$\displaystyle \dfrac{x+5}{4} <= (\dfrac{x-7}{2})^{2}$
$\displaystyle \dfrac{x+5}{4} <= \dfrac{x^{2}-14x+49}{4}$
$\displaystyle -x^{2}+15x <= 44$
How do I continue?
Hi,
the one on the right is correct, because the graph is below the x-axis between x=1 and x=2.
$\displaystyle 1\ \le\ x\ \le\ 2$
For the equation on the left, $\displaystyle -\infty\ < x\ \le4$ and $\displaystyle 11\ \le\ x\ < \infty$
The values you have written is the range of x on or below the x-axis for that graph.
If you want to find the solutions without reference to the graphs, you can reason thus...
$\displaystyle (x-4)(x-11)$ is positive if $\displaystyle x>11$
as that causes both factors to be positive
or $\displaystyle (x-4)(x-11)$ is positive if $\displaystyle x<4$
as that causes both factors to be negative.
If x is in between 4 and 11, one factor is negative and the other positive.
For $\displaystyle (x-1)(x-2)$ both factors are positive if $\displaystyle x>2$
and both factors are negative if $\displaystyle x<1$
Hence, the factors have opposite sign if x is between 1 and 2.