Download 1 Solving Equations. 2 Taking the Square Root of Both Sides

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MA 1165 - Lecture 12
2/16/09
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Solving Equations.
Whenever you are solving an equation algebraically, you want to get x by itself on one side of the equation. In
general, you want to perform a sequence of moves, which usually is doing the same thing to both sides. Most
of the time, you are undoing something. For example, if an expression includes adding 5, then subtracting
5 from both sides of the equation, will remove the plus 5. Over the next week, or so, and throughout the
semester, I want to look at some of the things we can do.
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Taking the Square Root of Both Sides
If we have an equation like
x2 = 9,
(1)
we would like to get the x by itself on one side of the equation. We need, therefore, to undo the square, and
this means taking the square root of both sides, since these are inverse operations. Be careful to remember,
however, that since squaring negatives produces positives, we have
√
x = ± 9,
(2)
or
x = ±3.
(3)
Equations involving squares will typically have two solutions. There are two exceptions, which we saw when
we were graphing. One is
x2 = 0,
(4)
where there is only one solution
x = ±0 = 0.
(5)
x2 = −4.
(6)
The other is when you have something like
Squares are always positive, so this can’t happen with real numbers. We’ll say that this equation has no real
solutions or just no solutions. We can throw in extra numbers, the imaginary and complex numbers, and
then we have solutions
x = ±2i,
(7)
but these are not real numbers.
For the most part, we’ll see things like
and the exact solutions are
x2 = 5,
(8)
√
x = ± 5.
(9)
I may also ask you, for example, to round your answer to four decimal places, and you would give me
x ≈ ±2.2361
(10)
Whenever we round, we’re only going to get an approximate answer, but most of the time, I’ll just write
equal, since we all know what we mean.
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3 QUIZ 12A
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Quiz 12A
Find all solutions to the following equations, and round to four decimal places, if the answers are not whole
numbers.
1.
x2 = 4
2.
x2 = 6
3.
x2 = −1
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Solving General Quadratic Equations
When we were looking at quadratic functions, we used a technique called completing the square. We can use
this to solve a quadratic equation. For example, suppose we wanted to solve the equation
x2 + 6x − 2 = 0.
(11)
This doesn’t factor easily, because the −2 doesn’t go with the x2 + 6x very well, so let’s move it to the other
side of the equation.
x2 + 6x = 2.
(12)
To complete the square (with a coefficient of 1 on the x2 ), we look at the middle coefficient, divide by two,
and then square. In this case 6/2 = 3, and 32 = 9. We need to add 9 to both sides.
x2 + 6x + 9 = 2 + 9.
(13)
The left side is now a perfect square trinomial, so it’s easy to factor. The equation simplifies to
(x + 3)2 = 11.
This is now essentially the same situation as x2 = 11, and so
p
√
(x + 3)2 = ± 11.
This gives us
(14)
(15)
√
x + 3 = ± 11,
(16)
√
11.
(17)
and subtracting 3 from both sides give us
x = −3 ±
I’ll call this the exact solution. We can get an approximate solution from our calculators. We’ll get two
numbers, one goes with the + and one goes with the −.
x ≈ 6.31662479, −0.31662479.
(18)
Rounding to four decimal places, we would get
x = 6.3166, −0.3166
(19)
x2 − 10x + 3 = 0.
(20)
x2 − 10x = −3,
(21)
x2 − 10x + 25 = −3 + 25,
(22)
Example: Let’s do another one.
Subtract 3 from both sides
add 25 to both sides
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5 QUIZ 12B
simplify
take the square root of both sides
(x − 5)2 = 22,
(23)
√
x − 5 = ± 22,
(24)
and then add 5 to both sides
x =5±
√
22 ≈ 9.6904, 0.3096
(25)
rounded to four decimal places
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Quiz 12B
QA1.
Solve x2 − 2x − 4 = 0 by completing the square.
QA1a.
After you’ve completed the square, you have . . .
QA1b.
Your exact solution was . . .
QA2.
Solve the equation x2 + 8x + 2 = 0 by completing the square.
QA2a.
After you’ve completed the square, you have . . .
QA2b.
Your approximate solution, rounded to four decimal places, was . . .
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Homework 12
Solve the following equations. If your solutions are not whole numbers, round correctly to four decimal
places. Enter your answers like x=3,-3 or x=0.2314,-2.5531 with no spaces, if there are two solutions. List
just one solution, if there is only one, and if there are no real solutions, enter no solutions
1.
x2 = 4
2.
x2 = 3
3.
x2 = −16
4.
x2 = 0
5.
x2 + 4x + 4 = 0
6.
x2 − 4x + 5 = 0
7.
x2 − 4x + 3 = 0
8.
x2 + 6x + 5 = 0
9.
x2 − 2x − 7 = 0
10.
x2 − 10x − 5 = 0