Factorization of quadratic trinomial Can ax ^ + BX + C be decomposed into the form of a (x-x1) (x-x2)

Factorization of quadratic trinomial Can ax ^ + BX + C be decomposed into the form of a (x-x1) (x-x2)

If ax ^ + BX + C = 0 equation has real roots, then it is OK
When B ^ 2-4ac > = 0, it can be divided into a (x-x1) (x-x2)
X1, X2 are the two roots of the equation
Factorization of quadratic trinomial
1. If the square of x-ax-8 (a is an integer) can be factorized in the range of integers, find the value of A
2. If the square of x-ax-8 (a is an integer) can be factorized in the range of real number, find the value of A
One is an integer and the other is a real number
Question 1: a = - 7,7,2, - 2
The second question: a = 7, - 7,2, - 2, the root of plus or minus 3 times 2,0
The method is to use the principle of cross method
Well, the teacher will say
A problem of elementary binary equations
x²-2xy+y²=1
A kind of
2x²-5xy-3y²=0
If X & sup2; - 2XY + Y & sup2; = 1, then (X-Y) & sup2; = 1, then X-Y = 1 or X-Y = - 1
If 2x & sup2; - 5xy-3y & sup2; = 0, then (2x + y) (x-3y) = 0, then 2x + y = 0 or x-3y = 0
If X-Y = 1,2x + y = 0, the solution is x = 1 / 3, y = - 2 / 3
Simultaneous X-Y = 1, x-3y = 0, the solution is x = 3 / 2, y = 1 / 2
Simultaneous X-Y = - 1,2x + y = 0, the solution is x = - 1 / 3, y = 2 / 3
Simultaneous X-Y = - 1, x-3y = 0, the solution is x = - 3 / 2, y = - 1 / 2
The above four groups are the solutions of the equations
2X & sup2; - 5xy-3y & sup2; = (2x + y) (x-3y) = 0, so x-3y = 0 or 2x + y = 0, the solution is x = 3Y or x = - Y / 2, substituting the first equation (X-Y) & sup2; = 1, y = 1 / 2, x = 3 / 2, y = - 1 / 2, x = - 3 / 2, or y = 2 / 3, x = - 1 / 3, y = - 2 / 3, x = 1 / 3
x²-2xy+y²=1
(x+y)^2=1
X + y = 1 or x + y = - 1
And 2x & sup2; - 5xy-3y & sup2; = 0
(2x+y)(x-3y)=0
Y = - 2x or x = 3Y
When y = - 2x is substituted by X + y = 1, x = - 1, y = 2
When y = - 2x is substituted by X + y = - 1, x = 1, y = - 2
When x = 3Y is substituted by X + y = 1, x = 3 / 4 and y = 1 / 4 are obtained
When x = 3Y is substituted by X + y = - 1, x = - 3 / 4, y = - 1 / 4
X & sup2; - 2XY + Y & sup2; = 1 is (X-Y) ^ 2 = 1, so x = y + 1 or x = Y-1
Substituting x = y + 1 into equation 2, we get 6y ^ 2 + Y-2 = 0, so y = 1 / 2 or y = - 2 / 3, then x = 3 / 2 or 1 / 3
Substituting x = Y-1 into equation 2, 6y ^ 2-y-2 = 0, so y = - 1 / 2 or y = 2 / 3, then x = - 3 / 2 or - 1 / 3
Question 1
X＾2-6*X+5
——-- - (fractional line) < 0
12+4*X-X＾2
In △ ABC, cosa = root 5 of 5, CoSb = root 10 of 10
(1) Find angle c
(2) Let AB = radical 2, find the area of △ ABC
Question 3. A2 = 8, A5 = 512 in known equal ratio sequence {an}
(1) The general term formula of {an}
(2) Let BN = log2an find the first n terms and s of {BN}
1> The original formula can be written as (X-5) (x-1) -- > 0 (x + 2) (X-6) by the number axis root marking method: X ＜ - 2 or 1 ＜ x ＜ 5 or X ＞ 62 > 1
Binary linear simultaneous equations, and to solve
