Download 1 The (positive) square root symbol 2 Squaring Both Sides

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MA 1165 - Lecture 13
02/18/09
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The (positive) square root symbol
We’ve used the square
root symbol already, but before we move on, I want to emphasize something about
√
it. When we say 4, this specifically is the positive square root of 4, so
√
4 = 2.
(1)
√
We don’t need a special symbol for the negative square root, since we can write − 4 = −2.
2
Squaring Both Sides
Taking a (positive) square root and squaring are inverse operations. They undo each other. Actually, as
we’ve just discussed, they aren’t exact inverses, since squaring will take a negative to a positive number,
and the (positive) square root takes us back to a positive number. As a result, squaring both sides of an
equation will sometimes introduce an incorrect solution (because of the wrong sign).
A basic rule of thumb is to always check your solutions, and throw out the bad ones. This way, if you forget
which techniques can introduce bad solutions, you’ll still be OK.
Consider the equation
√
x + 3 = 5.
(2)
We want to find an x so that the (positive) square root is 5. We want to get x by itself on one side, so we
need to get rid of the square root. We can do this by squaring both sides
√
2
x + 3 = 52 ,
(3)
which gives us
x + 3 = 25.
(4)
x = 22.
(5)
Subtracting 3 from both sides gives us
We should always check our solutions, especially in this case.
p
√
(22) + 3 = 25 = 5. OK
(6)
The bad solutions come from the fact that two unequal numbers, like 2 and −2, can be squared resulting in
equal numbers, 22 = 4 = (−2)2 .
For example, consider the equation
√
x + 2 = −3.
(7)
This equation is asking for a positive square root that equals −3. This can’t happen, of course, and there
should be no solutions. But, if we go ahead and square both sides, we get
x + 2 = (−3)2 ,
(8)
x = −2 + 9 = 7.
(9)
and
When we check the solution, we get
p
√
(7) + 2 == 9 = 3 6= −3, Bad Solution
and we just throw this solution out bringing us back to concluding that there are no solutions.
(10)
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3 QUIZ 13A
Example Here’s one more example. Consider the equation
√
x − 7 + 4 = 5.
(11)
We want to get x by itself, so we’ll need to square both sides eventually. But first we should subtract 4 from
both sides, since the main thing happening on the left side at this point.
√
x − 7 = 1.
(12)
Now we square both sides to get
x − 7 = 1,
(13)
x = 8.
(14)
and
Checking this solution, we see that
p
3
1.
2.
4
(8) − 7 + 4 =
√
1 + 4 = 1 + 4 = 5. OK
(15)
Quiz 13A
Solve
√
Solve
√
x + 2 = 3.
3x + 2 − 3 = 2.
Other Powers
What we’ve done with squares and square roots, we can do with other powers (although completing the cube
is much harder). Let’s look at a few examples.
Example Consider the equation
√
5
x + 2 = 3.
(16)
To undo the fifth root, we raise both sides to the fifth power.
√
5
x+2
5
= 35 .
(17)
This gives us
x + 2 = 243,
(18)
x = 241.
(19)
and
Since numbers only have one fifth root, we don’t have to worry about extra solutions. It’s not a bad idea to
check anyway.
Example Consider the equation
( x − 4 )6 = 2.
(20)
Here we should take the sixth root of both sides. There are two sixth roots, one positive and one negative,
so
q
√
6
6
6
( x − 4 ) = ± 2,
(21)
and
√
6
x − 4 = ± 2,
and
x=4±
√
6
2 ≈ 5.122462048, 2.877537952.
(22)
(23)
5 QUIZ 13B
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Most calculators don’t have a sixth-root button, but you can compute 21/6 using the exponent key. Be sure
to use parentheses around the (1 ÷ 6).
Don’t forget that since even powers of negatives are positive, we have to treat even powers and even roots the
same as we do with squares and square roots (which are even). Odd powers and odd roots are no problem.
We can take odd roots
√ of negatives, and every number has exactly one particular odd root (e.g., the only
cube root of −8 is 3 −8 = −2.
5
1.
2.
3.
4.
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Quiz 13B
Solve (2x − 7)3 = 12. Give solution in approximate decimal form.
√
Solve 5 x + 1 = 2.
√
Solve 4 x − 3 = 3.
Solve (x + 1)5 = 17.
Homework 13
Find all (correct) solutions to the following equations. If your solutions are not whole numbers, round
correctly to four decimal places. Enter your answers like x=3 or x=2.0031,-4.1136 etc. with no spaces. If
there are no real solutions, enter no solutions
√
1.
x + 2 = 1.
√
2.
2x − 3 = 2.
√
3.
5x + 7 = −2.
√
4.
3x + 2 − 2 = 3.
5.
6.
7.
8.
9.
10.
(2x − 5)3 = 27.
√
5
x + 3 = 2.
√
3
x + 2 = 3.
(x + 2)5 = 7.
√
8
3x − 5 = −5.
(2x − 3)4 = −11.