# Solve inequality $\sqrt{\frac{x^2-22x+121}{x^2-24x+140}}\geq 50x-2x^2-309$

Solve inequality $$\sqrt{\frac{x^2-22x+121}{x^2-24x+140}}\geq 50x-2x^2-309$$

$$\sqrt{\frac{x^2-22x+121}{x^2-24x+140}}\geq 50x-2x^2-309=x^2-22x+121-3\left ( x^2-24x+140 \right )-10$$

$$\begin{cases} x^2-22x+121=a \\ x^2-24x+140=b \end{cases}\Rightarrow \sqrt{\frac{a}{b}}\geq a-3b-10$$

I think I've found a pretty nice double replacement for the right side, but I can't implement it. If I want to find the zeros of the function and solve the equation, not the inequality, then after squaring there will be no beautiful polynomial and the roots there cannot be found or calculated

If this is so, then this method does not lead to anything, or you need to express the right side of the inequality in some other way through $$a$$ and $$b$$, so that later you get a beautiful polynomial (after squaring)

Question: how to express the right-hand side beautifully in terms of $$a$$ and $$b$$? Maybe it would be easier to use another solution? I don’t really want to solve this inequality by considering two cases when the right side is less than zero or greater than or equal to zero

• You do realise that $x^2-22x+121 = (x-11)^2$? :-) Commented Jul 10 at 11:22

Your idea is fine but as a better strategy we can firstly check the domain for the LHS which seems to be $$x<10$$ or $$x=11$$ or $$x>14$$ and then noting that the RHS:

• is negative for $$x<10$$ or $$x>14$$, where the LHS is positive;
• is positive at the isolated point $$x=11$$, where the LHS is equal to zero;

so the solution is $$x<10$$ or $$x>14$$.

First, determine the domain of the left hand side:

You want $$L = \dfrac{x^2 - 22 x + 121 }{x^2 - 24 x + 140} \ge 0$$

Factoring both the numerator and denominator, we get

$$L = \dfrac{ (x - 11)^2 }{ (x - 12)^2 - 4 } = \dfrac{ (x-11)^2 }{ (x - 14)(x - 10) }$$

Since the numerator is always non-negative, then $$L = 0$$ at $$x=11$$ otherwise, $$L \gt 0$$ if $$x \lt 10$$ or $$x \gt 14$$.

The right hand side of the inequality is

$$R = 50 x- 2 x^2 - 309 = - 2 (x^2 - 25 x) - 309 \\= - 2 (x - 12.5)^2 + 2(12.5)^2 - 309 = - 2 (x - 12.5)^2 + 3.5$$

So $$R \le 0$$ iff $$| x - 12.5 | \ge \sqrt{1.75}$$.

This interval is $$x \in (-\infty , 11.17712] \cup [13.82287, \infty)$$

Intersecting this interval with the interval $$(-\infty, 10) \cup (14, \infty)$$ gives the interval

$$(-\infty, 10) \cup (14, \infty)$$

Now if $$R \gt 0$$, then $$x \in (11.17712 , 13.82287)$$. In this interval the left hand side is not defined. So we can ignore this case.

And the net result is that the solution to the given inequality is

$$(-\infty, 10) \cup (14, \infty)$$

• Thanks for pointing this out. I'll correct shortly. Commented Jul 10 at 10:44
• Guys, why do we need to make a negative denominator if the numerator goes to zero at $x=11$? I understand that the denominator must be strictly positive, but an important point! You can’t just split one big root for a fraction into two small ones for the numerator and denominator, otherwise the solution is lost Commented Jul 10 at 10:55

The LHS is well defined only for $$x<10$$ and $$x>14$$ and at $$x=11.$$ The RHS is negative at $$x=10,$$ at $$x=14$$ and positive at $$x=11.$$ Therefore the RHS is negative for $$x<10$$ and for $$x>14,$$ so these two intervals are the solutions.

• @user Thanks. I should not answer questions while watching the EURO CUP in soccer. Corrected Commented Jul 10 at 20:51
• Yes I think it is not allowed by the rules! ;)
– user
Commented Jul 10 at 20:55