Question

При каких значениях параметра а квадратное уравнение AX^2+X-A-2=0 не имеет корней?

Answer 1

$ax^2+x-a-2=0\\ D=1+4(a+2)=0\\ 1+4a+8=0\\ 4a+9=0\\ a=-\frac{9}{4}$

Answer 2

Quadratic Equation and its Roots

A quadratic equation in the form of \(AX^2 + BX + C = 0\) can have different numbers of real roots based on the discriminant, which is given by the expression \(B^2 - 4AC\). When the discriminant is negative, the quadratic equation does not have real roots.

The discriminant can be expressed in terms of the parameter \(a\) as follows: - For the equation \(AX^2 + X - A - 2 = 0\), the coefficients are \(A = 1\), \(B = 1\), and \(C = -A - 2\). - The discriminant is then \(1^2 - 4(1)(-A-2) = 1 + 4A + 8 = 4A + 9\).

Conditions for No Real Roots

The quadratic equation \(AX^2 + X - A - 2 = 0\) does not have real roots when the discriminant \(4A + 9\) is negative. Therefore, the condition for the equation to have no real roots is when:

4A + 9 < 0.

This condition ensures that the discriminant is negative, indicating that the quadratic equation does not have real roots.

Conclusion

The quadratic equation \(AX^2 + X - A - 2 = 0\) does not have real roots when the parameter \(a\) satisfies the condition 4A + 9 < 0.