How do I solve a second-degree inequality with two variables

[Guest]

Find the solution for m∈ℝ such as [tex]f(x)=(m+3)x^2-(m-4)x+m+4>0, ∀x∈(-\infty;0)[/tex]

Can anyone thoroughly explain the steps required to solve this?

[Guest]

Strictly speaking that doesn't[b] have "two variable", it has one variable, x with a parameter, m. The simplest way to solve an inequality, often, is to first solve the equation. The equation, [tex](m+ 3)x^2+ (m+ 4)x+ (m+ 4)= 0[/tex], can be solved using the quadratic formula: [tex]x= \frac{-(m+ 4)\pm\sqrt{(m+ 4)^2- 4(m+ 3)(m+ 4)}{2(m+ 3)}[/tex]. Depending on the value of m, in particular on the discriminant, [tex](m+ 4)^2- 4(m+ 3)(m+ 4)[/tex], that equation may have two real solutions, or one, or none.

Looking at the discriminant, [tex](m+ 4)^2- 4(m+ 3)(m+ 4)= (m+ 4)(m+ 4- 4m- 12)= -(m+ 4)(3m+ 8 )[/tex], it is 0 at m= -4 and m= -8/3. The discriminant is negative for m< -4, positive for m between -4 and -8/3, and negative again for m greater than -8/3. So the original expression, [tex](m+ 3)x^2+ (m+ 4)x+ (m+ 4)[/tex] is never 0 for m less than -4, has a single zero if m is -4, has two roots for x between -4 and -8/3, has a single 0 if m= -8/3, and is never 0 for m greater than -8/3.

That means that [tex](m+ 3)x^2+ (m+ 4)x+ (m+ 4)[/tex] is either [b]always positive or always[b] for m< -4, changes sign once for m= -4, changes sign twice for m between -4 and -8/3, changes sign once for m= -8/3, is either always positive or always negative for m> -8/3.

Finally, we can determine when [tex](m+ 3)x^2+ (m+ 4)x+ (m+ 4)[/tex] is greater than 0 and when it is less. For x= 0, that has value [tex]m+ 4. That is negative for m< -4, 0 at m= -4, and positive for m> 4, Looking at the previous paragraph, that means that, for m< -4, the expression is negative for all x. If m< -4, the given inequality is not true for any x.

If m is -4, the expression is [tex]-x^2[/tex] which is never positive. If m is -4, the inequality is not true for any x.

If m is between -4 and -8/3, by the previous paragraph, the expression changes sign twice, so it has two zeros, and is positive at x= 0. The expression is negative for x less than the first zero, is positive for x between the two zeros, and negative again for x larger than the second zero.

If m is -8/3, the expression is [tex](1/3)x^2+ (4/3)x+ 4/3= (1/3)(x+ 4x+ 4)= (1/3)(x+ 2)^2[/tex]. That is positive for all x except x= -2.

Finally, for m greater than -8/3, the expression is again of one sign for all x, At x= 0 it is m+ 4= -8/3+ 4= 4/3, which is positive. The inequality is true for all x.