Let's take a look at solving radical equations.
A radical equation is an equation where the variable appears under the radical sign.
Here is an example, the square root of x minus 7 equals 3.
So x appears under the square root sign.
To solve a radical equation, you need to raise both sides of the equation to a power
so that you can eliminate the square root sign.
In this example, we will raise both sides to the second power to eliminate the square root.
This will give us the square root of 8 x minus 7, that entire quantity squared, equals 3 squared.
On the left side when you simplify this you will get 8 x minus 7.
On the right side 3 squared equals 9.
And now we solve this the usual way for solving a linear equation.
I will add 7 to both sides and get 8 x minus 7 plus 7 equals 9 plus 7.
This gives 8 x equals 16. And then to solve for x we will divide both sides by 8,
giving us 8 x over 8 equals 16 divided by 8.
Make that an 8.
So this will simplify to x equals 2.
You have to remember when you are solving radical equations that you need to check the answers
because squaring both sides could introduce extra answers.
So for the check, go back to the original problem, substitute 2 in for x,
and check to see if when you simplify the left side that you get 3.
So for the check you will have the square root of 8 times 2 minus 7, and the question is does it equal 3?
Simplifying, we will have square root of 16 minus 7. Is it 3?
So the square root of 9, is it 3?
And you end up with a true statement 3 equals 3.
Since we have a true statement on the check, x equals 2 is a valid answer.
Now we will take a look at another radical equation.
Let's solve the square root of 10 minus 3 x equals x.
To eliminate the radical, we will square both sides,
giving us the square root of 10 minus 3 x squared equals x squared.
This will eliminate the square root and simplify to 10 minus 3 x equals x squared.
At this point, I need to get 0 on one side of the equation,
and so that I do not have a negative coefficient in front of x squared I am going to choose to get 0 on the left side.
This will give me 0 equals x squared plus 3 x minus 10.
This problem can be factored. It factors into x plus 5 times x minus 2,
and when you set each factor equal to 0, you get what looks like two answers.
x equals negative 5 or x equals positive 2.
But you must remember, because we squared both sides, that you have to check these answers.
So we will check the first one, x equals negative 5.
Plug negative 5 back into the original equation.
Simplify it, and see if you end up with a true statement.
This will give us the square root of 10 minus 3 times negative 5.
We want to know, does this equal negative 5?
When you simplify the left side, you will have the square root of 10 plus 15. Is it negative 5?
So, the square root of 25. Is it negative 5?
As it turns out we end up with 5 equals negative 5.
This is a false statement that we have on the check.
This means that negative 5 is not an answer; it is not a valid answer.
It is an extraneous answer.
I just put a mark through it right here to indicate that it is not a valid answer.
For the check on x equals two, plug 2 into the original equation.
And you have 10 minus 3 times 2, the square root of that quantity. Is it 2?
When you simplify this, you will get the square root of 4. Is it 2?
And simplifying we get 2 equals 2, a true statement.
So x equals 2 is the only answer to this problem.
Practice solving radical equations and remember to check your solutions.
Let's take a look at solving quadratic type equations.
We will do this problem 7 x to the fourth minus 6 x squared minus 1 equals 0.
Now notice the exponents on these terms.
The first term has an exponent of 4, the second term has an exponent of 2, and the third term is a constant term.
The exponents, the two being exactly half of 4, characterizes this as a quadratic type equation.
A lot of these equations can be solved using either factoring or substitution.
I will work this one both ways. First by factoring.
We will just treat it as a regular problem where we factor an expression.
On the left side I can factor 7 x to the fourth into 7 x squared times x squared.
Then for the 1, I will have to put 1 times 1.
Then with a little bit of thinking, you will figure you need a minus sign.
x squared minus one, and this factor is 7 x squared plus 1.
Now that it is factored, put each factor equal to zero, and solve for x.
So 7 x squared plus one equals zero or x squared minus 1 equals 0.
Let’s work the left side first.
So I have 7 x squared equals negative 1. And when you divide by 7, x squared is negative 1/7.
We could continue solving this for x by taking the square root of both sides,
however you will notice you cannot substitute a real number for x, square it, and get a negative 1/7.
So this factor, 7 x squared plus 1, does not give us a valid solution to this problem.
For the other factor, we will add one to both sides to get x squared equals one.
At this point, we can say that x is either plus or minus one.
So the solution to this problem is right here is plus and minus one.
Let’s take this one, put an x through it, just to say that is not a solution.
So, the answer is plus or minus one.
Now I would like to do the same problem but by using substitution.
So for substitution I will recopy 7 x to the fourth minus 6 x squared minus 1 equals 0.
What I want to do is substitute for x squared, and it may be helpful to rewrite this in this form:
7 x squared, it is x squared squared to get x to the fourth, minus 6 x squared minus 1 equals 0.
At this point we will substitute u for x squared.
So, I am going to write let u equal x squared, giving us 7u squared minus 6u minus 1 equals 0.
Now we will use factoring for 7 u squared minus 6u minus 1 equals 0.
This will factor into 7 u plus 1 times u minus 1,
and then putting each factor equal to 0. 7u plus 1 is 0 and u minus 1 equals 0.
We will have u equals negative 1/7, when I solve this one for u.
Now the problem here is that I want to know what x is equal to, not what u is equal to, so I have to go back and resubstitute.
Since u was equal to x squared, I will get x squared equals negative 1/7.
Again I cannot plug in or substitute a number for x, a real number for x, and square it and get negative 1/7.
