Two x1.x2 of 7x-6x-2 = 0. Find | x1-x2 | = x1-x2 (x1 + x2) = X1 of x2= Every equal sign is followed by a question

Two x1.x2 of 7x-6x-2 = 0. Find | x1-x2 | = x1-x2 (x1 + x2) = X1 of x2= Every equal sign is followed by a question

According to the question, X1 + x2 = 6 / 7, X1 * x2 = - 2 / 7
So | x1-x2 | = √ [(x1 + x2) ^ 2-4x1 * x2] = 2 √ 23 / 7
Because we don't know which is positive and which is negative, so (x1 + x2) / (x1-x2) = ± (6 / 7) / (2 √ 23 / 7) = ± 3 √ 23 / 23
X 1 / x 2 = x 1 ^ 2 / (x 1 * x 2) = [(6 x 1 + 2) / 7] / (- 2 / 7) = - 3 x 1-1. Because we don't know the positive and negative of x 1, it seems that we can't simplify the calculation by taking both roots into account
How to get X1 = 1, X2 = 1 / 2 from the square of 2x - 3x + 1 = 0
2x²-3x+1=0
x²-3/2x+1/2=0
x²-3/2x+1/2+9/16-9/16=0
（x-3/4) ²=1/16
x-3/4=±1/4
x1=1
x2=1/2
Use cross multiplication!
2x²-3x+1=0
(2x-1)(x-1)=0
X = 1 / 2 or x = 1
Application of cross multiplication:
2x -1
X
x -1
X^2-3X+1=(2X-1)(X-1)=0;
So X1 = 1, X2 = 1 / 2
The factorization is (2x-1) (x-1) = 0
We get X1 = 1, X2 = 1 / 2?
The known equation (k2-1) x2 + (K + 1) x + (k-7) y = K + 2=______ When k, the equation is linear=______ &The equation is a quadratic equation of two variables
Since the equation does not indicate whether it is about X or Y, we should refer to the equation of first degree with respect to X or y. when k2-1 = 0 and K + 1 = 0, the equation is the equation of first degree with respect to y, and the solution is k = - 1. When k2-1 = 0 and K + 1 ≠ 0, the equation is the equation of second degree with respect to y, and the solution is k = 1. So the answer is: - 1,1
When the parabola passes through point (4, - 3) and x = 3, the maximum value of y = 4, the analytical expression of the parabola is obtained
Let y = a (x-3) ^ 2 + 4
If you take point a in, you can get a
Because when x = 3, y max = 4
So the vertex of parabola (3,4)
So let y = a (x-3) ^ 2 + 4
Substitute (4, - 3) into a = - 7
So y = - 7 (x-3) ^ 2 + 4
The equation (k ^ 2-9) x ^ 2 + (K + 3) x + (K-2) y = 2k-1, when k = () it is a linear equation of one variable, when k = () it is a linear equation of two variables?
1. When the equation is a linear equation of one variable
K ^ 2-9 = 0 gives k = + - 3, and K-2 is not equal to 0 no matter k = 3 or - 3
So K + 3 = 0, that is, only k = - 3
2. Similarly, K ^ 2-9 = 0, k = + - 3
If it is a quadratic equation of two variables, then K + 3 and K-2 are not equal to 0
So k = 3
-3 3
The equation (k ^ 2-9) x ^ 2 + (K + 3) x + (K-2) y = 2k-1, when k = (- 3), it is one variable linear equation, when k = (3), it is two variable linear equation
If a parabola passes through a point (1,1), and if x = 2, y has a maximum value of 3, find the analytical expression of the parabola
Century studies,
Let the parabola be y = - A (X-2) ² + 3
Substituting point (1,1) to get
1 = - A + 3, the solution is a = 2
So the analytical formula of parabola is y = - 2 (X-2) ² + 3
That is y = - 2x & # 178; + 8x-5
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The second semester of the seventh grade mathematics problem 2 yuan 1 degree equation
1. It is known that {x = 1, y = 2 is a solution of the equation x + KY = 7. Find the value of K, and check whether {x = 3, y = 4 / 3 is the solution of the equation. 2. It is known that the binary linear equation 2x-3y-1 = 0. (1) express y with the algebraic expression of X; (2) express x with the algebraic expression of Y. 1 bring x = 1, y = 2 into the equation x + KY = 7 with k = 3, so x + 3Y = 7 will
There is a problem with wood
y=ax+b
x+y=c
There are two kinds of electric rice cookers. The sum of the original unit price is 200 yuan,
Given that the parabola passes through the point (3,0) and when x = 1, y has the maximum value of 3, find the analytic expression of the function
When x = 1, y has a maximum value of 3
Know that the opening of the parabola is downward and the vertex is (1,3)
So let the parabolic equation be
y=a（x-1）^2+3
Passing through point (3,0) by parabola
That is, a (3-1) ^ 2 + 3 = 0
The solution is a = - 3 / 4
So the parabolic equation is
y=(-3/4)x^2+3x/2+9/4
Because when x = 1, y has a maximum of 3,
So let y = a (x-1) ² + 3, a
The distance between the enemy and our army is 42km. If the enemy invades our army, we can meet in two hours. If the enemy runs back, it will take us 14 hours to catch up. What's the speed of our army and the enemy?
Suppose the speed of our army is XKM / h and that of the enemy is YKM / h. according to the meaning of the question, there are 2x + 2Y = 4214x-14y = 42, and the solution is x = 12Y = 9. A: the speed of our army is 12km / h, and that of the enemy is 9km / h
If the parabola passes through point (1,1) and x = 2, y has a maximum value of 3
When x = 2, y has a maximum value of 3
Know that the vertex of parabola is (2,3)
So let the parabola be y = a (X-2) ^ 2 + 3
Then the parabola passes through the point (1,1)
We get a + 3 = 1
The solution is a = - 2
So the parabola is y = - 2 (X-2) ^ 2 + 3
That is y = - 2x ^ 2 + 8x-5