MA 1128: Lecture 17 – 11/14/16
Adding Radicals
Radical Equations
Adding/Subtracting Radicals
Be very careful when you see addition or subtraction mixed with radicals.
For example,
and
9  16
9  16
Are not the same!!
You can see this by simplifying each of them.
9  16  25  5
9  16  3  4  7
These are clearly not equal.
Next Slide
Special Situations
We can simplify in a few special situations (and otherwise, we’ll leave them as
they are).
Both of the situations mentioned here involve factoring out.
This first example is like combining like terms and uses the distributive property.
The radical parts have to be exactly the same.
73 2  53 2  23 2
This next example factors inside the radical, and depends on exponents and
radicals distributing over multiplication or division.
8  12x  4(2  3x)  4 2  3x  2 2  3x
By factoring inside, we get multiplication under the radical,
and radicals split with multiplication.
Next Slide
Practice Problems
Simplify if you can.
1.
2.
3.
4.
7 3 5 3 2 3
x3  x 2
x7
5 2
Answers on next slide.
Next Slide
Answers
Simplify if you can.
1.
2.
3.
4.
7 3  5 3  2 3  10 3
x 3  x 2  x 2 ( x  1)  x x  1
x7
5 2
There isn’t much we can do with the expressions in problems 3 and 4
Next Slide
Radical Equations
We have a little more freedom when we’re working with equations.
With an equation, we can do the same thing to both sides.
For example, we can square both sides of this next equation.
x 1  3


2
x  1  32
Since taking a square root and squaring are inverse operations,
they undo each other.
x 1  9
x 8
Next Slide
Be sure you check your answers
If you check x = 8 in this last equation, you’ll see that it works.
You do have to be careful with squaring both sides of an equation, however.
Squaring can make unequal things equal.
For example, 3  -3, but (3)2 = (-3)2.
Note that the right side of this next equation is negative.
x  1  3
x  1  (3) 2
x 1  9
x8
(8)  1  9  3  3
Squaring both sides of an equation
will sometimes introduce wrong answers.
You should always check your answers,
and throw out the bad ones.
Next Slide
The radical should be by itself
In order to get a simpler equation,
you need to be sure that the radical is by itself on one side.
For example, note what happens if you don’t do this.



x 3 5  6

x  3  5  x  3  5  36
x  3   x  3  5  5  x  3  25  36
2
x  3  5  62
2
( x  3)  10 x  3  25  36
There’s nothing really wrong with this,
But we still have a radical, so we’re no better off. In fact, things got worse.
Next Slide
Example (cont.)
WE SHOULD HAVE DONE THE FOLLOWING.
x3 5  6

x 3 1

2
x  3  12
x 3 1
x4
Check:
(4)  3  5  1  5  1  5  6
Next Slide
This solution is fine.
Practice Problems
1.
2.
3.
x2 5
x3 2  4
x  2  3
Answers:
1) x = 23
2) x = 7
3) No solutions. If you square both sides, you’ll end up with x = 7, but if you
plug this back in, this solution doesn’t work. You can see this from the beginning,
since the square root symbol is defined to indicate the positive square root, and the
positive square root can’t be negative 3.
Next Slide
More Examples
Be sure to square (or cube etc.) each side as a whole.
You should square the entire left side and square the entire right side, not the
individual terms. Look at the right side in the second line.
x2  3  x 1
 x  3   x  1
2
2
2
x2  3  x2  2x 1
3  2x 1
2  2x
1 x
Next Slide
More Examples
You might need to cube (or something else).
3

6x  3  3

3
6 x  3  33
6 x  3  27
6 x  30
x5
3
Next Slide
Practice Problems
Solve the following equations.
1.
2.
3 x  12  x  2
4
3x  1  2
Answers:
1) After squaring, you get a quadratic equation. Move everything to the right side,
and you get 0 = x2 – 7x – 8. This factors to 0 = (x + 1)(x – 8), so the solutions are
x = -1,8.
2) x = 5
Next Slide
Two Radicals in an Equation
Radical equations can be really hard, or even impossible, to do.
The worst that we’ll consider are like the equations in the last two quiz problems,
or equations with a single radical on both sides of the equation.
The idea is basically the same, and they’re maybe even a bit easier.
Just be sure that each radical is by itself on one side of the equation.

8x  4  7 x  2
 
2
8x  4  7 x  2
8x  4  7 x  2
x6

2
4
x  8  4 2x  0
4
x  8  4 2x

  2x 
4
x 8 
x  8  2x
8 x
4
Next Slide
4
4
Practice Problem
Solve.
10.
3
5x  10  3 3x  8
Answer
1) x = 1.
End