0

In algebra I, they taught us that when solving equations with radicals, you just had to plug the solutions you received back into the original equation to check if they were legit solutions or extraneous solutions. However, this can be time consuming. This is my method.

Take an equation like $\sqrt{10-x}$ = $3x$. If you square both sides, you get $10-x$ = $9x^2$. Then, you get the polynomial $9x^2+x-10$. Solving this polynomial, you get solutions $x=1$ & $x= -\frac{10}9$. However, if you plug the negative solution, the answers do not match up. Here's my way to solve this equation without checking for extraneous solutions.

First, note that $\sqrt{10-x}$ $\geq$ $0$ for all real x because square root is positive. $3x$ $\leq$ $0$ for all negative x. Since $\sqrt{10-x}$ $\geq$ $0$ for all real x, but $3x$ $\lt$ $0$ for all negative x, that means our solution can't be negative or the sides won't be equal. Thus, we can eliminate all negative solutions easily without plugging them in to check.

This method will save a lot of time for higher degree polynomials where you have to check more solutions. So, is this method mathematically valid? If so, why don't we learn it in school?

Some Guy
  • 2,687
  • Your logic is indeed valid. For higher degree polynomials, though, it becomes correspondingly harder to decide when the functions are positive or negative. – Greg Martin Jan 06 '21 at 08:17
  • "So, is this method mathematically valid?" Sure, but it's highly specialized. If you had, say, $\sqrt{10-x}=3x^7-1x^5+41x^2-5$, it'd take a lot more effort to articulate conditions that make the right-hand side negative than it would just to substitute-in candidate values and see what happens. Computers can make (in)validation-by-substitution instantaneous, so that this isn't a hassle. That said, it can be worthwhile to give some thought to (in)validity conditions that let you rule-out bad values; doing so might well provide insights into the problem that gave rise to the equation at hand. – Blue Jan 06 '21 at 08:18
  • 1
    Incidentally, I propose calling extraneous roots "weeds". (I'm probably not the first.) – Blue Jan 06 '21 at 08:23

1 Answers1

1

Yes, it is valid. However, it is a lot harder to use this in a lot of cases than to simply plug things in, because the equation may be way more complicated, so you can't tell if the functions are positive or negative. Also, $x^2$ stays positive everywhere.