The equation x-ax-b = 0 and X + BX + a = 0 have only one equal root of real number

The equation x-ax-b = 0 and X + BX + a = 0 have only one equal root of real number

x2-ax-b=0 ① x2+bx+a=0 ②
From the meaning of the title, we can get ② - ① (B + a) x + A + B = 0
The solution is x = - 1

It is known that the equations x ^ 2 - (K + 1) X-2 = 0 and x ^ 2-2x-k (K + 1) = 0 about X have only one same real root, so we can find the value of K and the common root

With the same root a, then:
a^2-(k+1)a-2=0
a^2-2a-k(k+1)=0
The difference between the two methods is as follows
(k-1)a=(k+2)(k-1)
When k = 1
x^2-2x-2=0,x^2-2x-2=0
There are two identical real roots
So K is not equal to 1
So the common solution is K + 2
Substituting K + 2 into the equation
(k+2)^2-(k+1)(k+2)-2=k=0
The equation here is
x^2-x-2=0,x^2-2x=0
In conclusion, k = 0 and the common root is 2

It is known that X1 and X2 are the two roots of the equation 3x-2x-2 = 0. Solve the equation and find the absolute value of x1-x2

X1 + x2 = B / a = - 2 / 3, X1 * x2 = C / a = - 2 / 3 (two basic formulas)
(x1-x2) square = (x1 + x2) square - 4 * X1 * x2 = 4 / 9 - (- 8 / 3) = 28 / 9
The absolute value of x1-x2 = 2 / 3 times the root sign 7