Algebra: Solving Radical Equations

Solving Radical Equations

Did you ever read Alice in Wonderland? It had a (slightly less popular) sequel called Through the Looking Glass, in which Alice actually passes into a mirror, and the world she encounters is a weird reflection of her own. Some physicists have likened the relationship between these worlds to the relationship between matter and antimatter.

Without getting too nerdy or discussing Star Trek propulsion systems (which are theoretically powered by matter/antimatter engines), I'll suffice it to say that you should never have matter and antimatter at the same social gathering, because if they come in contact with one another, there'll be a cataclysmic explosion. The two kinds of matter are the exact antitheses of one another, so much so that when they touch, they cancel one another out in the most permanent way possible: Detonation.

A slightly less violent, but similar, relationship exists between me and the girl I dated in high school, but that's irrelevant. More important for our purposes, the same relationship exists between radicals with index n and exponential expressions with power n. If they meet, they'll say, "It wasn't me that changed, it was you" (actually, that sounds more like my ex-girlfriend) and promptly explode, leaving behind only the contents of the radicand, a pile of smoke, and some embarrassing old love letters you can't remember writing.

This explosive property can be used both at heavy metal concerts to power the pyrotech- nics of groups like Limozeen and Taranchula, or to solve equations containing radicals. In the interest of space, I'll focus on the latter.

Example 6: Solve the equation

Solution: Since the variable is trapped inside that radical sign, you'll need to free it with some specially designed explosives. Before you set the charge, get everything but the radical away for safety. In other words, isolate the radical by subtracting 3 from both sides of the equation.

To destroy a radical with index 3, you'll raise it to the third power. (If it had been index 5, you'd raise both sides to the fifth power to eliminate it; just match the exponent with the index.) To keep the equation balanced, you should raise its other side to the third power as well.

All that remains on the left side of the equation are the smoldering contents of the radicand; the resulting equation is very simple to solve.

• 2x - 1 = 27
• 2x = 28
• x = 14

If you'd like, check your answer by substituting 14 for x in the original problem; you'll get

which is a true statement, which means you got the solution correct.

Excerpted from The Complete Idiot's Guide to Algebra © 2004 by W. Michael Kelley. All rights reserved including the right of reproduction in whole or in part in any form. Used by arrangement with Alpha Books, a member of Penguin Group (USA) Inc.

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