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On the relationship between two real roots of a quadratic equation of one variable Let a and B be equation X ²+ Two real roots of x-2012 = 0, then a ²+ What is the value of 2A + B
A is the root of the equation
Then a ^ 2 + a-2012 = 0
a^2+a=2012
Weida theorem, a + B = - 1
Original formula = a ^ 2 + A + A + B = 2012-1 = 2011
Please write a binary quadratic equation with two real roots with opposite symbols
Univariate quadratic equations such as (x-a) (X-B) = 0 (the symbols of a and B are opposite) are consistent with the meaning of the question,
For example, x2-x-6 = 0
If the two real roots of a univariate quadratic equation are larger than x ²- Find the univariate quadratic if both real roots of X-5 = 0 are greater than 1
Let the two real roots of the equation be x1, x2
Then the equation x ²- The real root of X-5 = 0 is x1-1, x2-1
Bring x1-1, x2-1 into equation X ²- X-5 = 0, get
（x1-1) ²- (x1-1) - 5 = 0, i.e. x1 ²- 3x1-4=0
（x2-2) ²- (x2-1) - 5 = 0, i.e. x2 ²- 3x2-4=0
Therefore, the univariate quadratic equation is: X ²- 3x-4=0
It is known that the univariate quadratic equation x2-3x + M-1 = 0. If the equation has two equal real roots, find the solution of the equation
∵ the univariate quadratic equation x2-3x + M-1 = 0 has two equal real roots,
∴△=b2-4ac=0，
Namely: (- 3) 2-4 (m-1) = 0,
Solution: M = 13
4．
When m is what value, there are two equal real roots for the univariate quadratic equation with x ^ 2-4x + M-1 / 2 = 0? What are the roots of these two real numbers?
If a quadratic equation of one variable has equal real roots, its discriminant is equal to zero
Discriminant
＝b ²- 4ac
＝(-4) ²- four × one × (m-1/2)
=16-4m+2
=18-4m
＝0
The solution is m = 4.5
The original equation is: X ²- 4x+4=0
(x-2) ²= 0
x-2=0
x=2
It is known that the quadratic equation x ^ 2-4x + M-1 = 0 has two equal real roots
There are two equal real roots
So the discriminant is 0
So 16-4m + 4 = 0
m=5
x ²- 4x+4=0
(x-2) ²= 0
So X1 = x2 = 2
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