# Inequalities question

I've been having trouble solving these kind of inequalities ;

$\sqrt { -7x+1 } - \sqrt {x+10} \gt \ {7}$

Attempt at a Solution; We first find the official boundaries of the inequality; x is limited to the section: $\ -10 \le x \le \frac{1}{7}$

Then, squaring both sides we get:

${-7x+1} -2 \sqrt{(x+10)}\sqrt{(-7x+1)} +x+10 \gt 49$

$-2 \sqrt{(x+10)}\sqrt{(-7x+1)} \gt 38+6x$

squaring again eventually yields:

$-28x^2-276x+40\gt 1444+456x+36x^2$

$-64x^2 -732x -1404 \gt 0$

binomial roots are as specified below.

since the parabola is concave and in order this inequality is to be satisfied, x must assume the values:$\ -9 \lt x \lt -2.4375$ upon intersecting this with the range we get the same result and this is the (apparent) final solution.

We then resume to the usual manipulation of the inequality; squaring both sides, factoring etc. We get two roots; $-9$,$-2.4375$ . The parabola we get after these manipulations is concave, so we get the solution: $\ -9 \lt x \lt -2.4375$ . We then intersect this solution with the range for x, yielding the final solution $\ -9 \lt x \lt -2.4375$. Upon testing it we find it does not fulfill the inequality we tried to solve, so we cancel it as a solution, leading to that there are no solutions to this inequality, as a final answer. Official answers state otherwise.

Any help?

• The inequality is certainly true for $x=-10$, so there must be some solution.. – Mårten W Sep 12 '13 at 16:13
• You're saying I should test the entire range even if it does not come up as a solution by it's own? – Bak1139 Sep 12 '13 at 16:14
• No, testing all numbers is not a viable approach, but it is usually a good idea to test some numbers in order to get a feeling for the problem. Since there are many possible ways to do the manipulations you describe, could you show exactly how you do? – Mårten W Sep 12 '13 at 16:24
• Note that for $x=-9$, the LHS will evaluate to 7 exactly, thus there may be something to be said for testing values on each side to see if as x is increased or decreased does it pass by this point. – JB King Sep 12 '13 at 16:27
• This is correct. However I decided to take a more direct approach perhaps? After I get the solution and rule it out, I check the remaining sections of the range of x to see if they fit my inequality. If they do I take them, if they dont I discard them, and then I unite all those that were ok together for the final solution. Is this OK? – Bak1139 Sep 12 '13 at 18:31

From $\sqrt{-7x+1} - \sqrt{x+10} \gt 7$, your assessment of the interval $x \in [-10, \frac{1}{7}]$is clearly right.

Let us make sure we have positive quantities on both sides first, so rewrite as $\sqrt{-7x+1} \gt 7 + \sqrt{x+10}$

Squaring, $- 7x + 1 > 49 + x + 10 + 14\sqrt{x+10}$
or $-4x -29 > 7\sqrt{x+10}$

Now it is easier to see that the LHS is positive only when $x < -\frac{29}{4}$, so the allowable values have shrunk to $x \in [-10, -\frac{29}{4}]$. In this interval, we have assurance of both sides being positive, so squaring again, we have

$16x^2 + 232x + 841 > 49(x+10) \implies 16x^2 + 183x + 351 > 0$ So we have $$16\left(x+\dfrac{183}{32}\right)^2 > \dfrac{11025}{64} \implies x +\dfrac{183}{32} \not \in \left[- \frac{105}{32}, \frac{105}{32} \right]$$

Thus the allowable values are further shrunk to $x \in [-10, -9)$.

• And how about this approach? After I get the solution and rule it out, I check the remaining sections of the range of x to see if they fit my inequality. If they do I take them, if they dont I discard them, and then I unite all those that were ok together for the final solution. Is this OK? – Bak1139 Sep 12 '13 at 18:32
• If you have been careful each step (in this case being careful about squaring only positives), the "remaining" sections should not have any case which satisfy the inequality. Perhaps you should post the steps you did also as an addendum to the question, it may be easier to spot if any mistake is being made. – Macavity Sep 12 '13 at 18:40
• @Macavity...I added a few more details.. – Bak1139 Sep 12 '13 at 19:16

First of all, before squaring the initial inequality, it seems appropriate to note that $\sqrt{-7x+1}-\sqrt{x+10}$ has to be greater that zero (i don't know, perhaps this is redundant, but i'd do that just to make sure...)

Then, after squaring both parts, i got $3x+19 + \sqrt{(-7x+1)(x+10)} < 0$, which gives us $\sqrt{(-7x+1)(x+10)} <-(3x+19)$, if i'm not mistaken. This yelds the condition $3x+19 < 0$, or $x < -\frac{19}{3}$. Then i squared everything, and got almost what you got - $x < -9$ or $x > -\frac{78}{32}$. So now it is enough to intersect all inequalities properly... Will the answer be correct?

And my final answer seems to be $[-10,-9)$ (after calculations)... Is it correct?

• Yes - that looks good. I posted my working before seeing your answer, matches exactly. – Macavity Sep 12 '13 at 16:38
• This is correct. However I decided to take a more direct approach perhaps? After I get the solution and rule it out, I check the remaining sections of the range of x to see if they fit my inequality. If they do I take them, if they dont I discard them, and then I unite all those that were ok together for the final solution. Is this OK? – Bak1139 Sep 12 '13 at 18:30
• frankly speaking, i've never heard of checking solutions of inequalities of such types before... Well, the inequalities of this kind: $(x-x_1)...(x-x_n) < 0$ are done like that, the real line is divided onto the respective number of segments, and the way to solve the inequality is to check one point from the leftmost segment (or any of them, whatever) and the rest of the segments are ruled out or included automatically - this is what i know... but here you can have some extra solutions due to the fact that you square both parts of the inequality, and who knows how it can behave... – W_D Sep 12 '13 at 20:17
• for instance, the solution after squaring might turn out to be, say, $[0,2]\cup [3,5]$, but to maintain equivalence, you had to add the inequality which would give you the restriction, say, $[1,4]$, So the solution would be $[1,2]\cup[3,4]$, but with this method of ruling out, if you chose points 0 and 5 to check the segments, you'd rule out everything... – W_D Sep 12 '13 at 20:21
• Have i got your method right? – W_D Sep 12 '13 at 20:21