1. ## Another algebra exercise

$\displaystyle \dfrac{x+5}{4} <= (\dfrac{x-7}{2})^{2}$

$\displaystyle \dfrac{x+5}{4} <= \dfrac{x^{2}-14x+49}{4}$

$\displaystyle -x^{2}+15x <= 44$

How do I continue?

2. Originally Posted by klik11
$\displaystyle \dfrac{x+5}{4} <= (\dfrac{x-7}{2})^{2}$

$\displaystyle \dfrac{x+5}{4} <= \dfrac{x^{2}-14x+49}{4}$

$\displaystyle -x^{2}+15x <= 44$ no need to bring the terms to the left

How do I continue?
$\displaystyle x^2-15x+44\ \ge\ 0$

$\displaystyle (x-11)(x-4)\ \ge\ 0$

This is a U-shaped quadratic, which lies above the x-axis to the left of the smaller root
and to the right of the larger root.

3. Thanks!!!!!

4. I'm sorry but I still can't solve the whole exercise.

The whole exercise is this:
"Solve the next systems:" (sorry for the bad translation)

The solution is
$\displaystyle 1 \ge\ x \ge\ 2$

What am I doing wrong?

5. Originally Posted by klik11
I'm sorry but I still can't solve the whole exercise.

The whole exercise is this:
"Solve the next systems:" (sorry for the bad translation)

The solution is
$\displaystyle 1 \ge\ x \ge\ 2$

What am I doing wrong?
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.

6. I see, thanks again!

7. 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.