Lower grade 8, 12 quadratic equations, 8th grade 1 quadratic equation practice problems

1.Knowing a=2+3,b=2-3, try to find the value of a b-b a.

a/b=(2+√3)/(2-√3)=(2+√3)²=7+4√3, b/a=7-4√3.

a/b-b/a=(7+4√3)-(7-4√3)=8√32.If a 0, then 1 b ab a (b a)1 b ab a (b a).

ab+√ab=0

3.In each of the following equations, the correct calculation is ( b ).

a.m m = m to the sixth power.

b.√(16 1/3)=√16×√1/3=（4√3）/3c.The cube root of (2 +3) = 2 + 3 = 5

d.(a-1)×√1/（1-a））= -（1-a）² 1/（1-a））=-√（1-a） （a＜0）

1. Determine whether the following equation is a quadratic equation

1) 1-2x 2=x Yes.

2) 3x 2- 2 x (x outside the root number) = 7 Yes.

3) x 2+1 2x 2=0 Yes If it's 1 (2x 2) then it's not.

4) (3x-2)(x+6)=3x^2-7.No.

2. Convert the following quadratic equation into a general formula, and say the coefficients and constant terms of the quadratic terms.

Format: Equation General Quadratic Coefficients Primary Coefficients Constant Terms.

1) 5-3x+2x 2=0 2x 2-3x+5=0 quadratic term coefficient 2 primary term coefficient -3 constant term 5

2) x(x-2)=1 x 2-2x-1=0 quadratic coefficient1 primary coefficient-2 constant term-1

3) (x-2) 2=(3x+2)(x-5) 2x 2-9x-14=0 quadratic coefficient2 primary coefficient-9 constant term-14

4) (1-2) x 2=(1+ 2)x (1-2)x 2-(1+ 2)x=0 Quadratic coefficients 1-2 Primary coefficients-1-2 Constant terms 0

3. Fill in the blanks. 1) Knowing that the equation (m2-4) x 2-(m+2)x+3=0 is a quadratic equation about x, then the value range of m is (m≠ 2).

2) It is known that a root of the equation for x 2x 2 + mx-3 = 0 is 3, then m=(-5).

4. Is x= - 3 2 the root of the unary quadratic equation 2x 2-(2a-3)x-3a=0? Why? (Process).

Solution: Substituting x= - 3 2 into the equation yields, 2(- 3 2) 2-(2a-3) (3 2)-3a=0

9/2+3a-9/2-3a=0

Left: 3a-3a=0 Right: 0 Left = Right.

So x= - 3 2 is the root of the unary quadratic equation 2x 2-(2a-3)x-3a=0.

5. Write a one-dimensional quadratic equation with a root of 1 and a coefficient of -1 in its primary term; And write out your way of making up the equations.

Solution: Let this unary quadratic equation be: ax 2-x=0 (you don't need to write c because the constant term can be zero).

So substituting the root x=1 into the equation yields: a-1=0

a=1, so the equation is x 2-x=0

After writing it, I hope it will help you

1=4-4(1+m)<0 m>0

2=(m-2) 2-4(-m-3)=m 2+16>0, so the second equation must have two unequal real roots.

4(b-a)^2-4(c-b)(a-b)=4(a-b)(a-c)=0

a=b or a=c

So the triangle is an isosceles triangle.

What is S?

1) Your first problem consists of "The equation x2+2x+1+m has no real roots.""It can be obtained that m>0, and from the second equation, the discriminant is equal to m2+7 greater than zero, so there must be two unequal real roots.

2) Isosceles triangle, write the discriminant formula to get a=c, so it is an isosceles triangle.

And then that is also a discriminant formula, which can be solved by simplifying the system of constituent equations that follow it.

ioqsijoiwjdoiawd child, don't copy it, it's useless to copy!

It's not clearly stated. Is the square of y on the fraction line or below? 8 (y 2-y*(y-3))=3y-2 or y 2-8 y*(y-3)=3y-2

（1）：x-3y=2 ①

y=x will be substituted for x-3x=2

2x=2x=-1

2）： x+y=5 ①

2x+y=8 ②

Get x=3

Substitute x=3

y=2（3）： 4x+3y=5 ①

x-2y=4 ②

4, get 4x-8y=16

Get 11y=-11

y=-1 substitutes y=-1 into ,, and x=2

4）：m-n/2=2 ①

2m+3n=12 ②

2, get 2m-n=4

Get 4n = 8

n=2 substitutes n=2 into m=3

Use substitution to eliminate the element, or add or subtract the elimination element, turn the binary system of equations into a unary equation, find an unknown, and then find the other;

These are the most basic questions, but you have to pass them.

