2x-3x-5 = 0, solve the equation, the absolute value of x1-x2 If the quadratic equation KX + 1 = x-x2 with respect to X has real roots, the value range of K is obtained

2x-3x-5 = 0, solve the equation, the absolute value of x1-x2 If the quadratic equation KX + 1 = x-x2 with respect to X has real roots, the value range of K is obtained

x1+x2=3/2
x1x2=-5/2
So (x1-x2) & sup2; = (x1 + x2) & sup2; - 2x1x2
=9/4+5
=29/4
So | x1-x2 | = √ 29 / 2

Square of B + 4ac
Do you want to bring in the negative sign before the square of B? Will it be positive if it is brought in? Will the negative number be removed if the square of B is - (4K + 1)?

No, it's all positive

After the square = 0 of equation (X-2 + radical 3) is transformed into general form, the square of b-4ac =?

Because the equation: (X-2 + radical 3) ^ 2 = 0 is already a complete square expression, the quadratic equation with one variable has a unique solution, so the discriminant = 0, that is: B ^ 2-4ac = 0