Explanation of the lesson solving the equation from the
second degree in one of the most interesting lessons in
algebra. We will provide a detailed explanation of this lesson
and we will learn about the solution steps in detail and
illustrative examples all of this and more in the geniuses of the
Definition of the equation of the second degree in one variable
The equation of the second degree is written in the form a x 2
+ b x + c = zero, provided that A is not equal to zero, and it is
called a quadratic equation * the quadratic equation * like 2 x
2 - x - 6 = zero quadratic equation in x and 25 y - 10 + 15 p. 2
= zero A quadratic equation on p
Steps for solving a quadratic equation
- The quadratic equation must be placed on the zero form
- Decipher the different parentheses, if any
- We analyze the right-hand side using one of the previous methods of analysis ( taking out the common factor, the simple three expression , the non-simple three expression, the perfect three square expression, and the difference between two squares)
Application examples for solving quadratic equations
solve the following equations
1. x 2 - 5 x = zero Solve the equation
Solve the quadratic equation x (x - 5) = zero
Either or
X = zero x - 5 = zero, and of which x = 5
Equation solution set = <0, 5>
2. x 2 - 25 = zero Solve the equation
Solve the quadratic equation (x - 5) (x + 5) = zero
Either or
X-5 = zero, of which x = 5 x + 5 = zero, and of which x = -5
Equation solution set = <5 and -5>
3. x 2 - x - 20 = zero Solve the equation
Solve the quadratic equation (x + 4) (x - 5) = zero
Either or
X + 4 = zero, of which x = -4 x - 5 = zero, and of which x = 5
Equation solution set = <-4 and> 5
4.2x2 + 7x - 4 = zero Solve for the equation (2x - 1) 1x
(X + 4) 8x by subtracting 7x
Solve the quadratic equation ( 2x - 1) (x + 4) = zero
Either or
2x - 1 = zero, of which x = 1/2 x + 4 = zero, and of which x = -4
<Equation solution set = <-4 and 1/2
5.2 x 2 + 10 x = -12 Solve the equation
Solving the quadratic equation 2x2 + 10x + 12 = zero take out
the common factor
2 (x2 + 5x + 6) = zero simple triple expression
2 (x + 3) (x + 2) = zero
Either or
X + 3 = zero, including x = -3x + 2 = zero, and of which x = -2
Equation solution set = <-3 and -2>
6. (x - 3) (x + 1) = 5 Expand the parentheses Find the solution to the equation
Solve the quadratic equation x 2 - 2 x - 3 = 5
X 2 -2 x - 3 - 5 = zero
X 2 - 2 x - 8 = zero
(X + 2) (x - 4) = zero
Either or
X + 2 = zero, including x = -2x - 4 = zero, and of which x = 4
Equation solution set = <-2 and 4>
Applications to solve the equation of the second degree
Important Notes
- The number we impose = x
- Multiply the number, and we apply it by twice
- three times the number given by 3s
- four times the number, we give it 4s
- half of the number is given by 1/2 h
- the additive inverse is imposed by b - c
- The multiplicative inverse is imposed by 1 / s
- The square of the number is imposed by
- Double the number, we assign it to 2 x 2
- Suppose three times a square by 3 x 2
- A square of its weakness is imposed by 4 x 2
- The square of three proverbs is dictated by 9 x 2
- Two numbers whose sum = 3 Suppose the first number x and the second number 3 - x
- Two numbers, the difference between them = 7 Suppose the first number x and the second number x - 7
- Two numbers with the ratio between 4: 5. Suppose the first number 4s and the second number 5s
- Two numbers, one more than the other by 6, suppose the first number x and the second number x + 6
- Two consecutive integers. Suppose the first number x and the second number x + 1
- The age of a man now is X one year old after 5 years = X + 5 his age after 7 years = Q - 7
Examples of applications of solving the second degree equation
- An integer if added to a square, the result was = 56, find the number
X 2 + x = 56
X 2 + x - 56 = zero
(X - 7) (x + 8) = zero
Either or
X - 7 = zero, of which x = 7 x + 8 = zero, and of which x = -8
The number 7 or -8
problem solution suppose this number = x squared = x 2 times it = 2x
X 2 - 2 x = 48
X 2 - 2 x - 48 = zero
(X + 6) (x - 8) = zero
Either or
X + 6 = zero, including x = -6, and this number is rejected. X - 8
= zero, and of which x = 8
The positive number is 8
To solve a math problem, assume that width = x and length = x +
4
Area of a rectangle = Length x Width
21 = (x + 4) x (x)
X 2 + 4 x - 21 = zero
(X - 3) (x + 7) = zero
Either or
X - 3 = zero, of which x = 3x + 7 = zero, and of which x = -7
Width = 3 cm length x + 4 = 3 + 4 = 7
- A B C is a triangle, s, an angle, a = x 2 + 61, and s, an angle, b = 110 - 11, x, and s, an angle, c = 90 - 7 x. Find the value of x and the measures of the angles of the triangle
Solve a math problem s angle a + s angle b + s angle c = 180
Q2 + 61 + 110 - 11x + 90 - 7x = 180
Q2 - 18x + 61 + 110 + 90 - 180 = zero
X 2 - 18 x + 81 = zero
(X - 9) (x - 9) = zero
Either or
X - 9 = zero, of which x = 9 x - 9 = zero, and of which x = 9
S angle A = x 2 + 61 = 9 x 9 + 61 = 142 degrees
S angle B = 110 - 11 x = 110 - 11 x 9 = 11 degrees
S angle, c = 90 - 7 x = 90 - 7 x 9 = 27 degrees
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