First of all, let the two roots be a and b respectively Since they are the cosine values of two acute angles of a right triangle, then according to the Pythagorean theorem, there must be a 2+b 2=1 (1), and then according to the unreached theorem in a quadratic equation, we can know that a+b=(m+1) 4 (2), a*b=m 4 (3).

In addition, according to the perfect flat method: (a+b) 2=a 2+2ab+b 2 knows: (combined with the equation).

m+1) 2 16=1+m 4 Solve m=5 or -3 according to this equation and then have a solution according to the original equation Know the discriminant equation 0

That is, there is (m+1) 2-16m>0 and so on, and the range of m is m>7+4*root number 3 or m<7-4*root number 3, and then according to the value of m, we can know that m=-3

This topic is too difficult for students in the second year of junior high school, and I personally think it is more appropriate to put it in the first year of high school In addition, if you haven't learned trigonometric functions, then you definitely don't know what the cosine value means

The cosine value refers to the ratio of the side length of the right triangle to the side length of the hypotenuse in a right triangle, which is the cosine value of the angle, for example, the cosine value of the 30-degree angle is 1 2

The two roots of the equation have the following relationship with each number in the equation: x1+x2= -b a, x1*x2=c a

The cosine value is the right side of the acute angle of the triangle divided by the hypotenuse, the cosine value of the two acute angles of the right triangle, and the sum of the two cosines satisfies the sum of squares as 1

So x1 +x2 = 1

So (-b a) 2c a = 1 i.e. [(m+1) 4] 2 m 4 = 1 solution gives m = 3+2 2, or, m = 3-2 2

The sum of squares of the two cosines is 1, with Weida's theorem, x1+x2=(m+1) 4, x1*x2=m 4, so (x1+x2) 2-2x1*x2=1, the answer is the same as "1137061417", pure hand, hope to adopt.

** In the calculation process, the calculation result is not counted, it is easy to calculate.

1, (1) is (2) is (3) is not (4) erect stupid is not 2, (1) general formula: 2x 0 5-3x+5=0, quadratic coefficient: 2, primary coefficient:

3. Constant term: 5 (2) General formula: x 0 5-2x-1=0, quadratic term coefficient:

1, the coefficient of the primary term: -2, the constant term: -1 (3) general formula:

2x 0 5-9x-14=0, quadratic coefficient: 2, primary coefficient: -9, constant term:

14 (4) General formula: (1-2) x 0 5-(1+ 2) x = 0, quadratic coefficient: 1-2, primary coefficient:

1+ 2, constant term: 03, (1) m≠ 2 (m 0 5-4≠0 is OK) (2) m = -5 (x = 3 can be substituted into the equation) residual liquid 4, x = -3 2 into the left side of the equation: 2 (-3 2) 0 5-(2a-3) (3 2) -3a = 9 2 + 3a-9 2-3a=0 x = -3 2 is a root of the original equation 5, let an equation that satisfies the condition be x 0 5-x + a=0 x = 1 is one of its roots, so substituting x = 1 into the original equation has 1-1 + a=0 a=0, so a quadratic equation that satisfies the condition is x 0 and 5-x=0

1, (1), (2), (3) is, (4) does not observe the Wu is 2, (1) 5-3x+2x 2=02x 2-3x+5=0 quadratic term coefficient 2, primary term -3, constant term 5 (2) x(x-2) = 1x 2-2x-1=0 quadratic term coefficient 1, primary term coefficient -2, constant term -1 (3) (x-2) 2=(3x+2)(x-5)-2x 2+11x+14=0 quadratic term envy coefficient -2, primary term coefficient 11, constant term 14 (4) (1- Brother Qi2) x 2=(1+ 2)x(1- 2)x 2-(1+ 2)x=0 quadratic coefficient 1-2, primary term coefficient - (1+ 2), constant term 0

Solution: b+(root2)a=(root2) 2: (root2)b +2a=12a-1+(root2)b=0

4a^2+1-2b^2-4a

4a^2-4a +1) -2b^2

2a-1)^2-2b^2

2a-1+(root2)b][2a-1-(root2)b].

Two positive numbers represent x1+x2>0, x1*x2>0, and the Huai Huchun is -a b>0, a lead is c>0

So a c has the same name as a and b has a different sign.

a,c>0 b0

Hello! There are two positive roots means that there are two positive solutions, as long as there is a solution, the formula is greater than or equal to 0, if the two solutions are the same, it is equal to 0, and the difference is greater than 0. For example, solutions from 2x 2-4x+2=0 are all 1.

As for the symbols, it depends on the situation. If B 2 is greater than 4AC, then AC can be the same or different sign, anyway, B 2-4AC "0. If the remaining acorn b 2 is less than 4ac, then b 2 should be +4ac, otherwise b 2-4ac < 0 will not be solved.

Because there is already a negative sign in front of 4, the AC different sign, that is, a positive and a negative number is destroyed.