It is known that 2 + AI and B + I are two quadratic equations with real coefficients x times x + PX and q = 0, then the values of P and Q are?

It is known that 2 + AI and B + I are two quadratic equations with real coefficients x times x + PX and q = 0, then the values of P and Q are?

p=-4,q=5

If the difference between the square of equation x + PX + 1 = 0 (P > 0) is 1, then the value of P is 1

Let two roots be X1 x2
Then: X1 + x2 = - P
x1-x2=1
x1=(-P+1)/2
x2=(-P-1)/2
Substituting into the original equation
P = positive and negative root 5
Remove the negative value
The result of P is root 5

If the square of the equation x + PX + 1 = 0 (P > 0) has a real root and its two differences are 1, what is the value of P?

Using Vader's law
a=1 b=p c=1
X1+X2=-b/a=-p/1=1 X1*X2=c/a=1/1=1
X1-x2 = 1 can be regarded as the square of (x1-x2) = 1 and the square of (x1 + x2) - 4x1 * x2
So the square of P-4 = 1 or the negative root 5
And because P > 0, P = root 5

The two parts of the equation x & # 178; + 3px + q = 0 are equal to the square of the two parts of the equation x & # 178; + PX = 0 plus 1 respectively to find the value of P and Q?

X & # 178; + PX = 0 solution x = 0 or x = P, then 1 and P ^ 2 + 1 are two of X & # 178; + 3px + q = 0, so 1 + 3P + q = 0 (P ^ 2 + 1) ^ 2 + 3P (P ^ 2 + 1) + q = 0 solution P = 0 or - 1 or - 2 q = - 1 or 2 or 5. It is verified that P = - 1, q = 2 or P = - 2 q = 5 satisfies the equation x & # 178; + 3px + q = 0, respectively

Given that X1 and X2 are the two real roots of the equation x ^ 2 + PX + q = 0, and X1 + x2 = 6, X1 ^ 2 + x2 ^ 2 = 20, find the values of P and Q

p=-(x1+x2)=-6
2x1x2=(x1+x2)^2-(x1^2+x2^2)=36-20=16
q=x1x2=8

Given the quadratic function y = x2 + PX + Q, when x = 1, the value of the function is 4, when m = 2, the value of the function is - 5

According to the meaning of the problem, we get 1 + P + q = 44 + 2p + q = − 5, the solution is p = − 12q = 15, so the analytic expression of quadratic function is y = x2-12x + 15

Given the quadratic function y = x ^ 2 + PX + Q, when x = 1, the function value is 4; when x = 2, the function value is - 5, find the analytic expression of the quadratic function

Substituting x = 1 and y = 4, we get the following results
P + Q + 1 = 4
p＋q=3 ---------------------(1)
By substituting x = 2 and y = - 5, we get the following results
4 + 2p + q = - 5
2p＋q=－9 -----------------(2)
By solving (1) and (2), it is obtained that:
p=－12、q=15
Then:
y=x²－12x＋15

Given the quadratic function y = x2 + PX + Q, when x = 1, the value of the function is 4, when m = 2, the value of the function is - 5

According to the meaning of the problem, we get 1 + P + q = 44 + 2p + q = − 5, the solution is p = − 12q = 15, so the analytic expression of quadratic function is y = x2-12x + 15

Students a and B solve the quadratic equation of one variable. A misread the coefficient of the first term and get solutions 2 and 7. B misread the constant term and get roots. 1, - 10. The original equation is

Let the original equation be ax ^ 2 + BX + C = 0
Then according to a, the equation can be written as a (X-2) (X-7) = 0, that is, ax ^ 2-9ax + 14a = 0. A misread B, so a and C are right
According to B, the equation can be written as a (x-1) (x + 10) = 0, that is, ax ^ 2 + 9ax-10a = 0. B misread C, so a and C are right
So the equation is ax ^ 2 + 9ax + 14a = 0
That is x ^ 2 + 9x + 14 = 0

There is a quadratic equation with one variable. A misread the constant term and got two roots 3 and 4. B misread the linear term, so the two roots are - 8 and - 1. The original equation is ()

aX^2+bX+c=0
A misread the constant term and get two roots 3 and 4
Ax ^ 2 + BX + C1 = 0 (C1 is a new unknown)
Substitute x = 3 and 4 into the equation
9A + 3B + C1 = 0 and 16A + 4B + C1 = 0
So we can get b = - 7a
B is wrong once, so the two roots are - 8 and - 1
Ax ^ 2 + b1x + C = 0 (B1 is a new unknown)
Substitute x = - 8 and - 1 into the equation
64a-8b1 + C = 0 and a-b1 + C = 0
So we can get C = 8A
So the original equation is ax ^ 2-7ax + 8A = 0
And because the equation is quadratic in one variable, so a must not be zero, so we can get a by omitting a
X^2-7X+8=0