The square of X 1 0 finds the value of X solution 5

Without a solution, the square of x must be positive, and the equation is reduced to the square of x -1, so there is no solution, and it is tested to be a root.

A method for solving a quadratic equation.

1) Root Finding Formula Method.

For the unary quadratic equation ax 2+bx+c=0(a≠0), it can be solved according to the root finding formula x=(-b (b 2-4ac)) (2a).

2) Factorization.

First, the equation is shifted so that the right side of the equation is reduced to zero, then the left side of the equation is converted into the product of two unary linear equations, and finally the value of x is obtained for each factor to zero. The value of x is the solution of the equation.

3) Kaiping method.

If the unary quadratic equation is of x 2=p or (mx+n) 2=p(p 0), the unary quadratic equation can be solved by the direct open-flat method. You can get x=p, or mx+n=p.

There is no solution to this problem in the range of real numbers.

So mathematicians introduced a new kind of number: complex numbers.

In the plural, it is stated: i=(-1 under the root number).

i is an imaginary unit (i.e., -1 open root).

So the solution to this problem is x = plus or minus i

Of course, if you haven't learned complex numbers, you will encounter this problem and say: the equation has no solution in the range of real numbers.

Solution: x +1=0

x = -1 discussion: In the range of real numbers, there is no solution.

In the complex range, x=i or x=-i

x squared is greater than or equal to 0

x squared +1 evergrande at 0

So there is no solution to x.

Within the scope of reality, there is no solution.

Two solutions in the complex range are positive and negative i.

x squared + 1" is an unsolvable problem.

The square of x is equal to -1, then x is equal to plus or minus i

There is no solution in the range of real numbers.

The original formula can be turned into x 2+x+1 4-1 4-1=0

(x+1 2) 2=5 4

x+1 2= 5 2 or x+1 2=- 5 2 is obtained, so the solution of x=( 5-1) 2 or x=-( 5+1) 2 is solved:

First, the direct leveling method.

For example, (x+a) 2=b, when b is greater than or equal to 0, x+a=positive and negative root number b, x=-a plus or minus root number b; When b is less than 0. Equations have no real roots.

Second, the matching method.

1. The coefficient of the quadratic term is 1.

2. Shift terms, the left side is the quadratic term and the primary term, and the right side is the constant term.

3. Formula, add half of the square of the primary term coefficient on both sides, and form the form of (x=a) 2=b.

4. Use the direct leveling method to find the solution of the equation.

Using the root finding formula, substituting a=1, b=1, c=-1 into the root finding formula, x=(-1 5) 2

The answer process is as follows.,You can refer to it.,If there's something you don't understand, you can ask me.,I hope it helps you.,If there's no other question.,Can you give a like?,Thank you(

a b c is the coefficient of each item of the equation, a=1 b=1 c=-1

4ac-b 2) 4a=[4*1*(-1)-1 2] 4*1=-5 4 is less than 0

There is no solution to the equation.

The original formula can be reduced to x 2+x+1 4-1 4-1=0 to get (x+1 2) 2=5 4

We get x+1 2= 5 2 or x+1 2=- 5 2, so we get x=( 5-1) 2 or x=-( 5+1) 2

Make use of the root-finding formula. According to the title.

x=(-1±√5)/2

x1=(-1+√5)/2

x2=(-1-√5)/2

No solution.

x²+x+1=0

x²+x=-1

x²+x+1/4=-3/4

x+1/2)²=-3/4

In the range of real numbers, the square term is constant and non-negative, so there is no real solution to the equation.

An integral equation that contains only one unknown (unary) and the highest order of the unknown term is 2 (quadratic) is called a quadratic equation. The unary quadratic equation can be formed into a general form ax+bx+c=0(a≠0). where ax is called the quadratic term, and a is the quadratic coefficient; bx is called the primary term, and b is the coefficient of the primary term; c is called a constant term.

For a quadratic equation to be true, three conditions must be met at the same time:

1. It is an integer equation, that is, both sides of the equal sign are integers, and if there is a denominator in the equation; And the unknown number is on the denominator, then this equation is a fractional equation, not a one-dimensional quadratic equation, if there is a root number in the equation, and the unknown number is in the root number, then this equation is not a one-element quadratic equation (it is an irrational equation).

2. Contains only one unknown.

3. The maximum number of unknown items is 2.

x²+x+1=0

x²+x=-1

x²+x+1/4=-3/4

x+1/2)²=-3/4

In the range of real numbers, the square term is constant and non-negative, so there is no real solution to the equation.

