If the sum of the two roots of the equation x ^ 2 + 2kx + K + 1 = 0 is 2 root sign 3, then the value of K is

If the sum of the two roots of the equation x ^ 2 + 2kx + K + 1 = 0 is 2 root sign 3, then the value of K is


Because the sum of the two roots of X & sup2; + 2kx + K + 1 = 0 is 2 √ 3
From the relationship between the root and coefficient of quadratic equation of one variable, we can get: - 2K / 1 = 2 √ 3
So k = (2 √ 3) / - 2 = - √ 3
The relation between the root and coefficient of quadratic equation with one variable (also called "Weida theorem"): if a quadratic equation with one variable is ax & sup2; + BX + C = 0, and the two solutions of the equation are x1, x2
Then X1 + x2 = - B / A; x1 × x2 = C / A



On the equation (2kx + a) △ 3 = 2 + (x-bk) △ 6 of X, no matter what the value of K is, its root is always 1. Find the value of A.B


No matter what the value of K is, its root is always 1, so let x = 1 be brought into the equation
(2k+a)÷3=2+(-bk)÷6
The results show that: (4 + b) k = 13-2a
When B = - 4, a = 13 △ 2, no matter what the value of K is, the left and right sides of the equation are 0,
So B = - 4, a = 13 △ 2



In the case of the root of the quadratic equation with one variable x + 2kx + k-1 = 0


The discriminant of root = (2k) ^ 2-4 (k-1) = 4K ^ 2-4k + 4 = (2k-1) ^ 2 + 3 > 0
So the equation has two unequal real roots



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English translation
ditto


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