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$\sqrt { -6x+10 } + \sqrt {-x+2} \gt \sqrt {4x+5}$

In the equality above I've tried to solve several times, several problems were discovered...solving it the regular way doesn't yield the correct solution, since apparently I need to consider during the process when does the inequality that is solved is true and when it's not...Can anyone elaborate?

$\frac {-34}{97}\lt x \lt 1 $ Is the solution I got however it was deemed wrong, since the

official answer is : $\frac {-5}{4}\le x \lt 1 $

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Also, another question; if I would like to check the correctness of the solution to $$2 \sqrt{10-6x} \sqrt{2-x} \gt 11x-7$$ aka $\frac {-5}{4}\le x \le \frac {7}{11} $ should I take a value of it and place it in $\sqrt { -6x+10 } + \sqrt {-x+2} \gt \sqrt {4x+5}$ or in $$2 \sqrt{10-6x} \sqrt{2-x} \gt 11x-7$$ ? –  Bak1139 Sep 14 '13 at 17:27

2 Answers 2

For the inequality to make sense, we seek real solutions. Thus, we require $$-6x+10 \geq 0, \qquad -x+2 \geq 0, \qquad \text{and} \qquad 4x+5 \geq 0.$$ These are equivalent to $$x \leq \tfrac{10}{6}, \qquad x \leq 2, \qquad \text{and} \qquad x \geq -\tfrac{5}{4}.$$ So $x \in [-\tfrac{5}{4},2]$ for this inequality to make sense.

The function $f:[-\tfrac{5}{4},\tfrac{10}{6}] \rightarrow [-\tfrac{5}{4},\tfrac{10}{6}]$ defined by $$f(x):=\sqrt { -6x+10 } + \sqrt {-x+2}-\sqrt {4x+5}$$ is continuous and strictly decreasing with $x$.

From here, we find $a$ where $f(a)=0$. Since $f$ is continuous and strictly decreasing, the points in $[-\tfrac{5}{4},a)$ will satisfy the inequality (i.e., $f(x)>0$ for $x \in [-\tfrac{5}{4},a)$) and the points $[a,\tfrac{10}{6}]$ will not (i.e., $f(x) \leq 0$ for $x \in [a,\tfrac{10}{6}]$).

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I think this is the interval in which the inequality will make sense $x \in [-\frac{5}{4},\frac{10}{6}]$. One counterexample: Your interval allows 2 as solution, but for $x=2$, $\sqrt{10-6x}$, becomes $\sqrt{-2}$, which has a complex solution. –  Stefan4024 Sep 9 '13 at 15:42
Thanks for pointing that out! Fixed it. –  Rebecca J. Stones Sep 9 '13 at 15:49
$\frac {-34}{97}\lt x \lt 1 $ So this is, in fact, the solution? –  Bak1139 Sep 10 '13 at 10:31
My answer agrees with Stefan4204's answer: since $f(1)=\sqrt{4}+\sqrt{1}-\sqrt{9}=0$, the interval is $[-\frac{5}{4},1)$. –  Rebecca J. Stones Sep 10 '13 at 11:19
Ok, It looks pretty more question if you may; if I would like to check the correctness of the solution to $$2 \sqrt{10-6x} \sqrt{2-x} \gt 11x-7$$ aka $\frac {-5}{4}\le x \le \frac {7}{11} $ should I take a value of it and place it in $\sqrt { -6x+10 } + \sqrt {-x+2} \gt \sqrt {4x+5}$ or in $$2 \sqrt{10-6x} \sqrt{2-x} \gt 11x-7$$ ? –  Bak1139 Sep 14 '13 at 17:21

I think you should square both sides until you get rid of the square roots. First because probabily you are interested into real solution, we should found out the interval in which the solution we'll be defined. The expression under the square root should be greater than or equal to $0$.

$$10-6x \ge 0 \implies x \le \frac{10}{6}$$ $$2-x \ge 0 \implies x \le 2$$ $$4x+5 \ge 0 \implies x \ge -\frac{5}{4}$$

So the interval in which our solution should be in order to have a defined solution is $[-\frac{5}{4},\frac{10}{6}]$

Now we start squaring:

$$\sqrt { -6x+10 } + \sqrt {-x+2} \gt \sqrt {4x+5}$$ $$(\sqrt { -6x+10 } + \sqrt {-x+2})^2 \gt (\sqrt{4x+5})^2$$ $$12+2 \sqrt{10-6x} \sqrt{2-x}-7 x \gt 4x+5$$ $$2 \sqrt{10-6x} \sqrt{2-x} \gt 11x-7$$

Now we'll split the inequality in two parts. First we'll check when the RHS is negative.

RHS is non-positive for $x\le \frac{7}{11}$. While the LHS is always positive, so for the interval $[-\frac 54, \frac{7}{11}]$ the inequality holds.

Second we'll check when the RHS is positive or $x>\frac{7}{11}$

Now we continue squaring:

$$(2 \sqrt{10-6x} \sqrt{2-x})^2 \gt (11x-7)^2$$ $$80-88 x+24 x^2 \gt 121x^2-154x + 49$$ $$-97x^2 + 66x + 31 >0$$

This quadratic equation is concave down so this means that this inequality will hold for the interval between its two roots.

Solving the quadratic equation we found its roots $x_1 = 1$ and $x_2 = -\frac{31}{97}$. But we made restiriction to make the RHS positive so the inequality will hold for every real number in the following interval:

$$x \in \left(\frac{7}{11},1\right)$$

If we add up both intervals from both cases we'll get: The inequality holds when $x$ is in the following interval:

$$x \in \left[-\frac{5}{4}, 1\right)$$

Note that also all numbers in this interval are also in the interval for a defined solution we obtained earlier.

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Buddy, I got the same solution as you and it was wrong. Note that 11x-7 isn't always smaller than left side and so the inequality isn't always true. –  Bak1139 Sep 9 '13 at 15:42
For any number in this interval $x \in \left(-\frac{31}{97}, 1\right)$, $11x - 7$ is smaller than the lHS. –  Stefan4024 Sep 9 '13 at 15:44
It's not...test it.. –  Bak1139 Sep 9 '13 at 15:50
The problem is that $11x−7$ may be positive or negative. So we cannot simply square the inequality $2 \sqrt{10-6x} \sqrt{2-x} \gt 11x-7$. –  Rebecca J. Stones Sep 9 '13 at 15:55
I fix it. Is it alright now? –  Stefan4024 Sep 9 '13 at 16:31

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