In a mathematics midterm exam, there are 30 points for each multiple-choice question, 45 points for each filling question, and 25 points for each calculation question. Xiaowei's score is 83 points, and the number of multiple-choice questions is half of the number of filling questions, and all the questions are only 5 wrong. How many questions can Xiaowei answer correctly? (the full score is 100 points) to deal with the questions
If multiple-choice questions are answered correctly in total, then fill in the number of correct answers 2x
Let's answer the calculation question correctly
Then 10 + 15 + 5 - (x + 2x + y) = 5
3x+2x×3+5y=83
The solution is x = 7, y = 4
Therefore, Xiao Wei answered 7 multiple-choice questions correctly
Let f (x) = - 3A ^ 2 + a (B-A) x + B, solve the inequality about a: F (1) > 0
There is also the second question, if the solution set of inequality f (x) > 0 is (- 1,3), find the value of real number a and B
1. Substituting x = 1 into the function to get the formula of a and B, and making it greater than 0 is the inequality
Consider B as a constant and solve a
2. We know that the solution of F (x) = 0 is x = - 1 and x = 3, and then find a, B
A trial study on the linear simultaneous equations of two variables
There are 45 students in three classes in two years. In a math quiz, the average score of the whole class is 75. The average score of girls is 80. The average score of boys is 75. How many boys and girls are there in three classes in two years
Let X and y be male and female respectively
According to the theme
X+Y=45
75X+80Y=75*45
We get x = 45, y = 0
The solution of X square-x-6 < 0
Cross multiplication X & # 178; - X-6 ﹤ 0, let me tell you a little trick. It means that the sum of the two numbers of cross multiplication should be equal to the coefficient in the middle, and the multiplication should be equal to the constant at the end - 3. 2 just satisfies this, so (x-3) (x + 2) < 0, so - 2-2)
x²-x-6
If the solution of the binary linear simultaneous equation: 2x-y = 3; 3x-4y = 3 is x = a, y = B, then a + B is (?)
A.a=1 b=2
B.a=0 b=2
C.a=2 b=1
D.a=1 b=1
2a-b=3;
3a-4b=3
6a-3b=9
6a-8b=6
5b=3
b=3/5
a=(3+3/5)/2=9/5
3/5+9/5=12/5
I don't know if it's right... I can't bring in the second equation···
C
X = 9 / 5, y = 3 / 5, not in your options.
C
First, put the four answers into the above two equations, and you will find that none of the four answers is right. Then we will look at the title clearly, and we will find that it requires a + B, not the value of a and B.
2x-y = 3 = > 8x-4y = 12 (1)
Formula 3x-4y = 3 (2)
① Formula - 2, we get 5x = 9, that is, x = 9 / 5 = a
So y = 3 / 5 = b
Then a + B = 12 / 5
If x + 1 / x = 3, then xsquare - 1 / xsquare =?
x+1/x=3
(x+1/x)^2=x^2+2+1/x^2=9
x ^2+1/x^2=7
(x-1/x)^2=x^2-2+1/x^2=7-2=5
X-1 / x = (+ / -) radical 5
So x ^ 2-1 / x ^ 2 = (x + 1 / x) (x-1 / x) = (/ -) 3 root sign 5
X = 0.5, so the answer is - 3
±3√5
Original formula = (x-1 / x) (x + 1 / x)
=3（x-1/x）
（x+1/x）²=x²+1/x²+2=9
So (x-1 / x) &# 178; = x & # 178; + 1 / X & # 178; - 2 = 5
So (x-1 / x) = ± √ 5
So the original formula = ± 3 √ 5
Hope to adopt (it's hard to type such characters)
∵X+1/X=3
∴（X-1/X）²=（X+1/X）²-4=3²-4=5
So X-1 / x = ± root 5
Xsquare - 1 / xsquare = (x + 1 / x) (x-1 / x) = 3 × (± root 5) = ± 3 root 5
（X+1/X）²
=X²+2+(1/X)²
=3²
=9
Then x & # 178; - (1 / x) &# 178; = 9-2 = 7