Week 5 – Math 11 – Solving Radical Equations

Solving radical equations is very similar to normal algebra, but it includes square roots which make things fairly interesting.

Radicals make things interesting because most times, a radical will be an irrational number, which is generally ugly and confusing, so it’s best top get rid of them when you can.

Getting rid of square roots or any root in general is easy as it follows the same rules as everything else in algebra. You take the inverse to cancel it out and what you do to one side, you do to the other. The opposite of square roots is the power of 2, so all you need to do is so that to both sides.

Like so…

\sqrt {5x} = 10

 

(\sqrt {5x})^2 = 10^2

 

5x = 100

 

And from here you just simplify like normal.

\frac {5x}{5} = \frac {100}{5}

 

x = 20

 

This usually won’t be as stand alone and simple as my previous example unfortunately. We will need to use our previously learned algebra skills to some some equations too.

Here’s an example:

4 \sqrt {x + 1} - 5 = 3

 

4 \sqrt {x + 1} = 8 First we start to isolate our radical by removing the -5 by adding +5 to both sides.

 

\sqrt {x + 1} = 2 We do the same here but with the x4 by dividing both sides by 4.

 

(\sqrt {x + 1})^2 = 2^2 Now that our radical is isolated, we square both sides.

 

x + 1 = 4 Then we isolate x.

 

x = 3 Too easy.

That’s the basics of solving radical equations. Not too hard right?

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