It is known that the solution sets of the equations X & sup2; - Px + 15 = 0 and X & sup2; - 5x + q = 0 are m and s, and m ∩ s = {3}. Find m ∪ s

It is known that the solution sets of the equations X & sup2; - Px + 15 = 0 and X & sup2; - 5x + q = 0 are m and s, and m ∩ s = {3}. Find m ∪ s

From m ∩ s = {3}, we know that two equations have a common root x = 3, then 3 & # 178; - 3P + 15 = 0, the solution is p = 83 & # 178; - 3 * 5 + q = 0, the solution is q = 6, so from X & # 178; - 8 + 15 = 0 (x-3) (X-5) = 0, the solution is x = 3 or 8x & # 178; - 5x + 6 = 0 (X-2) (x-3) = 0, the solution is x = 2 or 3, so m ∪ s = {2,3,8}

Let a be the solution set of the equation x ^ 2-px + 15 = 0, B be the solution set of the equation x ^ 2-5x + q = 0, and a ∪ B = {2,3,5}, a ∩ B = {3} find the value of P, Q and the set a, B
If the solution sets of equation x ^ 2-px + 15 = 0 and equation x ^ 2-5x + q = 0 are m and N respectively, and m ∩ n = {3}, the value of P-Q is obtained

If the solution sets of equation x ^ 2-px + 15 = 0 and equation x ^ 2-5x + q = 0 are m and N respectively, and m ∩ n = {3}, the value of P-Q is obtained
We can know that 3 is the root of two equations. Because the two products of equation x ^ 2-px + 15 = 0 are 15, then the other one is 5, two products and P = 8; the two products of equation x ^ 2-5x + q = 0 are 5, the other one is 2, two products q = 6
p-q=2

Given that the solution sets of the equations X & # 178; - Px + 15 = 0 and X & # 178; - 5x + q = 0 are a and B respectively, and a ∩ B = {3}, then the value of P + Q is

That is, x = 3 is the common root of the two equations
Substituting for the following:
9-3p + 15 = 0, i.e. P = 8
9-15 + q = 0, i.e. q = 6
So p + q = 8 + 6 = 14

Given that the solution sets of the equations X & # 178; - Px + 15 = 0 and X & # 178; - 5x + q = 0 are s and m respectively, and s ∩ M = {3}, then what is p + Q?
Why do I calculate 3 & # 178; - 3P + 15 = 0 as 2?

Let s = x1, X2, M = X3, X4
x1+x2=P
x1*x2=15
x3+x4=5
x3*x4=Q
And x1, X2; X3, X4 have a 3
Let X1 = X3 = 3
x2=5.x4=2
P=8,Q=6
P+Q=14

Given that the solution sets of the equations x ^ 2-px + 15 = 0 and x ^ 2-5x + q = 0 are s and m respectively, and s ∩ M = {3}, then p + q =? S ∪ M =?

p+Q=14 S∪M=[2,3,5]

Given the set u = {1,2,3,4,5}, a is a subset of u, and a = {x | x ^ 2 + PX + 4 = 0}, find the complementary set a
When a = an empty set, CUA = Cu Φ = u, how can CUA = Cu Φ = u be obtained
When a is not equal to an empty set, the two roots of the equation x ^ 2 + PX + 4 = 0, x 1 times x 2 = 4. Why do you want x 1 times x 2 = 4? What's the matter? Please explain in detail

When a = an empty set, the complement of a to u is Φ, and the complement of a to u is u
When a is not equal to an empty set, the two roots X1 and X2 of the equation x ^ 2 + PX + 4 = 0 are determined by Weida's theorem: X1 * x2 = 4,
So X1 = x2 = 2, then the complement of a is {1,3,4,5},
Or: X1 = 1, X2 = 4, then the complement of a is {2,3,5},

Given the complete set u = {1,2,3,4,5}, the set a = {x | x ^ 2 + PX + 4 = 0}, a is contained in U, find the complement of A
When a = an empty set, the equation x ^ 2 + PX + 4 = 0 has no real solution

(1) When a = an empty set, CUA = Cu, Φ = U
(2) When a is not equal to an empty set, the two roots X1 of the equation x ^ 2 + PX + 4 = 0 are multiplied by x2 = 4
When there is only one element in a, 2 * 2 = 4, that is 2, CUA = 1,3,4,5
When there are two elements in a, 1 * 4 = 4, that is, 1,4, CUA = 2,3,5
Yes, you don't need P

U = {1.2.3.4} a ∈ u and a = {x | x + PX + 4 = 0} to find the complement of a

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Given the complete set u = {1,2,3,4,5}, a = {XLX ^ 2 + PX + 4 = 0}, find the complementary set a
Why is x1x2 = 4?

Weida theorem:
For quadratic equation AX ^ 2 + BX + C = 0 (a ≠ 0)
Two x1, X2, with X1 + x2 = - B / A
x1x2=c/a
For x ^ 2 + PX + 4 = 0, there are x1x2 = 4, X1 + x2 = P
Proof: using the root formula
x1=(-b+√△)/2a
x2=(-b-√△)/2a
Multiply to get

There is only one element in the set M = {X / (A-1) x square + 3x-2 = 0}. What is the value of a
Can do the boss, thank you, please tell me I can't, another a has two solutions. What I need is the step of the answer

If A-1 = 0, a = 1, then it is a linear equation with one variable and has a solution
Then there is an element
If a is not equal to 1
This is a quadratic equation of one variable
Only one element has a solution
The discriminant is equal to 0
9+8(a-1)=0
a=-1/8
So a = 1, a = - 1 / 8