Solve (9x+63)^2>2025

$$( 9 x + 63 ) ^ { 2 } > 2025$$

Solve for x

$$x\in \left(-\infty,-12\right)\cup \left(-2,\infty\right)$$

Solution Steps

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.

$$\left(9x+63\right)^{2}=0$$

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 81 for a, 1134 for b, and 1944 for c in the quadratic formula.

$$x=\frac{-1134±\sqrt{1134^{2}-4\times 81\times 1944}}{2\times 81}$$

Do the calculations.

$$x=\frac{-1134±810}{162}$$

Solve the equation x=\frac{-1134±810}{162} when ± is plus and when ± is minus.

$$x=-2 x=-12$$

Rewrite the inequality by using the obtained solutions.

$$81\left(x+2\right)\left(x+12\right)>0$$

For the product to be positive, x+2 and x+12 have to be both negative or both positive. Consider the case when x+2 and x+12 are both negative.

$$x+2<0 x+12<0$$

The solution satisfying both inequalities is x<-12.

$$x<-12$$

Consider the case when x+2 and x+12 are both positive.

$$x+12>0 x+2>0$$

The solution satisfying both inequalities is x>-2.

$$x>-2$$

The final solution is the union of the obtained solutions.

$$x<-12\text{; }x>-2$$