Solving biquadratic equations. Online equations Checking the correctness of the solution

In this article we will learn to solve biquadratic equations.

So, what type 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 entry for a quadratic equation, so 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 example, let's solve the equation x 4 + 4x 2 ‒ 5 = 0.

Let's denote x 2 through at (x 2 = y ) and we 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 return to our variable x.

We found that x 2 = ‒ 5 and x 2 = 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 the negative root (most often they get the answer x = 1, but 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 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. Please note that none of these equations have a solution.

Answer: there are 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 out of 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, if you can solve quadratic equations, you can also solve biquadratic equations.

If you still have questions, sign up for my lessons. Tutor Valentina Galinevskaya.

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

Prove that there are no integers that increase by 5 times when the initial digit is moved to the end.

In a certain kingdom, every two people are either friends or enemies. Every person can at some point quarrel with all his friends and make peace with all his 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 side of this triangle is twice the size of the other.

Assignments for holding a regional (city) Olympiad for schoolchildren in mathematics.

In target shooting, the athlete scored only 8,9 and 10 points. In total, having fired more than 11 shots, he scored exactly 100 points. How many shots did the athlete take, 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 crossing out the middle digit.

In triangle ABC, bisectors are drawn from vertices A and B. Then, 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.

Assignments for holding a regional (city) Olympiad for schoolchildren in mathematics.

Solve the system of equations:

On 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 VD. Prove that the areas of triangles ALL and SDK are equal.

They 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 put one tourist on. When one bus left empty, all the tourists boarded the remaining buses equally. How many buses were there initially and how many tourists were in the group, if it is known that each bus can accommodate no more than 32 people?

Assignments for holding a regional (city) Olympiad for schoolchildren in mathematics.

Solve the system of equations:

Prove that four distances from a point on a circle to the vertex of a square inscribed in it cannot simultaneously be rational numbers.

Possible solutions to problems

1. Answer: x=1, x=0.5

Moving the starting digit to the end does not change the value of the number. In this case, according to the conditions of the problem, they should get a number that is 5 times larger than the first number. Therefore, the first digit of the required number must be equal to 1 and only 1. (since if the first digit is 2 or more, the value will change, 2*5=10). When you move 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 will not be reconciled). Let's take all the friends of person A. From what has been said it follows that they are all friendly with each other and are at enmity with the others. Now let A and his friends take turns quarreling with friends and making peace with enemies. After this 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 make peace with him, and the former enemies will remain friends. So, all people turn out to be friends of A, and therefore friends of each other.

The number 111 is divisible by 37, so the above sum is also divisible by 37.

According to the condition, the number is divisible by 37, therefore the sum

Divisible by 37.

Note that the indicated median and bisector cannot exit from the same vertex, since otherwise the angle at this vertex would be greater than 180 0. Now let in triangle ABC the bisector AD and the median CE intersect at point F. Then AF is the bisector and the altitude 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 for 8 points,

2 shots for 9 points,

1 shot for 10 points.

Let x the athlete made shots, 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:

From this system it follows that x+ y+ z=12

Let's 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– 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

Let us denote by x, y, z the first, second and third digits of the desired three-digit number, respectively. Then it can be written in the form . Crossing out the middle digit will result in a two-digit number. According to the conditions of the problem, i.e. unknown numbers x, y, z satisfy the equation

7(10 x+ z)=100 x+10 y+ x, which after bringing 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, due to the condition, has a unique solution x = 1, y = 0. Consequently, the condition of the problem is satisfied by the only number 105.

Let us denote by the letter F the point at which straight lines AB and CE intersect. Since lines DB and CF are parallel, then . Since BD is the bisector of angle ABC, we conclude that . It follows that , i.e. triangle BCF is isosceles and BC=BF. But from the condition it follows 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 leads to the required equality.

Possible solutions to problems

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 are convinced 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, let's construct a graph.

Using formula let's perform further transformations

Applying addition formulas and performing further transformations, we obtain

5. Answer: 24 buses, 529 tourists.

Let us denote by k initial number of buses. From the conditions of the problem it follows 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 must be divisible by k-1. Thus, the problem has been reduced to determining all integers for which the number

Is an integer and satisfies the inequality (the number n is equal to the number of tourists boarded on each bus, and according to the conditions 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 if k=2 and at k=24 .

If k=2 , That n=45.

And if k=24 , That n=23.

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

Therefore, initially there were 24 buses, and the number of all tourists is equal to n(k-1)=23*23=529

Possible solutions to problems

1. Answer:

Then the equation will take the form:

We have obtained 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

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Quadratic equations.

Quadratic equation- algebraic equation of general form

where x is a free variable,

a, b, c, are 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's factorize the left 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, then at least one of its factors is zero. Therefore, the left side of the equation becomes zero at x = 2, and also when x = - 12. This means that the number 2 And - 12 are the roots of the equation x 2 + 10x - 24 = 0.

2. METHOD : Method for selecting a complete square.

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

To do this, we 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 double 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.

Let us now transform the left side of the equation

x 2 + 6x - 7 = 0,

adding to it 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.

Let's multiply both sides of the equation

ax 2 + bx + c = 0, a ≠ 0

on 4a 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.

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

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

As is known, the reduced quadratic equation has the form

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

Its roots satisfy Vieta’s theorem, which, when a =1 looks like

x 1 x 2 = q,

x 1 + x 2 = - p

From this we can draw the following conclusions (from the coefficients p and q we can predict the signs of the roots).

a) If the half-member q given equation (1) is positive ( q > 0), then the equation has two roots of equal 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 example,

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 a free member q given equation (1) is negative ( q< 0 ), then the equation has two roots of different sign, and the larger root will be positive if p< 0 , or negative if p > 0 .

For example,

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) Let's solve the equation 345x 2 – 137x – 208 = 0.

Solution. Because a + b + c = 0 (345 – 137 – 208 = 0), That

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), That

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

IN. Reduced equation

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 is

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 plot the function y = x2 - 2x - 3

1) We have: a = 1, b = -2, x0 = = 1, y0 = f(1) = 12 - 2 - 3 = -4. This means that 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 symmetrical about the axis of the parabola, for example, points x = -1 and x = 3.

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

3) Through the points (-1; 0), (1; -4), (3; 0) we draw a parabola (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; This means that the roots of the equation are: x1 = - 1, x2 - 3.

Solving an equation means finding such values of the unknown for which the equality will be true.

Solving the equation

Let's present the equation as follows:

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 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. This means that we can write the equalities:

x = 0 or 2x - 3 = 0.

This means that one of the roots of the original equation is x 1 = 0.
Let's 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 a product. To find the unknown factor, you need to divide the product by the 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 whether an equation has been solved correctly, you need to substitute the numerical values of x into it and perform the necessary arithmetic operations. If, as a result of the calculations, it turns out that the left and right sides of the expression have the same value, then the equation has been solved correctly.

Let's check: