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Subtleties of Radical Equations

Date: 12/07/2003 at 20:55:16
From: Kristina
Subject: Radical Equations

I don't understand on how to solve radical equations. Example:
 \/5x = 1.5 

I always seem to lose a step, or skip a step. 

Date: 12/08/2003 at 12:31:11
From: Doctor Riz
Subject: Re: Radical Equations

Hi Kristina -

Thanks for writing Dr. Math. Radical equations are one of the trickier 
things in algebra because they have some subtle points that need to be 
considered while solving and after you get your answer.

The general rule for solving radical equations is to isolate the 
radical on one side of the equation, then raise both sides to whatever 
power you need to undo or cancel the radical.  

For example, if you have a square root, you will square both sides of
the equation.  If you have a cube root, you will cube both sides.

Here's an example:
  \/x  = 5

Square both sides, and get x = 25.

So in your problem, you have:
  \/5x  = 1.5

Squaring both sides gives 5x = 2.25, and then you just need to divide 
by 5 to find x.

Now, here are the two main subtleties that come into play on radical 
equations.  First, when you square both sides, you have to square the 
entire side, not piece by piece.  Suppose you have:
\/x = x + 2

When you square both sides, you can't just square the x and the 2 
separately on the right.  You have to take the whole side and square 
  \/x = x + 2
    x = (x + 2)^2
    x = x^2 + 4x + 4

Many students make the mistake of just squaring each piece and getting 
x^2 + 4 on the right rather than x^2 + 4x + 4.

The second and more common subtlety involves what happens when you 
square both sides.  Here's an untrue statement:

  -2 = 2

But when I square both sides, that untrue statement becomes:

   4 = 4

which is clearly true!  This means that sometimes when you square both 
sides of an equation you wind up with answers that don't really check 
in the original equation.  Therefore, you ALWAYS have to check your 
answer and make sure it works.  Here's an example:
  \/x = -3

Squaring both sides gives x = 9, but when we check that we have:
  \/9 = -3

Is that a true statement?  Remember that the square root of 9 could be 
3 or -3, but we agree that the way the radical is written will tell us 
which one we mean:
   \/9 = 3
  -\/9 = -3

So in this example, the answer of 9 does not work because it gives us 
  3 = -3 

when we check it.  That means there is no possible solution to the 
given equation.

Confusing?  Sometimes, yes.  The things to remember as you work on 
radical equations are:

  1) Isolate the radical and raise to a power to get rid of it.

  2) When you raise to that power, raise both sides of the equation.  
     You can't do it term-by-term.  Slap ( ) around the other side 
     of the equation and put the exponent outside the ( ) to remind
     you to raise the whole side to that power.

  3) Always check your answer.  You'll find that it will either 
     work or be off by a negative, in which case you have to throw 
     it out.

Hope that helps.  Good luck, and write back if you need more help on 

- Doctor Riz, The Math Forum 
Associated Topics:
High School Square & Cube Roots
Middle School Square Roots

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