2
$\begingroup$

Consider the following inequality: $\sqrt { 8x-7 } + \sqrt {2x+2} \lt 3$

Attempt a solution: after doing all the calculations I got: $x \gt 7$, or, $\frac 78 \le x \lt 1$. On checking this result, however applying greater than $7$ values to the inequality yielded absurdities. Can I cancel an entire part of the solution based on one result? Or were my calculations wrong?

$\endgroup$
  • 1
    $\begingroup$ Please make sure I got the formatting right. Thanks. $\endgroup$ – Ross Millikan Sep 6 '13 at 16:36
  • $\begingroup$ Everything is correct. Thank you. $\endgroup$ – Bak1139 Sep 6 '13 at 17:40
1
$\begingroup$

When solving an inequality with square roots like this, extraneous solutions should be expected. You have already clearly solved this, so I will not recreate all of that, but I know exactly what you did, so I will sketch that and the proper logic for the record.

Before solving this you set the bounds for the radicals themselves:

$$8x-7 \geq 0 \wedge 2x+2 \geq 0 \Rightarrow x \geq \frac{7}{8} \wedge x \geq -1 \Rightarrow x \geq \frac{7}{8}.$$

Ok good, one lower boundary. Next you solved the equation, and I am quite sure you squared both sides of an inequality twice. In the end you had

$$(x-7)(x-1)>0.$$

Your thinking here should have been that the product of two binomials being positive implies that they are either both positive, or they are both negative.

$$(x-7)(x-1)>0 \Rightarrow \left((x-7)>0 \wedge (x-1)>0\right) \vee \left((x-7)<0 \wedge (x-1)<0\right).$$

So if both binomials are positive you conclude

$$(x-7)>0 \wedge (x-1)>0 \Rightarrow x>7 \wedge x>1 \Rightarrow x>7.$$

If both are negative you conclude

$$(x-7)<0 \wedge (x-1)<0 \Rightarrow x<7 \wedge x<1 \Rightarrow x<1.$$

So you not only need to test the conditions $x>7$ and $x<1$, but you should expect extraneous solutions as a result of squaring square roots twice. Clearly as you noted $x \ngtr 7$, but $x <1$ works. You combine this with your initial lower bound $x \geq \frac{7}{8}$ to conclude as you have that

$$\frac{7}{8} \leq x < 1.$$

So to precisely answer your question, knowing precisely what you experienced, your calculations were correct, and yes you can and should cancel a part of your solution as it is not actually a solution. It is just an extraneous result of repeated squaring on square roots.

$\endgroup$
  • $\begingroup$ Many thanks for you detailed replay! How do I justify my abandonment of an entire part of the solution? Is it OK if I take a particular value of x , greater than 7 for instance, apply to my original equation and show that I get an absurd expression? $\endgroup$ – Bak1139 Sep 6 '13 at 18:18
  • 1
    $\begingroup$ The reason this happens in inequalities involving radicals like this is because when you rid the inequality with square roots by squaring, sometimes the resulting (in this case quadratic) inequality is not equivalent to what you started with. Think about going backward and taking square roots. As for your proper demonstration, yes I do think it looks really good to cite an example of why you are excluding a region. It shows your understanding and might even help a reader. $\endgroup$ – J. W. Perry Sep 6 '13 at 18:29
  • $\begingroup$ OK, I think I got it...Many thanks! $\endgroup$ – Bak1139 Sep 7 '13 at 11:32
  • $\begingroup$ Also, one more minor question, I hope you don't mind...Is it OK on the base of only one value which doesn't fit into the inequality to throw out an entire solution section? $\endgroup$ – Bak1139 Sep 7 '13 at 15:49
  • 1
    $\begingroup$ If you are solving one of these, and you find after dutifully checking that just one single value in a solution section does not fit, then you should absolutely consider that entire section suspect. If you truly have all critical values, and one value in a section does not fit, then none of them will fit. If you know you made no omissions, then yes just throw it out. I usually give it a double look though. If you find a value does not fit in a section, but others do fit, then you definitely missed a critical value and need to hunt for an error or omission in your process. $\endgroup$ – J. W. Perry Sep 7 '13 at 22:05
1
$\begingroup$

Clearly if $x$ is very large, the left side will be very large and greater than $3$, so the range cannot be unbounded above. You probably squared something to get there and introduced an extraneous solution, but we can't be sure unless you show your work. We must have $x \ge \frac 78$ for the first square root to make sense.

$\endgroup$
  • $\begingroup$ No, no it doesn't seem that way... $\endgroup$ – Bak1139 Sep 6 '13 at 17:50
1
$\begingroup$

In order to avoid negative square roots, we need $8x-7 \geq 0$ and $2x+2 \geq 0.$ This gives the conditions $$x \geq \tfrac{7}{8}$$ and $x \geq -1$ (this second inequality being redundant).

We also observe that $f:[\tfrac{7}{8},\infty) \rightarrow [\tfrac{7}{8},\infty)$, defined by $f(x)=8x-7$, is a non-negative, increasing and continuous function. The same is true for $g:[\tfrac{7}{8},\infty) \rightarrow [\tfrac{7}{8},\infty)$ defined be $g(x)=2x+2$. We conclude that $h:[\tfrac{7}{8},\infty) \rightarrow [\tfrac{7}{8},\infty)$ defined by $h(x)=\sqrt{f(x)}+\sqrt{g(x)}$ is a well defined function, and is increasing and continuous.

We observe that $h(1)=\sqrt{1}+\sqrt{4}=3$. Hence $h(x) \geq 3$ whenever $x \geq 3$ and $h(x)<3$ whenever $x \in [\tfrac{7}{8},1)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.