How to solve (x)^(2)-x-2 > 0

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

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Question

$${ x }^{ 2 } -x-2 > 0$$

Solve for x

$x\in \left(-\infty,-1\right)\cup \left(2,\infty\right)$

Steps Using the Quadratic Formula

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $ax^{2}+bx+c=0$.

$$x^{2}-x-2=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. Substitute $1$ for $a$, $-1$ for $b$, and $-2$ for $c$ in the quadratic formula.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\left(-2\right)}}{2}$$

Do the calculations.

$$x=\frac{1±3}{2}$$

Solve the equation $x=\frac{1±3}{2}$ when $±$ is plus and when $±$ is minus.

$$x=2$$ $$x=-1$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-2\right)\left(x+1\right)>0$$

For the product to be positive, $x-2$ and $x+1$ have to be both negative or both positive. Consider the case when $x-2$ and $x+1$ are both negative.

$$x-2<0$$ $$x+1<0$$

The solution satisfying both inequalities is $x<-1$.

$$x<-1$$

Consider the case when $x-2$ and $x+1$ are both positive.

$$x+1>0$$ $$x-2>0$$

The solution satisfying both inequalities is $x>2$.

$$x>2$$

The final solution is the union of the obtained solutions.

$$x<-1\text{; }x>2$$

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