"Association Method" Quadratic Equation of One Variable PPT Courseware 3

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"Association Method" Quadratic Equation of One Variable PPT Courseware 3

How to find the exact solution of a quadratic equation of one variable

We use the method of "first determine the approximate range; then calculate the value and gradually approximate" to obtain the approximate solution to the quadratic equation of one variable.

For example, the solution of equation 2x²-13x+11=0 is x=1; that is, the lace width is 1m.

For example, the solution of the equation x²+12x-15=0 is about 1.2; that is, the sliding distance at the bottom of the ladder is about 1.2m.

For example, the solution of the equation x²-8x-20=0 is x=10 or x=-2; that is, the five consecutive integers are -2,-1,0,1,2; or 10,11,12,13,14, 15.

1. Move the term: Move the constant term to the left side of the equation;

2. Formula: Add the square of half the absolute value of the linear term coefficient to both sides of the equation;

3. Deformation: factorize the left side of the equation and combine similar factors on the right side;

4. Square root: Factor the left side of the equation and combine the same types of factors on the right side;

5. Solve: Solve linear equations of one variable;

6. Definite solution: Write the solution of the original equation.

1. According to the meaning of the question, list the equations:

As shown in the figure, on a rectangular ground 35m long and 26m wide, two mutually perpendicular roads of the same width are built, and flowers and plants are planted in the remaining part. If the area of ​​the remaining part is 850m², what should be the width of the road?

Solution: Assume the width of the road is x m. According to the question, we get

(35-x) (26-x) =850.

That is, x² - 61x-60 = 0.

Solving this equation, we get x1 =1;

x2=60 (does not fit the meaning of the question, discard it).

2. Solve the following equations:

(1).x²+12x+25=0;

(2).x²+4x=10;

(3).x²–6x=11;

(4).x²–2x-4=0.

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Update Time: 2024-11-22

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