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I recently had an interesting discussion with my Algebra II teacher about solving the following inequality: $\sqrt{x}>-4$. As far as we could tell, squaring both sides results in $x>16$. However, common knowledge shows that any positive value would work. What is going on here?

I proceeded to generalize this to saying $1>-4$. By squaring both sides, we get $1>16$, which is clearly false. But there is no need to switch the inequality sign as we are not multiplying both sides by a negative, only squaring it.

Finally, I thought of the following problem: $\sqrt{x}>x-4$. Again, squaring both sides results in a solution of $\frac{9-3\sqrt{2}}{2}<x<\frac{9+3\sqrt{2}}{2}$. But again, 1 is a valid solution that does not fit this inequality.

Can someone please explain how this is happening?

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    $\begingroup$ When squaring $1>-4$, you are multiplying one side with a negative number and the other with a positive number. That changes the inequality sign in a completely unpredictable manner. $\endgroup$ – Arthur Feb 22 '18 at 21:59
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    $\begingroup$ squaring inequalities is very tricky. $-1>-2$ but $1\not > 4$, for example. $\endgroup$ – lulu Feb 22 '18 at 22:01
  • $\begingroup$ The inverse of square root is $+$, it is $\pm$, causing extraneous solutions. $\endgroup$ – user532449 Feb 22 '18 at 22:25
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Squaring is no equivalent-transformation.

But if both sides are known to be non-negative, it works , not only in the case of equations, but also in the case of inequalities.

To see the problem, just consider $x=3$ having obviously only solution $3$, whereas $x^2=9$ has the solutions $-3$ and $3$.

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Observe that $$ x^2>a^2\geq0 \iff \sqrt{x^2}>\sqrt{a^2}\iff |x|>|a|. $$ In particular $|1|<|-4|$ hence it does not follow that $1^2>(-4)^2$.

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$$\sqrt{x}>-4$$ it's $$x\geq0.$$

For solving $\sqrt{x}>x-4$ you need $0\leq x<4$ or $x\geq4,$ which gives $$x>(x-4)^2.$$ Finally, we obtain $$\left[0,\frac{9+\sqrt{17}}{2}\right).$$

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You should care that you can apply an operator to the both side an inequality if and only if this operator is injective. if the operator is ascending the direction of inequality will not change. if the operator is descending the direction will be reversed. Since $$y=x^2$$ is not injective you can not apply it to the both side. A similar discussion can be done in solving the equations. in which we will have the answers that are not the primary equation answer because of applying operators such $$y=x^2$$.

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The square function is not monotonic so it doesn't preserve inequalities.

$$a<b\not\Rightarrow a^2<b^2$$

Anyway it does so in the monotonic sections.

$$0<a<b\Rightarrow 0<a^2<b^2$$ and

$$a<b<0\Rightarrow a^2>b^2>0.$$

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  • $\begingroup$ Crossed with Salahamam_ Fatima. $\endgroup$ – Yves Daoust Feb 22 '18 at 22:12
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$\sqrt x>y$, where $y<0$ results in $y$ being positive after being squared, so you will get extraneous solutions.

This is because the inverse of $y^2$ is not $\sqrt y$, it is $\pm\sqrt{y}$.

If you take a look of the graph of $\sqrt x>-4$(red) and compare it to $x>16$(blue), you can notice what happens between the area of $0<x<16$.

enter image description here

For your second inequality, the same problem applies, those "valid" solutions are not actually valid, but are extraneous.

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$x\mapsto x^2$ is not increasing at $\Bbb R $ but it is increasing at $\Bbb R^+$ and decreasing at $\Bbb R^-$.

thus $$x>y\ge 0 \implies x^2>y^2$$

$$x <y \le 0 \implies x^2>y^2$$

$x <0 <y \;\;$ does not imply anything.

For your inequality, If $0\le x \le 4$ then it satisfies the inequation.

If $x>4$ the we can square to get

$$x>x^2+16-8x $$ $$x^2-9x+16 <0$$ $\delta=81-64=17$ $x_1=(9-\sqrt {17})/2 <4$ $x_2=(9+\sqrt {17})/2 >4$

$x \in (x_1,x_2) $

the final solution is $$[0,x_2) $$

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$A>B$ does not imply $A^2>B^2$ unless $B\geq 0.$ E.g. $-2>-3$ but $(-2)^2=4 \not >9=(-3)^2.$

The convention is that if $x\geq 0$ then $\sqrt x\;$ denotes the non-negative $y$ for which $y^2=x.$

And that $\sqrt x$ does not exist if $x<0.$ Usually. Some people use $\sqrt x\;$ when $x<0$ to denote a complex number $z$ such that $z^2=x,$ but since there are two allowable values for $z,$ it is ambiguous. If $x<0$ and you write $y=\sqrt x\;,$ it is not clear which of the two values is actually $y.$

So if we are restricted to only members of $\Bbb R $ then $\sqrt x\;\geq 0 >-4$ when $x\geq 0;$ while if $x<0$ then it is not true that $\sqrt x\;>-4$ because $-4$ cannot be less than something that doesn't exist.

Rather than trying to answer a Q by "applying a process" such as squaring both sides, apply logical inference to show that statements imply (or are equivalent to) another, and write it in full, with justifications,and without omitting the verbs, especially "$\implies$" and "$\iff$". (If a process like squaring both sides is done, justify why the resulting statement follows logically from the previous one.)

For example:

$(\sqrt x\;>-4)\iff$

$\iff (\;x\geq 0 \;...[$else $\sqrt x\;$ doesn't exist]... and $\sqrt x\;>-4\;)\iff$

$\iff (\;[x\geq 0$ and $\sqrt x\geq 0]\;$...(by def'n of $\sqrt x\;)...$ and $[\sqrt x\;>-4]\;)\iff$

$\iff (\;x\geq 0$ and $\sqrt x\;\geq 0\;)\iff (x\geq 0)$...(by def'n of $\sqrt x\;).$

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