Solve the equation X 2 + (1-x) 2 = x

Prove that there are no integers that increase by 5 times from permutation of the initial digit to the end.

In a certain kingdom, every two are either friends or enemies. Each person 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 as large as the other.

Tasks for the regional (city) Olympiad for schoolchildren in mathematics.

In target shooting, the athlete knocked out only 8.9 and 10 points. In total, having made more than 11 shots, he knocked out exactly 100 points. How many shots did the athlete fire and what were the hits?

Prove the truth of the inequality:

3. Solve the equation:

Find a three-digit number that decreases by 7 times after crossing out the middle digit in it.

In triangle ABC, bisectors are drawn from vertices A and B. Then, straight lines parallel to these bisectors are drawn from vertex C. Points D and E of intersection of these lines with bisectors are connected. It turned out that straight lines DE and AB are parallel. Prove that triangle ABC is isosceles.

Tasks for the regional (city) Olympiad for schoolchildren in mathematics.

Solve the system of equations:

On the sides AB and HELL of the parallelogram AVSD, points E and K are taken, respectively, so that the segment EK is parallel to the diagonal VD. Prove that the areas of triangles ALL and SDK are equal.

It was decided to seat the group of tourists on 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 seat one tourist. When one bus left empty, all the tourists got on the remaining buses equally. How many buses were there initially and how many tourists were there in the group, if it is known that each bus can accommodate no more than 32 people?

Tasks for the regional (city) Olympiad for schoolchildren in mathematics.

Solve the system of equations:

Prove that the four distances from a point on a circle to the top of a square inscribed in it cannot be rational numbers at the same time.

Possible solutions to problems

1. Answer: x = 1, x = 0.5

From rearranging the starting digit to the end, the meaning of the number will not change. At the same time, according to the condition of the problem, a number should be obtained that is 5 times greater than the first number. Therefore, the first digit of the required number should be equal to 1 and only 1. (since if the first digit is 2 or more, then the value will change, 2 * 5 = 10). When rearranging 1 at 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 will not be reconciled). Let us take all the friends of man A. It follows from what has been said that they are all friendly with each other and are at enmity with the rest. Now let A and his friends 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 reconcile with him, and former enemies will remain friends. So, all people turn out to be friends of A, and, therefore, friends with each other.

The number 111 is divisible by 37, so the named amount is also divisible by 37.

By condition, the number is divisible by 37, therefore the sum

Divisible by 37.

Note that the indicated median and bisector cannot go out of one vertex, since otherwise the angle at this vertex would be greater than 180 0. Now let 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 this triangle is isosceles (AC = AE), and since CE is the median, then AB = 2AE and, therefore, AB = 2AC.

Possible solutions to problems

1. Answer: 9 shots of 8 points,

2 shots of 9 points,

1 shot for 10 points.

Let x shots were fired by an athlete, knocking out 8 points, y shots of 9 points, z shots of 10 points. Then you can compose 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 to the first. We get that y+2 z=4 , where y=4-2 z, y=2(2- z) ... Hence, at- an even number, i.e. y = 2t, where .

Hence,

3. Answer: x = -1/2, x = -4

After reducing the fractions to one denominator, we get

4. Answer: 105

Let us denote by x, y, z respectively, the first, second and third digits of the desired three-digit number. Then it can be written as. Crossing out the middle digit will result in a two-digit number. By 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 reducing similar terms and abbreviations takes the form 3 z=15 x+5 y.

It follows from this equation that z must be divisible by 5 and must be positive, as 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 is satisfied by the only number 105.

Let us denote by the letter F the point at which lines AB and CE intersect. Since straight lines DB and CF are parallel, then. Since BD is the bisector of the angle ABC, we conclude that. Hence it follows 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 is proved in a similar way that AC = DE. This results in 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. Hence the required equality of the areas of triangles CDF and BCE follows.

Considering the domain of the function, let's build a graph.

Using the formula perform further transformations

Applying the addition formulas and performing further transformations, we obtain

5. Answer: 24 buses, 529 tourists.

Let us denote by k the initial number of buses. It follows from the problem statement that and that the number of all tourists is 22 k +1 ... After the departure of one bus, all the tourists were seated in the remaining (k-1) buses. Therefore, the number 22 k +1 should be divisible by k-1... Thus, the problem was reduced to determining all integers for which the number

It is an integer and satisfies the inequality (the number n is equal to the number of tourists seated in each bus, and by the condition of the problem the bus can accommodate no more than 32 passengers).

The number will be whole only when the number is whole. The latter is possible only when k=2 and at k=24 .

If k=2 , then n = 45.

And if k=24 , then n = 23.

From this and the condition, we obtain that only k=24 satisfies all the conditions of the problem.

Therefore, there were originally 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:

We got a quadratic equation with respect to R.

2. Answer: (0; 1), (2; -1), (-1; 0), (1; -2)

Adding the equations of the system, we obtain, 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 obtain

b) x + y = -1... After substitution x = -1-y into the first equation of the system, we obtain either

Solving an equation means finding 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 the parentheses and write it down:

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. Hence, 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 decreasing, 3 is the subtracted, 0 is the difference. To find the subtracted, it is necessary to add the subtracted to the difference:

In the last expression, 2 and x are factors, 3 is the product. To find an unknown factor, you need to divide the product by a known factor:

Thus, we found the second root of the equation: x 2 = 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 sides of the expression have the same value, then the equation is solved correctly.

