Solve 6+(3-x)^2/x+2-1geq2-x^2/-x-2

$$\frac { 6 + ( 3 - x ) ^ { 2 } } { x + 2 } - 1 \geq \frac { 2 - x ^ { 2 } } { - x - 2 }$$

Solve for x

$$x\in (-2,\frac{15}{7}]$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-x\right)^{2}.

$$\frac{6+9-6x+x^{2}}{x+2}-1\geq \frac{2-x^{2}}{-x-2}$$

Add 6 and 9 to get 15.

$$\frac{15-6x+x^{2}}{x+2}-1\geq \frac{2-x^{2}}{-x-2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x+2}{x+2}.

$$\frac{15-6x+x^{2}}{x+2}-\frac{x+2}{x+2}\geq \frac{2-x^{2}}{-x-2}$$

Since \frac{15-6x+x^{2}}{x+2} and \frac{x+2}{x+2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{15-6x+x^{2}-\left(x+2\right)}{x+2}\geq \frac{2-x^{2}}{-x-2}$$

Do the multiplications in 15-6x+x^{2}-\left(x+2\right).

$$\frac{15-6x+x^{2}-x-2}{x+2}\geq \frac{2-x^{2}}{-x-2}$$

Combine like terms in 15-6x+x^{2}-x-2.

$$\frac{13-7x+x^{2}}{x+2}\geq \frac{2-x^{2}}{-x-2}$$

Subtract \frac{2-x^{2}}{-x-2} from both sides.

$$\frac{13-7x+x^{2}}{x+2}-\frac{2-x^{2}}{-x-2}\geq 0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x+2 and -x-2 is x+2. Multiply \frac{2-x^{2}}{-x-2} times \frac{-1}{-1}.

$$\frac{13-7x+x^{2}}{x+2}-\frac{-\left(2-x^{2}\right)}{x+2}\geq 0$$

Since \frac{13-7x+x^{2}}{x+2} and \frac{-\left(2-x^{2}\right)}{x+2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{13-7x+x^{2}-\left(-\left(2-x^{2}\right)\right)}{x+2}\geq 0$$

Do the multiplications in 13-7x+x^{2}-\left(-\left(2-x^{2}\right)\right).

$$\frac{13-7x+x^{2}+2-x^{2}}{x+2}\geq 0$$

Combine like terms in 13-7x+x^{2}+2-x^{2}.

$$\frac{15-7x}{x+2}\geq 0$$

For the quotient to be ≥0, 15-7x and x+2 have to be both ≤0 or both ≥0, and x+2 cannot be zero. Consider the case when 15-7x\leq 0 and x+2 is negative.

$$15-7x\leq 0 x+2<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when 15-7x\geq 0 and x+2 is positive.

$$15-7x\geq 0 x+2>0$$

The solution satisfying both inequalities is x\in \left(-2,\frac{15}{7}\right].

$$x\in (-2,\frac{15}{7}]$$

The final solution is the union of the obtained solutions.

$$x\in (-2,\frac{15}{7}]$$