Solve Radical Equations


Uploaded by MATHRoberg on 03.12.2010

Transcript:
Hi, I'm Kendall Roberg, and today we're going to solve radical equations.
The first step is to isolate the radical.
So, in this first problem, we have the cube root of x minus 9 equals -1.
We want to isolate this part of the equation.
So by adding 9 to both sides of the equation, we now have the cube root of x equals 8.
Now, another way of writing the cube root of x is x to the 1/3.
We want x to the first power, not x to the 1/3 power, so by cubing both sides...3 times 1/3 is just one so we get x is equal to whatever 8 cubed is.
8 times 8 times 8.
And I'll leave you to do that on your calculator.
Let's try another one.
On this one, the radical is already isolated.
We have the square root of (x + 25) = 4.
By squaring both sides, we're left with x + 25 = 16.
The square root went away because it's really the exponent of 1/2.
1/2 times 2 would give us the exponent of 1.
So now we have x + 25 = 16.
If we subtract 25 from both sides, we find out that x = -9.
Let's try another one.
This time we have a fractional exponent.
We have 3 times x to the 3/2 equals 375.
We need to isolate the x and the 3/2; in other words, we're isolating the x with its exponent.
If we divide both sides by 3, we're left with x to the 3/2 equals 125.
So, x to the 3/2, we want the exponent to be one.
So we should raise both sides to the 2/3 power.
Because 3/2 times 2/3 is 1.
So now we know that x = 125 to the 2/3 power, and our challenge is figuring out what is 125 to the 2/3.
Using the properties of exponents, we can rewrite 125 to the 2/3 as (the cube root of 125) squared.
The cube root of 125 is 5, 5 squared is 25.
So we end up with x = 25.
Alright, let's do one last problem.
In this problem, we have the square root of (10x + 9) equals x + 3.
By squaring both sides, we get rid of the square root on this side, and we have to FOIL this, (x + 3) times (x + 3) is x squared + 6x + 9.
Now, if we subtract 10x from both sides, and if we subtract 9 from both sides, we get 0 equals x squared - 4x and plus 9 minus 9, that goes away.
Now let's factor this side.
So either this or this has to be 0 for the product to be 0.
So our two solutions here are x = 0 or x = 4.
Now what we need to do is actually check to see if either or both of these solutions is correct.
We have to check for extraneous solutions.
Anytime you raise both sides of an equation to an exponent, you're going to need to check for extraneous solutions.
So here's how we check.
Let's see, if x = 0, if our original equation would actually be true.
So I plug in 0 where the x's are...
And I get...well 10 times 0 is 0 plus 9...so I get the square root of 9, and I want to know does that equal 3.
And yes, 3 equals 3.
So, x does equal 0.
Let's check this one to see if it works.
I plug in the 4, so I get 10 times 4 plus 9 equals 4 plus 3...
So this would be 40 plus 9 is the square root of 49.
And we want to know does that eqaul 7?
Well the square root of 49 is 7, 7 equals 7, so x = 4 is also a solution.
So this time we had no extraneous solutions, but that won't always be the case.