Question

При каких значениях параметра p уравнение 2x^2+px-p=0 не имеет корней? :3

Answer 1

Квадратное уравнение не имеет корней, если дискриминант этого уравнения меньше нуля.
2x²+px-p<0
D=p²-4·2·(-p)=9p²>0 при всех p≠0
при p=0 уравнение принимает вид
2х²=0
есть корень х=0
Ответ. Уравнение имеет корни при любом р.

Answer 2

Finding the Values of Parameter p for which the Equation 2x^2 + px - p = 0 Has No Roots

The equation 2x^2 + px - p = 0 represents a quadratic equation of the form ax^2 + bx + c = 0, where a = 2, b = p, and c = -p. For this equation to have no real roots, the discriminant (b^2 - 4ac) must be less than zero.

The discriminant for the given equation is: b^2 - 4ac = p^2 - 4(2)(-p) = p^2 + 8p

Therefore, for the equation 2x^2 + px - p = 0 to have no real roots, the discriminant p^2 + 8p must be less than zero.

Solving the inequality p^2 + 8p < 0: p(p + 8) < 0

This inequality has solutions when p < 0 and p + 8 > 0, which simplifies to p < 0 and p > -8.

Therefore, the values of the parameter p for which the equation 2x^2 + px - p = 0 has no roots are when p is less than 0 and greater than -8.

I hope this helps! If you have any further questions, feel free to ask.