Let's check:

• We calculate the value of the original expression at x 1 = 0 and get:

2 * 0 2 - 3 * 0 = 0,

0 = 0, correct.

• We 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, correct.

• This means that the equation is solved correctly.

Answer: x 1 = 0, x 2 = 1.5.

Quadratic equations.

Quadratic equation- general algebraic equation

where x is a free variable,

a, b, c, - coefficients, and

Expression called a square trinomial.

Methods for solving quadratic equations.

1. METHOD : Factoring the left side of the equation.

Let's solve the equation x 2 + 10x - 24 = 0... Let us factor the left-hand side:

x 2 + 10x - 24 = x 2 + 12x - 2x - 24 = x (x + 12) - 2 (x + 12) = (x + 12) (x - 2).

Therefore, the equation can be rewritten as follows:

(x + 12) (x - 2) = 0

Since the product is zero, at least one of its factors is zero. Therefore, the left-hand side of the equation vanishes at x = 2 and also for x = - 12... This means that the number 2 and - 12 are the roots of the equation x 2 + 10x - 24 = 0.

2. METHOD : Full square selection method.

Let's solve the equation x 2 + 6x - 7 = 0... Select a complete square on the left.

To do this, write the expression x 2 + 6x in the following form:

x 2 + 6x = x 2 + 2 x 3.

In the resulting expression, the first term is the square of the number x, and the second is the doubled product of x by 3. Therefore, to get a complete square, you need to add 3 2, since

x 2 + 2 x 3 + 3 2 = (x + 3) 2.

Now we transform the left side of the equation

x 2 + 6x - 7 = 0,

adding and subtracting 3 2. We have:

x 2 + 6x - 7 = x 2 + 2 x 3 + 3 2 - 3 2 - 7 = (x + 3) 2 - 9 - 7 = (x + 3) 2 - 16.

Thus, this equation can be written as follows:

(x + 3) 2 - 16 = 0, (x + 3) 2 = 16.

Hence, x + 3 - 4 = 0, x 1 = 1, or x + 3 = -4, x 2 = -7.

3. METHOD :Solving quadratic equations using the formula.

Multiply both sides of the equation

ax 2 + bx + c = 0, and ≠ 0

on 4а and sequentially we have:

4a 2 x 2 + 4abx + 4ac = 0,

((2ax) 2 + 2ax b + b 2) - b 2 + 4ac = 0,

(2ax + b) 2 = b 2 - 4ac,

2ax + b = ± √ b 2 - 4ac,

2ax = - b ± √ b 2 - 4ac,

Examples of.

a) Let's solve the equation: 4x 2 + 7x + 3 = 0.

a = 4, b = 7, c = 3, D = b 2 - 4ac = 7 2 - 4 4 3 = 49 - 48 = 1,

D> 0, two different roots;

Thus, in the case of a positive discriminant, i.e. at

b 2 - 4ac> 0, the equation ax 2 + bx + c = 0 has two distinct roots.

b) Let's solve the equation: 4x 2 - 4x + 1 = 0,

a = 4, b = - 4, c = 1, D = b 2 - 4ac = (-4) 2 - 4 4 1 = 16 - 16 = 0,

D = 0, one root;

So, if the discriminant is zero, i.e. b 2 - 4ac = 0, then the equation

ax 2 + bx + c = 0 has a single root,

v) Let's solve the equation: 2x 2 + 3x + 4 = 0,

a = 2, b = 3, c = 4, D = b 2 - 4ac = 3 2 - 4 2 4 = 9 - 32 = - 13, D< 0.

This equation has no roots.

So, if the discriminant is negative, i.e. b 2 - 4ac< 0 , the equation

ax 2 + bx + c = 0 has no roots.

Formula (1) for the roots of a quadratic equation ax 2 + bx + c = 0 allows you to find the roots any quadratic equation (if any), including reduced and incomplete. Formula (1) is expressed in words as follows: the roots of a quadratic equation are equal to a fraction, the numerator of which is equal to the second coefficient, taken with the opposite sign, plus minus the square root of the square of this coefficient without the quadruple product of the first coefficient by the free term, and the denominator is twice the first coefficient.

4. METHOD: Solving equations using Vieta's theorem.

As you know, the given quadratic equation has the form

x 2 + px + c = 0.(1)

Its roots satisfy Vieta's theorem, which for a = 1 has the form

x 1 x 2 = q,

x 1 + x 2 = - p

Hence, the following conclusions can be drawn (the signs of the roots can be predicted from the coefficients p and q).

a) If the consolidated term q given equation (1) is positive ( q> 0), then the equation has two roots of the same sign and this depends on the second coefficient p... If R< 0 , then both roots are negative if R< 0 , then both roots are positive.

