The discussion focuses on solving the rational inequality ##\sqrt{x} < 2x## and the implications of squaring both sides. It highlights that squaring is not a one-to-one operation, which can lead to extraneous solutions. For instance, squaring the inequality results in ##x(4x - 1) > 0##, yielding solutions ##x > \frac{1}{4}## and ##x < 0##, where the latter is extraneous due to the square root constraint. The key takeaway is to always verify solutions after squaring to avoid incorrect results.
PREREQUISITESStudents, educators, and anyone interested in mastering algebraic inequalities and understanding the nuances of squaring operations in mathematical problem-solving.
It's not so much the square root -- it's the simple fact of squaring both sides of an inequality.Mr Davis 97 said: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?
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.Mr Davis 97 said: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?
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 said: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.
Yes, this is something you should do. This would also include raising both sides to the fourth or other even power.Mr Davis 97 said: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?