When solving a quadratic equation of one variable, if a misprints the constant term and gets roots of - 2 and - 3, and B misprints the primary term and gets roots of 6 and - 1, then the correct equation is

When solving a quadratic equation of one variable, if a misprints the constant term and gets roots of - 2 and - 3, and B misprints the primary term and gets roots of 6 and - 1, then the correct equation is

x²+bx+c=0
A C wrong, B correct
Then - 2 + (- 3) = - B
b=5
B wrong, C right
Then 6 × (- 1) = C
c=-6
x²+5x-6=0
(x+6)(x-1)=0
So the correct ones are - 6 and 1

The sufficient conditions for two real coefficient equations X & # 178; + PX + q = 0 to become one RT ⊿ two acute angles are given
P & # 178; - 2q = 1,0 < Q ≤ 1 / 2, P < 0

The sufficient condition is: P ^ 2-2q = 1.0

There is a point P (m, n) on the image of the inverse scale function y = K / x, whose coordinates are two of the unary quadratic x ^ 2-3x + k = 0 of X, and the distance between P point and the origin is the root sign 13. The analytic expression of the inverse scale function is obtained

From the question: m ^ 2 + n ^ 2 = 13
From the relationship between root and coefficient: M + n = 3
mn=k
（m+n)^2=m^2+2mn+n^2=9
So 2Mn = - 4
mn=-2
Then k = - 2
Easy to get X1 = 1, X2 = 2
So the analytic formula is y = 2 / X

Mathematics -- common equations and the relationship between the root and coefficient of linear equation with one variable
If the real numbers a, B and C satisfy a = 6-b and the square of C = ab-9, then a=______ ,b=____________ ,c=_________

a=6-b =>a+b=6
c^2=ab-9 =>ab=c^2+9
So a, B are the roots of the equation x ^ 2-6x + C ^ 2 + 9 = 0
△=36-4(c^2+9)=-4c^2
So C = 0
So a = b = 3

When the Dragon Boat Festival is coming, a shop bought a batch of zongzi in boxes for 2400 yuan. During the festival, each box was sold at the price of 20% increase. After the festival, each box was sold at the price of 5 yuan lower than the purchase price, and the rest of the zongzi was sold out. The whole business process made a profit of 350 yuan

Let the price of each box of rice dumplings be x yuan. From the meaning of the title, we can get 20% X × 50 - (2400x-50) × 5 = 350, simplify to get x2-10x-1200 = 0, and solve the equation to get X1 = 40, X2 = - 30. Through the test, x1 = 40, X2 = - 30 are all the solutions of the original equation, but x2 = - 30 is not the meaning of the title, so it is omitted. Answer: the price of each box of rice dumplings is 40 yuan

Write a linear equation with one variable, the coefficient of exponent is - 3, and the solution of the equation is 1 / 2

(1)-3x=-3/2
(2)-3x+3/2 =0
(3)-3x+1.5=0
(4)-3x=-1.5

The parabola y = x2 + PX + Q intersects the x-axis at two points a and B, intersects the negative half axis of the y-axis at point C, the angle ACB = 90 ° and 1 / 2 of OA minus 1 / 2 of OB equals 2 / 2 of OC

1/OA-1/OB=2/OC
(OB-OA)/OA*OB=2/OC
OC=2OA*OB/(OB-OA)
Angle ACB = 90 ° OA * ob = OC square = | Q | = OC q = - 1
OC = 2oC / (ob-oa) 2oC = ob-oa
(ob-oa) square = (x1-x2) square = (x1 + x2) square - 4x1x2 = P square - 4q = 4q square = 4
4q = p-4q P = 4q + 4q
The length of AB = 2

Function y = x ^ 2 + PX + Q, set a = {x | x = f (x)}, set B = {x | f [f (x)] = x}. If there is only one element in a, judge the relationship between a and B
What I want to ask is if there is only one in a... their relationship!

Take any element x0 in the set a, because x0 ∈ a,
If x0 = f (x0), then f (f (x0)) = f (x0) = x0
That is, x0 = f (f (x0)), so x0 ∈ B,
That is, any x ∈ B in a, so a ∈ B
Even if there is only one in a, is it still a ∈ B?

Let a = {x ∈ Z X & sup2; - Px + 15 = 0}, B = {x ∈ Z X & sup2; - 5x + q = 0}, if a ∪ B = {2,3,5} find a, B

∵ the elements of set a and B are the solutions of quadratic equation with one variable. It can be seen that a and B have two elements
Let the elements of set a be X1 and X2, and the elements of set B be X3 and X4,
Then a ∪ B = {x1, X2, X3, X4},
If X1 and X2 are two of x2 PX + 15 = 0, then X1 + x2 = P, X1 * x2 = 15
From X1 * x2 = 15, we can see that X1 = 3, X2 = 5 or X1 = 1, X2 = 15,
And 1,15 ￠ a ∪ B = {2,3,5}, 2,3 ∈ a ∪ B = {2,3,5}
That is, a = {3,5}
Then B must have an element of 2, let X4 = 2
We substitute X4 = 2 into X & sup2; - 5x + q = 0, so q = 6
And X3 + X4 = 5, X3 = 3
So X1 = 3, X2 = 5, X3 = 3, X4 = 5,
That is, a = {3,5}, B = {2,3}

Let a = {x ∈ Z | X & sup2; - Px + 15 = 0}, B = {x ∈ Z | X & sup2; - 5x-q + 0}, if a ∪ B = {2,3,5}, a and B are ()
（A）{3,5}、{2,3} （B）{2,3}、{3,5}
（C）{2,5}、{3,5} （D）{3,5}、{2,5}
A = {x ∈ Z | X & sup2; - Px + 15 = 0}, B = {x ∈ Z | X & sup2; - 5x-q = 0}. Same after

15 = 1 * 15 = 3 * 5, so in a, either (x-1) (X-15) = 0 or (x-3) (X-5) = 0, but a ∪ B = {2,3,5}, so it can only be (x-3) (X-5) = 0, x = 3 or x = 5A = {x | 3,5} B: x ^ 2-5x-q = 0 (x-m) (x-n) = 0, because a ∪ B = {2,3,5}, and a = {x | 3,5}, so there must be 2 in B, assuming n = 2 (X-2) (x-m) = 0