Solve the inequality 9x^2 + 48x - 576 >= 0.

Step-by-step solution

1. 1

   Pull out the common factor. All three coefficients are divisible by $3$:

   $9x^2+48x-576=3\left(3x^2+16x-192\right)$

   Since $3>0$, dividing the inequality by it changes nothing:

   $3x^2+16x-192\ge0$

2. 2

   Test whether it factors, and find that it does not. Factoring would need integers with product $3\times(-192)=-576$ and sum $16$. Running through the factor pairs of $576$ — $(1,576),(2,288),(3,192),(4,144),(6,96),(8,72),(9,64),(12,48),(16,36),(18,32),(24,24)$ — the achievable signed sums are $\pm575,\pm286,\pm189,\pm140,\pm90,\pm64,\pm55,\pm36,\pm20,\pm14,0$. None is $16$, so the trinomial is irreducible over the integers. A commonly quoted factorisation $(x-6)(3x+32)$ expands to $3x^2+14x-192$, not $3x^2+16x-192$ — its middle term is wrong, and indeed $9(6)^2+48(6)-576=36\neq0$, so $x=6$ is not a root at all.

3. 3

   Use the quadratic formula instead. For $3x^2+16x-192=0$:

   $\Delta=16^2-4(3)(-192)=256+2304=2560$

   $x=\frac{-16\pm\sqrt{2560}}{6}$

4. 4

   Simplify the radical. $2560=256\times10$, and $\sqrt{256}=16$, so $\sqrt{2560}=16\sqrt{10}$:

   $x=\frac{-16\pm16\sqrt{10}}{6}=\frac{-8\pm8\sqrt{10}}{3}$

   Numerically, with $\sqrt{10}=3.16228$:

   $x_1=\frac{-8+25.2982}{3}=5.76607,\qquad x_2=\frac{-8-25.2982}{3}=-11.09941$

5. 5

   Choose the intervals from the direction the parabola opens. The leading coefficient $9$ (or $3$ after reduction) is positive, so the parabola opens upward and is $\ge0$ outside the two roots:

   $x\le\frac{-8-8\sqrt{10}}{3}\qquad\text{or}\qquad x\ge\frac{-8+8\sqrt{10}}{3}$

6. 6

   Include the endpoints and verify. The inequality is non-strict, so the roots themselves — where the expression is exactly $0$ — belong to the solution set. Checking: both roots give $9x^2+48x-576=0$ to within $10^{-9}$ ✓, and scanning $6001$ points from $-30$ to $30$ matches this solution set at every point ✓. In interval form: $\left(-\infty,\tfrac{-8-8\sqrt{10}}{3}\right]\cup\left[\tfrac{-8+8\sqrt{10}}{3},\infty\right)$.