Quadratic inequality

[Rbfht]

Why do I get different answers using this method instead of interval method?

 

[skeeter]

Why do I get different answers using this method instead of interval method?

because the original function is undefined at [MATH]x = -3[/MATH]
[MATH]x-7+\frac{25}{x+3} \ge 0[/MATH]
[MATH]\frac{(x-7)(x+3)+25}{x+3} \ge 0[/MATH]
[MATH]\frac{x^2-4x+4}{x+3} \ge 0[/MATH]
[MATH]\frac{(x-2)^2}{x+3} \ge 0[/MATH]
critical values are [MATH]x=2 \text{ and } x=-3[/MATH]
check a value of x in each interval formed by the critical values to determine where the inequality is true

 

[lex]

@Rbfht
You multiplied an inequality by something which could be negative (reversing the inequality sign) or positive, so you must consider both cases.

Firstly, as observed, [MATH]x\neq-3[/MATH]
Then [MATH](x-7)+\frac{25}{(x+3)}\geq0[/MATH]Multiply across by [MATH](x+3)[/MATH] and distinguish 2 cases:

1. When [MATH]x+3>0[/MATH] (i.e. [MATH]x>-3[/MATH])
[MATH](x-7)(x+3)+25\geq0[/MATH][MATH]x^2-4x+4\geq0[/MATH][MATH]\left(x-2\right)^2\geq0[/MATH]True for all [MATH]x\in\mathbb{R}[/MATH]
So combining the restrictions, x must satisfy both [MATH]x>-3[/MATH] and [MATH]x\in\mathbb{R}[/MATH]i.e. [MATH]x>-3[/MATH]
2. When [MATH]x+3<0[/MATH] (i.e. [MATH]x<-3[/MATH])
[MATH](x-7)(x+3)+25\le0[/MATH] inequality sign reverses, since multiplying by a negative number
[MATH]x^2-4x+4\le0[/MATH][MATH]\left(x-2\right)^2\le0[/MATH]True only for [MATH]x=2[/MATH].

So combining the restrictions, x must satisfy both [MATH]x<-3[/MATH] and [MATH]x=2[/MATH] [MATH]\qquad\mathcal{X}[/MATH] i.e. no solution

Final answer: [MATH]x>-3[/MATH]