However, it can be solved in the range of complex numbers:

x=-1 2 +(3 2)i or x=-1 2 -(3 2)i

The answer process is as follows.,You can refer to it.,If there's something you don't understand, you can ask me.,I hope it helps you.,If there's no other question.,Can you give a like?,Thank you(

The correct answer is:

Because a=1, b=1, c=1, the b-4ac=-3 0 equation has two imaginary roots:

x=（-b±√b²-4ac）/2

That is, x1=(-1+ 3) 2, x2=(-1- 3) 2 junior high school is unsolvable!

You probably read equations like this in some physics literature or high school math textbooks.

When you get to high school, you can easily solve this problem if you learn imaginary numbers.

x ^ 2 = - 1

Solvable: x = i

If you want to have an intuitive understanding of "imaginary numbers", you can read extracurricular books on the philosophy of mathematics (find domestic mathematicians to write, those books start from natural numbers, zeros, and integers, and gradually talk about imaginary numbers).

I have the impression that the first volume of Fichjingolz's "Course in Calculus" also talks about imaginary numbers. That is also simple and vivid.

The answer process is as follows.,You can refer to it.,If there's something you don't understand, you can ask me.,I hope it helps you.,If there's no other question.,Can you give a like?,Thank you(

There is no solution in the range of real numbers.

This is a question about imaginary numbers.

x^2+1=0

x^2=-1

Since the square of positive and negative i = -1, then x=-i or x=i note that i is not 1

That is, the complex function is not the range you are talking about, x= i; i is an imaginary unit.

There is no solution in the range of real numbers, there is a solution in the range of complex numbers, x=i or -i.

x^2+1=0

x^2=-1

There is no real solution. But there are complex solutions.

x=-i or x=i

This is a question about imaginary numbers.

x squared = -1 then x = 1!

There is no solution to this question.

Because x2 0's.

x²+x+1=0

x+1/2)²+3/4=0

x+1/2)²=3/4

x+1/2)²=3 i /2)²

x=(-1±√3 i) /2

The general steps for solving a system of binary linear equations using addition, subtraction, and elimination

1. Transformation coefficient: use the basic properties of the equation to multiply both sides of one or two equations by the appropriate number, so that the coefficient of an unknown number in the two equations is the opposite number or equal to each other;

2. Addition, subtraction, and elimination: add or subtract the two sides of the two equations respectively, remove an unknown number, and obtain a unary one-dimensional equation;

3. Solve this unary equation to obtain the value of an unknown number;

4. Retrogression: Substitute the value of the unknown into any equation of the original equation system to find the value of another unknown.

The process is as follows, you can refer to it, if you don't understand anything, you can ask me, dear.

There is no solution to this problem in the range of real numbers.

So mathematicians introduced a new type of coarse number: complex numbers.

In the plural, it is stated: i=(-1 under the root number).

i is an imaginary unit (i.e., -1 open root).

So the solution to this problem is x = plus or minus i

Of course, if you don't learn complex numbers, you will say that the equation has no solution in the range of real numbers.

Using the formula method, x = 2 parts of a -b (b -4ac), where a (quadratic coefficient) = 1, b (primary term coefficient) = 1, c (constant term) = -1, bring in the formula to get x = -1 (1 + 4), so x1 = -1 + 5 out of 2, x2 = -1 - 5 out of 2

To use the matching method, first move the constant term to the right, and add (1 2) 2 to both sides of the x 2-x=1 equation, i.e., 1 4, x 2-x+1 4=1+1 4

Carry out the recipe. x-1/4）^2=5/4

The time x-1 4 = 5 2

x=1/4±√5/2

x1=1/4+√5/2,x2=1/4-√5/2。

5 2 is the root number 5).

Use the root finding formula:

where a is the quadratic pre-term coefficient, b is the primary pre-term coefficient, and c is the constant term.

Direct formula method.

Or with a method.

x=(-b±√b^2-4ac)/2a

a, b, and c are quadratic coefficients, respectively.

Primary term coefficients.

Constant: In this problem, a=1

b=-1c=-1

Just bring it in.

x squared - x-1 = 0

x squared - x+1 4=1+1 4

x-1 2) squared = 5 4

x=1/2+√5/2

x=1/2-√5/2

With the words of the recipe.

x^2-x-1=0

x^2-x+1/4=5/4

x-1/2)^2=+-

Root number 2/5

x1 = 1 + 2/5 of the root number

x2 = 1 - 2/5 of the root number

The calculation process according to the topic is as follows:

x²+x+1＝0

With the formula method, b -4ac

Equations do not have real roots.

When x2=0, x=0, when x2=1, x=1, x=-1x2=x, x=0, x=1

Due to the suspicion of regression in the set, 0,1

Therefore, Qin carries the World Bank x=-1