Solving Irrational Inequality: Why Square Root Matters

[Mr Davis 97]

So I am trying to solve a simple rational inequality: $\sqrt{x} < 2x$. Now, why can't I just square the inequality and go on my way solving what results? What precisely is the reason that I need to be careful when squaring the square root?

 

[Mark44]

So I am trying to solve a simple rational inequality: $\sqrt{x} < 2x$. Now, why can't I just square the inequality and go on my way solving what results? What precisely is the reason that I need to be careful when squaring the square root?

It's not so much the square root -- it's the simple fact of squaring both sides of an inequality.

Example:
-2 < 1
If you square both sides, you get 4 on the left side and 1 on the right side.
Obviously, the direction of the inequality needs to change, since 4 > 1.

 

[pwsnafu]

In this particular question, we know that both sides are non-negative, so there is nothing to worry about.

 

[Mr Davis 97]

Well, the signs are not an issue in this scenario, but if I don't pay attention to the fact that I am squaring, I get an extraneous result. For example, if we go from $\sqrt{x} < 2x \implies x < 4x^2 \implies x(4x - 1) > 0$, which then gives us $x > \frac{1}{4}$ or $x < 0$. However, $x < 0$ is obviously extraneous because of the square root in the beginning of the problem, which implies that x must be positive. Thus, what is it about squaring that causes the extraneous solution set to occur?

 

[Mark44]

Well, the signs are not an issue in this scenario, but if I don't pay attention to the fact that I am squaring, I get an extraneous result. For example, if we go from $\sqrt{x} < 2x \implies x < 4x^2 \implies x(4x - 1) > 0$, which then gives us $x > \frac{1}{4}$ or $x < 0$. However, $x < 0$ is obviously extraneous because of the square root in the beginning of the problem, which implies that x must be positive. Thus, what is it about squaring that causes the extraneous solution set to occur?

Squaring both sides of an equation or inequality is not a one-to-one operation, so isn't reversible. Once you have squared something, you can no longer determine whether the thing that was squared was positive or negative.
Leaving aside inequalities for the moment, consider $\sqrt{x} = -2x$. Squaring both sides gives $x = 4x^2$ or x(4x - 1) = 0, with solutions x = 1/4 or x = 0. x = 1/4 is not a solution of the original equation. This extraneous solution arose because of the squaring operation.

 

[Mr Davis 97]

Squaring both sides of an equation or inequality is not a one-to-one operation, so isn't reversible. Once you have squared something, you can no longer determine whether the thing that was squared was positive or negative.
Leaving aside inequalities for the moment, consider $\sqrt{x} = -2x$. Squaring both sides gives $x = 4x^2$ or x(4x - 1) = 0, with solutions x = 1/4 or x = 0. x = 1/4 is not a solution of the original equation. This extraneous solution arose because of the squaring operation.

So would a hard and fast rule be to just proceed as normal (i.e. square both sides and solve for x), and then after all is said done, check back to see whether you have an extraneous answers?

 

[Mark44]

So would a hard and fast rule be to just proceed as normal (i.e. square both sides and solve for x), and then after all is said done, check back to see whether you have an extraneous answers?

Yes, this is something you should do. This would also include raising both sides to the fourth or other even power.