For instance,

x 2 - 3x + 2 = 0; x 1 = 2 and x 2 = 1, because q = 2> 0 and p = - 3< 0;

x 2 + 8x + 7 = 0; x 1 = - 7 and x 2 = - 1, because q = 7> 0 and p = 8> 0.

b) If the free term q given equation (1) is negative ( q< 0 ), then the equation has two roots different in sign, and the root with a larger absolute value will be positive if p< 0 , or negative if p> 0 .

For instance,

x 2 + 4x - 5 = 0; x 1 = - 5 and x 2 = 1, because q = - 5< 0 and p = 4> 0;

x 2 - 8x - 9 = 0; x 1 = 9 and x 2 = - 1, because q = - 9< 0 and p = - 8< 0.

Examples.

1) Solve the equation 345x 2 - 137x - 208 = 0.

Solution. Because a + b + c = 0 (345 - 137 - 208 = 0), then

x 1 = 1, x 2 = c / a = -208/345.

Answer: 1; -208/345.

2) Solve the equation 132x 2 - 247x + 115 = 0.

Solution. Because a + b + c = 0 (132 - 247 + 115 = 0), then

x 1 = 1, x 2 = c / a = 115/132.

Answer: 1; 115/132.

B. If the second coefficient b = 2k Is an even number, then the root formula

Example.

Let's solve the equation 3x2 - 14x + 16 = 0.

Solution... We have: a = 3, b = - 14, c = 16, k = - 7;

D = k 2 - ac = (- 7) 2 - 3 16 = 49 - 48 = 1, D> 0, two different roots;

Answer: 2; 8/3

V. Equation reduced

x 2 + px + q = 0

coincides with a general equation in which a = 1, b = p and c = q... Therefore, for the reduced quadratic equation, the root formula

It takes the form:

Formula (3) is especially convenient to use when R- even number.

Example. Let's solve the equation x 2 - 14x - 15 = 0.

Solution. We have: x 1.2 = 7 ±

Answer: x 1 = 15; x 2 = -1.

5. METHOD: Solving equations graphically.

Example. Solve the equation x2 - 2x - 3 = 0.

Let's build a graph of the function y = x2 - 2x - 3

1) We have: a = 1, b = -2, x0 = = 1, y0 = f (1) = 12 - 2 - 3 = -4. Hence, the vertex of the parabola is the point (1; -4), and the axis of the parabola is the straight line x = 1.

2) Take two points on the x-axis that are symmetric about the parabola axis, for example, points x = -1 and x = 3.

We have f (-1) = f (3) = 0. Let us construct points (-1; 0) and (3; 0) on the coordinate plane.

3) Draw a parabola through the points (-1; 0), (1; -4), (3; 0) (Fig. 68).

The roots of the equation x2 - 2x - 3 = 0 are the abscissas of the points of intersection of the parabola with the x axis; hence, the roots of the equation are as follows: x1 = - 1, x2 - 3.

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In this article, we will learn how to solve biquadratic equations.

So, what kind of equations are called biquadratic?
Everything equations of the form ah 4 + bx 2 + c = 0 , where a ≠ 0 that are square with respect to x 2, and are called biquadratic equations. As you can see, this notation is very similar to writing a quadratic equation, therefore, we will solve biquadratic equations using the formulas that we used to solve 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 instance, let's solve the equation x 4 + 4x 2 - 5 = 0.

We denote x 2 across 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 = 4 2 - 4 (- 5) = 16 + 20 = 36, √D = √36 = 6.

y 1 = (- 4 - 6) / 2 = - 10/2 = - 5,

y 2 = (- 4 + 6) / 2 = 2/2 = 1.

Let's go back to our variable x.

We got that x 2 = - 5 and x 2 = 1.

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 the negative root (most often the answer is x = 1, which is not correct).

Answer:- 1 and 1.

For a better understanding of the topic, we will analyze a few examples.

Example 1. Solve the equation 2x 4 - 5 x 2 + 3 = 0.

Let x 2 = y, then 2y 2 - 5y + 3 = 0.

D = (- 5) 2 - 4 2 3 = 25 - 24 = 1, √D = √1 = 1.

y 1 = (5 - 1) / (2 2) = 4/4 = 1, y 2 = (5 + 1) / (2 2) = 6/4 = 1.5.

Then x 2 = 1 and x 2 = 1.5.

We get x 1 = ‒1, x 2 = 1, x 3 = - √1.5, x 4 = √1.5.

Answer: ‒1; 1; ‒ √1,5; √1,5.

Example 2. Solve the equation 2x 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 have a solution.

Answer: no solutions.

Incomplete biquadratic equations- it is when b = 0 (ax 4 + c = 0) or c = 0

(ax 4 + bx 2 = 0) are solved like incomplete quadratic equations.

Example 3. Solve the equation x 4 - 25x 2 = 0

Let's factorize, put x 2 outside the brackets and then x 2 (x 2 - 25) = 0.

We get x 2 = 0 or x 2 - 25 = 0, x 2 = 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 (has no solutions)

x 2 = √9, x 1 = - 3, x 2 = 3.

As you can see, knowing how to solve quadratic equations, you can cope with biquadratic equations.

If you still have questions, sign up for my lessons. The tutor is Valentina Galinevskaya.

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