So this is not a valid solution for this problem.
For this side u equals 1, but you need to resubstitute.
u equals x squared, so x squared equals 1, and here you will get the solution plus or minus one.
Now let's take a look at another example.
Let's take a look at the next example.
3 x to the 2/3 minus 2 x to the 1/3 minus 1 equals 0.
This is another equation that could be classified as quadratic type.
The first exponent, the first term, has an exponent of 2/3,
the second term has an exponent of 1/3, and the third time is a constant term.
The two to one ratio on exponents fits with the 2/3 and the 1/3.
This one I will solve by factoring.
To factor 3 x to the 2/3 it is simply this: 3 x to the 1/3 times x to the 1/3.
And then for the 1, of course that will be 1 times 1 is 1.
Placing in the plus and the minus sign, I will have x to the 1/3 minus 1
and this one is 3 x to the 1/3 plus 1 for the factors.
Then put equal to zero on the right side of this equation.
Now we will put each factor equal to zero.
3 x to the 1/3 plus 1 equals 0 or we have x to the 1/3 minus 1 equals 0.
I will work the one on the left side first.
Subtract 1 from both sides. 3x to the 1/3 equals negative 1.
When you divide by 3, you have x to the 1/3 equals negative 1/3.
To solve for x, I will need to raise both sides to the third power.
This will give us x equals negative 1/3 to the third power. So, x will be negative 1/27.
So, the factor 3 x to the 1/3 plus 1 equals zero has given us x equals negative 1/27.
For the other factor, x to the 1/3 minus 1, we have that one equal to 0.
Add 1 to both sides, x to the 1/3 equals 1.
When you raise both sides to the third power, you will have x equals 1 to the third power, which is simply x equals 1.
So the two solutions to this problem are negative 1/27 and positive 1.
Now we will take a look at another example.
Let's take a look at the next example.
This example is quadratic type.
I have a quantity raised to the second power within the first term.
In the second term, I have the same quantity and it is raised to the first power
although we normally do not write the 1 for the exponent.
This equation you could solve by multiplying x plus 2 times x plus 2,
distributing the 8, and distributing negative 23, then finding like terms.
Or we can use substitution. Substitution is the more efficient way to go, so we are going to work with substitution.
I am going to say let u equal x plus 2, and then with substitution you have 8 u squared minus 23 u minus 3 equals 0.
Now factoring 8 u squared minus 23 u minus 3 is much easier
than what I would have gotten had I multiplied all of this out at the beginning.
To factor 8 u squared minus 23 u minus 3, I will factor it into 8 u times u and of course the 3 will be 3 times 1.
I will have to put the 3 in the second set of parentheses so that when I multiply I will get the negative 23 u and the middle.
This would force the negative sign in front of the 3 and a positive sign in front of the 1.
At this point, I will set each factor equal to 0.
8 u plus 1 equals 0 or u minus 3 equals 0.
Solving for u, we will have that u equals negative 1/8 and here u equals 3.
The problem with this again is that I was supposed to solve for x and not u.
So I have to resubstitute the value for u. u is x plus 2.
So instead of u equals negative 1/8, x plus 2 equals negative 1/8.
And to solve for x subtract 2 from both sides.
This will give me x equals negative 2 minus 1/8.
Getting a common denominator, it will be 8. I will have negative 16/8 minus 1/8.
So one possible solution for x is negative 17 / 8.
For this one u equals 3, we will put the same thing over here.
u equals 3, and then resubstitute x plus 2 in for u.
Subtract 2 from both sides and get x equals 1.
So the solutions to this equation will consist of negative 17/8 and x equals 1.
Practice solving these equations until you have mastered the skill.
Now we will look at other equations that we could solve by factoring.
We will do this example, x cubed equals 81 x.
You may be tempted to divide by x on this problem,
but remember you cannot divide by zero, and we have no guarantee that x is not zero at this point.
To solve this equation what you need to do instead is subtract 81x from both sides to get x cubed minus 81x equals 0.
And now we will factor. Start by factoring out an x.
This will give you x times x squared minus 81 equals 0.
You could put each factor equal to 0 at this point, or you can continue factoring.
x squared minus 81 will factor into the difference of two squares, x plus 9 times x minus 9.
So I have x times x plus 9 times x minus 9 equals 0.
Set each factor equal to 0, you will get x equals 0, x plus 9 equals 0, and x minus 9 equals 0.
Solving each one, we will have x equals negative 9, x equals positive 9, or x equals 0.
The three solutions are 0, negative 9, and 9.
And now we will look at another example.
Let's take a look at another example.
This equation we will solve by grouping.
The first two terms have a common factor of x squared, so I will factor out the x squared.
And the last two terms have a common factor of negative 25. Factor out negative 25.
Notice with the grouping I now have x plus 3 and x plus 3 with each of these terms, so I will factor out x plus 3.
And then we will have x squared minus 25 for the second factor.
I could factor x squared minus 25, or at this point I could put both factors equal to 0.
I am going to do the second one, I am going to set each factor equal to 0.
x plus 3 equals 0 and x squared minus 25 equals 0.
For this one move the 3 over, subtract 3 from both sides, and get x equals negative 3.
For the second one, x squared minus 25 equals 0, add 25 to both sides, and get x squared equals 25.
When you square root both sides, you will get x is plus or minus 5.
So, the three solutions to this problem are x equals negative 3 and x equals plus or minus 5.
Practice solving these problems until you have mastered the skill.