+ x, which after reduction of similar terms and abbreviations takes the form 3 z=15 x+5 y.
From this equation it follows that z must be divisible by 5 and must be positive, since by condition . Therefore, z = 5, and the numbers x, y satisfy the equation 3 = 3x + y, which, by virtue of the condition, has a unique solution x = 1, y = 0. Therefore, the condition of the problem satisfies singular 105.
Let F denote the point at which the lines AB and CE intersect. Since the lines DB and CF are parallel, then . Since BD is the bisector of angle ABC, we conclude that . It follows from here that , i.e. triangle BCF is isosceles and BC=BF. But it follows from the condition that the quadrilateral BDEF is a parallelogram. Therefore BF = DE, and therefore BC = DE. It can be proved similarly that AC = DE. This leads to the required equality.
Possible solutions to problems
1.
From here (x + y) 2 = 1 , i.e. x + y = 1 or x + y = -1.
Let's consider two cases.
A) x + y = 1. Substituting x = 1 - y
b) x + y = -1. After substitution x=-1-y
So, only the following four pairs of numbers can be solutions to the system: (0;1), (2;-1), (-1;0), (1;-2). By substituting into the equations of the original system, we make sure that each of these four pairs is a solution to the system.
Triangles CDF and BDF have a common base FD and equal heights, since lines BC and AD are parallel. Therefore, their areas are equal. Similarly, the areas of triangles BDF and BDE are equal, since line BD is parallel to line EF. And the areas of triangles BDE and BCE are equal, since AB is parallel to CD. This implies the required equality of the areas of triangles CDF and BCE.
Considering the domain of definition of the function, we will build a graph.
Using the formula 
perform further transformations
Applying addition formulas and performing further transformations, we obtain
5. Answer: 24 buses, 529 tourists.
Denote by k initial number of buses. From the condition of the problem it follows that and that the number of all tourists is equal to 22 k +1 . After the departure of one bus, all tourists were seated in the remaining (k-1) buses. Therefore, the number 22 k +1 should be divided by k-1. Thus, the problem was reduced to the determination of all integers for which the number
Is an integer and satisfies the inequality (the number n is equal to the number of tourists seated in each bus, and according to the condition of the problem, the bus can accommodate no more than 32 passengers).
A number will only be an integer if the number is an integer. The latter is only possible with k=2 and at k=24 .
If k=2 , That n=45.
And if k=24 , That n=23.
From this and from the condition, we obtain that only k=24 satisfies all conditions of the problem.
Therefore, initially there were 24 buses, and the number of all tourists is n(k-1)=23*23=529
Possible solutions to problems
1. Answer:
Then the equation will take the form:
Got a quadratic equation for R.
2. Answer: (0;1), (2;-1), (-1;0), (1;-2)
Adding the equations of the system, we get , or
From here (x + y) 2 = 1 , i.e. x + y = 1 or x + y = -1.
Let's consider two cases.
A) x + y = 1. Substituting x = 1 - y into the first equation of the system, we get
b) x + y = -1. After substitution x=-1-y into the first equation of the system, we get or
In this article, we will learn to solve bi quadratic equations.
So, what kind of equations are called biquadratic?
All equations of the form ah 4+
bx
2
+
c
= 0
, Where a ≠ 0, which are square with respect to x 2 , and are called biquadratic equations. As you can see, this entry is very similar to the quadratic equation, so we will solve biquadratic equations using the formulas that we used when solving the quadratic equation.
Only we will need to introduce a new variable, that is, we denote x 2 another variable, for example, at or t (or any other letter of the Latin alphabet).
For example, solve the equation x 4 + 4x 2 - 5 = 0.
Denote x 2
through at
(x 2 = y
) and get the equation y 2 + 4y - 5 = 0.
As you can see, you already know how to solve such equations.
We solve the resulting equation:
D \u003d 4 2 - 4 (- 5) \u003d 16 + 20 \u003d 36, √D \u003d √36 \u003d 6.
y 1 = (‒ 4 - 6)/2= - 10 /2 = - 5,
y 2 \u003d (- 4 + 6) / 2 \u003d 2 / 2 \u003d 1.
Let's go back to our variable x.
We got that x 2 \u003d - 5 and x 2 \u003d 1.
We note that the first equation has no solutions, and the second gives two solutions: x 1 = 1 and x 2 = –1. Be careful not to lose negative root(most often they get the answer x = 1, and this is not correct).
Answer:- 1 and 1.
To better understand the topic, let's look at a few examples.
Example 1 Solve the Equation 2x4 - 5x2 + 3 = 0.
Let x 2 \u003d y, then 2y 2 - 5y + 3 \u003d 0.
D = (‒ 5) 2 - 4 2 3 = 25 - 24 = 1, √D = √1 = 1.
y 1 \u003d (5 - 1) / (2 2) \u003d 4 / 4 \u003d 1, y 2 \u003d (5 + 1) / (2 2) \u003d 6 / 4 \u003d 1.5.
Then x 2 \u003d 1 and x 2 \u003d 1.5.
We get x 1 \u003d -1, x 2 \u003d 1, x 3 \u003d - √1.5, x 4 \u003d √1.5.
Answer: ‒1; 1; ‒ √1,5; √1,5.
Example 2 Solve the Equation 2 x 4 + 5 x 2 + 2 = 0.
2y 2 + 5y + 2 = 0.
D = 5 2 - 4 2 2 = 25 - 16 = 9, √D = √9 = 3.
y 1 = (– 5 – 3)/(2 2) = – 8/4 = –2, y 2 = (–5 + 3)/(2 2) = – 2/4 = – 0.5.
Then x 2 = - 2 and x 2 = - 0.5. Note that none of these equations has a solution.
Answer: there are no solutions.
Incomplete biquadratic equations- it is when b = 0 (ax 4 + c = 0) or else c = 0
(ax 4 + bx 2 = 0) are solved like incomplete quadratic equations.
Example 3 solve the equation x 4 - 25x 2 = 0
We factorize, take x 2 out of brackets and then x 2 (x 2 - 25) = 0.
We get x 2 \u003d 0 or x 2 - 25 \u003d 0, x 2 \u003d 25.
Then we have roots 0; 5 and - 5.
Answer: 0; 5; – 5.
Example 4 solve the equation 5x 4 - 45 = 0.
x 2 = - √9 (no solutions)
x 2 \u003d √9, x 1 \u003d - 3, x 2 \u003d 3.
As you can see, knowing how to solve quadratic equations, you can cope with biquadratic ones.
If you still have questions, sign up for my lessons. Tutor Valentina Galinevskaya.
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