Download MA 0090 Section 26 - Solving Equations Involving Radicals Friday

MA 0090 Section 26 - Solving Equations Involving Radicals
Friday, April 21, 2017
Note: We have Mini-Test 5 on Wednesday (4/26/2017) and Test III on Monday (5/1/2017).
We will have a first try for the Final Exam on Wednesday (5/3/2017), and on Friday, we will have an optional
review of the Final. The second try for the Final will be at our scheduled time during Finals Week. I will
take the better score of your two tries.
Objectives: Review solving equations with radicals in them, or using radicals to solve equations.
Equations with radicals
Squaring and taking the square root are inverse operations. We use inverse operations when we solve
equations (e.g., to solve x + 3 = −7, to undo the “plus 3”, we subtract 3 from both sides). Consider the
following equation.
√
(1)
x + 3 = 6.
The variable x is inside the square root, and we want it by itself on one side of the equation. The square
root symbol is in the way, so we would like to get rid of it. We can undo it by squaring both sides.
√
2
2
x +3 = (6)
(2)
x + 3 = 36
x = 33
We can always check our solutions to make sure we did things right. Just plug the solution in for x.
p
√
(33) + 3 = 36 = 6.
(3)
Here’s another example.
(4)
√
2x − 7 = 2
√
2
2
2x − 7 = ( 2 )
2x − 7 = 4
2x = 11
11
x=
2
Roots other than square roots work the same way. Just raise both sides to the correct power.
√
5
3x − 2 = 3
√
5
5
5
3x − 2 = ( 3 )
3x − 2 = 243
3x = 245
245
x=
3
(5)
You should get in the habit of checking your answers, but that is especially important here. When raising
both sides of an equation to an even power, it is possible to make a false equation into a true one. As a
result, you may end up with a solution that doesn’t really work. You can see the problem in the following.
2 6= −2,
but
(6)
2
2
( 2 ) = ( −2 )
4=4
1
MA 0090 Section 26 - Solving Equations Involving Radicals
2
In the following equation, you can see that there are no solutions (if you’re paying attention), but look what
happens, if you square both sides.
√
x − 7 = −5
√
2
2
x − 7 = ( −5 )
(7)
x − 7 = 25
x = 32
This looks OK, but when we check the solution, we get
p
√
(8)
(32) − 7 = 25 = 5 6= −5.
The difficulty, of course, is that the square root symbol is asking for the positive square root, and this cannot
be negative. There are no solutions.
Quiz 26, Part I
Find solutions to the following equations. Enter your answers as fractions, if you need to (e.g., x=-2/3 or
x=17 or no solutions).
1.
√
2.
√
3.
√
4.
√
5.
6.
x+2= 3
3x − 6 = 7
2x − 5 = 0
x − 11 = −2
√
3
2x + 1 = 3
√
7
3x − 2 = 2
Using the square root to solve an equation
To get rid of a square root, we can square both sides of an equation. With the same relationship, we can
get rid of squares by taking the square root of both sides of an equation. For example,
(9)
p
(x − 3)2 = 3
√
(x − 3)2 = 3
√
x−3 = 3
√
x = 3 + 3 = 4.73205 . . . = 4.7321
rounded to four decimal places. There is another solution, however. Note that squaring makes numbers
positive, just like absolute values do. In other words, solving
p
√
(10)
(x − 3)2 = − 3
will give us another solution. Specifically,
(11)
√
x−3=− 3
√
x = − 3 + 3 = 1.267949 . . . = 1.2679
rounded to four decimal places. The solutions here would be x = 4.7321, 1.2679.
MA 0090 Section 26 - Solving Equations Involving Radicals
3
The way we’ve just done it is just like what we did with absolute values. Instead of solving two equations,
it’s common to do both at the same time. Consider the following example.
(2x + 1)2 = 5
p
√
√
√
(2x + 1)2 = ± 5
there are now two equations, one with + 5 and one with − 5.
√
2x + 1 = ± 5
(12)
√
2x = ± 5 − 1
√
± 5−1
x=
= 0.6180, −1.6180
2
rounded to four decimal
you’re going to do two computations. The
√ places. To work this in your calculator,
√
second one, using − 5, you might have to enter “ 5,
, +/−, − , 1, =, ÷, 2, =.” Those of you with
√
, 5, −, 1, =, ÷, 2, =.”
different calculators would have to enter “−,
Quiz 26, Part II
Find both solutions to the following equations, and round to four decimal places, if appropriate (in D2L,
your answers will look like x=-3.2215,5.3232 or x=-2,3).
7.
(x + 7)2 = 5
8.
(x − 3)2 = 7
Homework 26
Find all solutions to the following equations.
In problems 1-6, enter your answers as fractions, if you need to (e.g., x=-2/3 or x=17 or no solutions).
√
1.
x+3= 5
√
2.
3x − 2 = 4
√
3.
3x − 5 = 0
√
4.
5x − 11 = −10
√
4
5.
2x − 1 = 2
√
5
6.
2x − 5 = 5
In problems 7-10, round your answers to four decimal places, if you need to (e.g., x=-3.2215,5.3232 or
x=-2,3).
7.
(x + 3)2 = 2
8.
(x − 4)2 = 9
9.
(x + 1)2 = 4
10.
(x − 3)2 = 3