Given a ≠ 0, quadratic function f (x) = ax2-2x-2a, let the solution set of F (x) > 0 be a and B = {x | 1

Given a ≠ 0, quadratic function f (x) = ax2-2x-2a, let the solution set of F (x) > 0 be a and B = {x | 1

Let's set two X10
When a > 0, X1 = (1 - √ (1 + 2A ^ 2) / A0
The solution set is x > X2, or X6 / 7
When a (a + 2) > 0 --- > A6 / 7 or a
Solve the equation {5x + 3Y = 36,2x-7y = - 2
emergency
x=6,y=2
What is the formula of root sign (1 + K ^ 2) * | x1-x2 |?
It's like this:
A: (x1, Y1). B: (X2, Y2). Find | ab|
K is the slope of the line ab
|Ab | = radical (1 + K ^ 2) ×| x1-x2|
|Ab | = radical [1 + (1 / k) ^ 2] ×| y1-y2|
By the way, it seems to be called the distance formula between two points. Anyway, that's what we call it
Simplification: 7Y (x-3y) & # 178; - (3y-x) & # 178;
7y(x-3y)²-(3y-x)²=7y(x-3y)²-(x-3y)²=(7y-1)(x-3y)²
=(7y-1)(x-3y)²
How is ab = radical {(1 + K ^ 2) * [(x1 + x2) ^ 2-4x1x2]} derived?
Be specific. Oh, urgent, urgent,
Is this the chord length formula in analytic geometry
Let's assume that the equation of the line is y = KX + B
The equation of a line and a curve is ax ^ 2 + BX + C = 0, and its two points are the intersection points of the line and the curve
Let's think this way. Suppose a straight line requires the distance between two points on the line. We can first find the absolute value of the difference between the abscissa of two points. The inclination angle of the straight line is a, Tana = k, then the ratio of the difference between the ordinates of two points and the difference between the abscissa is K. if the difference between the abscissa is D, the difference between the ordinates is DK, and the distance between two points is calculated by Pythagorean theorem, which is the root sign of D * (k ^ 2 + 1)
And how to find D? It requires the difference between the two ordinates, known X1 + X2, x1x2, then
(x1-x2)^2=(x1+x2)^2-4x1x2
So the difference between the two ordinates is (x1-x2) ^ 2 under the root sign, which is the root sign [(x1 + x2) ^ 2-4x1x2]
D is solved, and then the chord length formula is obtained
Radical (k ^ 2 + 1) * radical [(x1 + x2) ^ 2-4x1x2]
5x+3y=28 x-7y=-2
x-7y=-2
x=7y-2
5x+3y=28
5(7y-2)+3y=28
35y-10+3y=28
38y=10+28
Y=1
x=7-2=5
(2-x / x + 2 - 1) divide by 2x / x2-4, where x = 2-radical 2 reduces 3Q first
Urgent need
[(2-x)/(x+2)-1]/[2x/(x2-4)] =(-2x)[(x2-4)/2x] =(2+x)(2-x) =(4-√2)√2 =4√2 -2
Factorization; X3 + 4x2-9
Who can help me
x3+4x2-9
=x^3+3x^2+x^2-9
=x^2(x+3)+(x+3)(x-3)
=(x+3)(X^2+x-3)
Given that x2 = 3-2 times the root 2, then x=
A:
x^2=3-2√2
=2-2√2+1
=(√2)^2-2*√2*1+1^2
=(√2-1)^2
So:
X = √ 2-1 or x = 1 - √ 2
Given X & # 178; = 3-2 √ 2, then x=
x=±√(3-2√2)=±√(1-√2)²=±∣1-√2∣=±[(√2)-1]
X & # 178; = 3-2 times root sign 2
X & # 178; = (radical 2-1) &# 178;
X = radical 2-1, or x = - radical 2 + 1
X2 = 2 ^ 2 times root sign 2 + 1
X2 = (square of root 2 – 1)
Square on both sides, so x = plus or minus (root 2 – 1)
Hope to help you, if you don't understand can continue to ask! Hope to adopt! thank you!
9y2-4x2-4x-1 factorization
9y²-4x²-4x-1
=9y²-(4x²+4x+1)
=9y²-(2x+1)²
=(3y+2x+1)(3y-2x-1)
Hope to adopt