Solving Painful Radical Equations
Purplemath
Every once in a while, they throw in an exercise which is meant primarily to catch out those careless students who tend to square all the terms in each side, rather than squaring the sides themselves. The following is a typical example:
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Solve:
This already has the square root by itself on one side, so I can proceed directly to squaring.
I should square both sides. But let's pretend for a minute that I'm being careless, and that I'm squaring all the terms on each side. Where does this lead? Let's see:
This is wrong!
Don't do this!
By (wrongly) squaring terms instead of (properly) squaring sides, I have arrived at a result which, technically speaking, means that every single value of x will work. (Zero is always equal to zero, regardless of the value of the variable, so the last line in my solving above means that "all real numbers are solutions".)
But, even if you didn't know that you're supposed to square sides rather than term, this result should leave you a little uneasy. Not because "all real numbers" is a "bad" solutions — sometimes it's the right solution — but because of what you know about graphs.
Looking at the original equation, and treating its sides as their own functions, I would have:
You know that the square-root function should graph as an arcing line, or some sort of curvaceousness; and that the other function is a straight line. No curved line can somehow be everywhere equal to any straight line.
There might be one solution (where the straight line cuts through the arc) or two solutions (if the straight line is angled correctly to cut off a portion of the arc), or even no solution (if the straight line is, say, always higher than the square root's arc). But the straight line can't possibly be always the same as the square root's curve. It makes no sense to think that it would be.
And the graph of the two sides of the original equaton for this exercise confirms what we already know from the graphing we've done in the past:
![y_1 = sqrt[9x^2 + 4] shown in blue, looking similar to a parabola, but with a sharper turn at the "vertext"; y_2 = 3x + 2 shown in green, being a straight line; a red dot at (x, y) = (0, 2) shows the intersection point](radicals/solve46.gif)
We must remember that we don't square terms; we square sides!
Returning to the exercise:
Remembering to square both sides of the equation they gave me, I get the following:
This matches the graph above, which is encouraging. Now, checking my solution, I get:
LHS = RHS
The solution checks, so my answer is:
x = 0
The rest of the exercises on this page require a lot of algebra to solve. I'll show you the algebra, so you can get a feel for what's required and how complete you'll want to be in your steps. I'll leave it to you to check the solutions (and, if you like, to do the graphs).
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Solve the equation:
I need to square both sides of this equation, being careful to write out the square on the right-hand side:
Then x = −8 and x = 2. Are both of these solutions valid? Check to see.
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Solve:
This equation will have to be squared twice in order to solve it:
It appears that the solutions are x = −5 and x = 0. However, only one of these solutions is actually valid. To find out which one, check them both.
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Solve:
This equation will also have to be squared twice. Don't forget to square that 3 in front of the square root on the right-hand side!
To finish solving this, use the Quadratic Formula. Then check your answers, because only one of the solutions is actually valid.
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Solve:
This equation is actually simpler than the two previous examples, because the two square roots on the left-hand side are multiplied together, rather than added or subtracted.
So, despite maybe looking a litle bit worse, this equation will need to be squared only once:
Then the solutions are x = −9 and x = 16. But x cannot equal −9, because this would put negatives inside both radicals in the original equation.
Now you check the other solution, to see if it works in the original equation.
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Solve:
Since there is a square root inside a square root, I'll have to square twice. But other than that, this equation really isn't that bad:
Using the Quadratic Formula, I get solutions of and x = 3. Check these, as only one is a valid solution.
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