Solve the inequality -4x² + 260x - 200 is less than or equal to 0.

Step-by-step solution

1. 1

   Divide by the negative leading coefficient and reverse the sign. Dividing every term by $-4$ makes the quadratic monic, but because $-4<0$ the inequality flips:

   $-4x^2+260x-200\le 0\quad\Longleftrightarrow\quad x^2-65x+50\ge 0$

   This is the single most error-prone step; keeping the original downward-opening form and reasoning about "below the axis" is an equally valid alternative.

2. 2

   Find the roots of the monic quadratic. With $a=1$, $b=-65$, $c=50$,

   $\Delta=(-65)^2-4(50)=4225-200=4025$

   and $4025=25\cdot 161$, so $\sqrt{4025}=5\sqrt{161}$ (with $161=7\cdot 23$ square-free, this is fully simplified).

3. 3

   Write the roots exactly.

   $x=\frac{65\pm 5\sqrt{161}}{2}$

   Since $5\sqrt{161}\approx 63.4429$, the roots are approximately $0.7786$ and $64.2214$.

4. 4

   Use the direction of opening to pick the region. The parabola $x^2-65x+50$ opens upward, so it is $\ge 0$ outside the two roots and negative strictly between them:

   $x\le \frac{65-5\sqrt{161}}{2}\qquad\text{or}\qquad x\ge \frac{65+5\sqrt{161}}{2}$

5. 5

   Include the endpoints and test the answer. The inequality is non-strict, so both roots belong to the solution (there the original expression equals $0$). Testing: at $x=0.5$ the original is $-71\le 0$ ✓; at $x=1$ it is $+56$, which fails, correctly excluding the middle; at $x=65$ it is $-200\le 0$ ✓.