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MA 1128: Lecture 18 – 4/14/11
Completing the Square
Equations and Square Roots
We’ll be looking at a technique that will allow us to solve any quadratic equation,
and this technique is useful in a number of other situations.
Before we do that, I would like to talk a little about imaginary and complex
numbers.
Recall that the square root of x is a number that you square to get x.
We use the symbols
x
 x
for the positive square root of x, and
for the negative square root of x.
Next Slide
(Cont.)
In an equation, we can take the square root of both sides,
and we generally will end up with two equations,
one for the positive square root and one for the negative square root.
For example,
2
x 4
x  4 and x   4
x  2 and x  2
We’ll usually write the two equations as one using .
x2  4
x 4
x  2
Next Slide
Imaginary Numbers
Any time we square a real number,
we either get 0 (from 02)
or a positive number (like (-2)2 = 4 or 72 = 49).
Therefore, the equation x2 = 4 cannot have a solution that is a real number,
since we can’t square anything that gives us a negative number.
Mathematicians long ago decided that it might be interesting to invent a number
whose square is negative.
They called it i.
In particular i2 = 1,
From this we can say things like
1  i
and
 4  (1)( 4)   1 4  i 4  2i
Next Slide
Complex Numbers
Any multiple of i is called an imaginary number (including i).
We can also have combinations like 7 + 3i and 1 – i,
which are called complex numbers.
The number i and all of the other imaginary numbers are also considered to be
complex.
This may seem a little silly, but complex numbers are surprisingly useful in realworld applications.
One example is in electronics.
In all but the simplest of electronic circuits, the currents and voltages don’t follow
any logical rules unless we use complex values for the currents.
There are many other examples of this, including Einstein’s theory of special
relativity.
I won’t say anything more about this, but it is not an exaggeration to say that most
substantive real-world applications involve complex numbers somehow.
Next Slide
Practice Problems
Find both solutions (real or complex) to each equation.
1.
x2 = 4.
2.
x2 = 1.
3.
x2 = 1.
4.
x2 = 9.
Answers:
1) x = 2,2 = 2
2) x = 1,1 = 1
3) x = i,i =  i
4) x = 3i,3i = 3i
Next Slide
Square Roots and Quadratic Equations
We can solve many quadratic equations by factoring.
For example,
x2  9  0
( x  3)( x  3)  0
x  3, 3
We can also solve this equation using square roots.
x2  9  0
x2  9
x 9
x  3
Next Slide
(Cont.)
Even if the solutions don’t come out as whole numbers,
it’s still pretty easy to use square roots.
x2  5  0
x2  5
x   5 (the exact solutions)
x  2.236 (approxima te solutions)
x2  4  0
x 2  4
x    4 (no real solutions)
x  2i (but two complex solutions)
Next Slide
Practice Problems
Find both solutions (real or complex).
1.
x2 – 16 = 0.
2.
x2 + 16 = 0.
3.
x2 + 1 = 0.
Answers:
1)
x = 4;
2)
x = 4i;
3)
x =  i.
Next Slide
More General Quadratic Equations
Consider the quadratic equation in the following special form.
(x – 2)2 = 9 [[Take the square root of both sides]]
x–2=3
x – 2 = 3 and x – 2 = 3
x = 5 and x = 1
x = 5, 1.
Next Slide
Practice Problems
Find both solutions (real or complex) by taking the square root of both sides.
1.
(x + 2)2 = 4.
2.
(x  2)2 = 9.
3.
(x + 1)2 = 4.
Answers:
1)
x + 2 = 2  x = 2  2  x = 0,4
2)
x  2 = 3  x = 2  3  x = 5,1
3)
x + 1 = 2i  x = 1  2i  x = 1 + 2i, 1  2i
Next Slide
More Quadratic Equations
Even if the solutions don’t come out as whole numbers, we can get the answers
the same way.
( x  3) 2  5
x3  5
x  3  5
x  3  5 and x  3  5 are exact solutions.
The calculator will give us approximate solutions
 3  5  3  2.236  0.764 and  5.236
Next Slide
Completing the Square
Note that if we can get to something like (x + 3)2 = 5, then we can solve the
equation, no matter how bad the numbers are.
Let’s multiply this out to see what the corresponding standard quadratic equation
looks like.
(x + 3)2 = 5
x2 + 6x + 9 = 5 (of course, this is a perfect square trinomial on the left)
x2 + 6x + 4 = 0.
How would we work this backwards?
The main thing is that we want a perfect square trinomial.
That is, since half of the x-coefficient is 3,
we need the constant term to be 32 = 9.
The 4 is not what we want, so move it to the other side, and then get a 9 in.
We’ll go through this on the next slide.
Next Slide
(Cont.)
We’re starting with
x2 + 6x + 4 = 0.
The 4 doesn’t go with a perfect square trinomial, so move it to the other side.
x2 + 6x = 4.
We want a 9 as the constant term, so add it to both sides.
x2 + 6x + 9 = 4 + 9
x2 + 6x + 9 = 5
And then the left side factors (as a perfect square).
(x + 3)2 = 5.
We know what to do from here.
Getting a perfect square trinomial into the equation is called completing the
square. We can solve any quadratic equation with this technique.
Next Slide
Example
Complete the square.
x2 – 10x + 3 = 0. [[Half of -10 is -5, and (-5)2 = 25, we need 25 in the
equation.]]
x2 – 10x = 3
x2 – 10x + 25 = 3 + 25
(x – 5)2 = 22
This completes the square, and solving the equation from here is easy.
Next Slide
Practice Problems
1.
2.
3.
4.
What do we need to add to complete the square in x2 + 6x = 2?
What do we need to add to complete the square in x2 – 4x = 3?
Rewrite x2 + 8x – 3 = 0 in the form (x + a)2 = d.
Rewrite x2 – 16x + 2 = 0 in the form (x – a)2 = d.
Answers:
1)
9;
2)
4;
3)
(x + 4)2 = 19;
4)
(x – 8)2 = 62.
Next Slide
Examples
Use completing the square to find the solutions for the following quadratic
equations.
x2  4x  2  0
x2  6x  7  0
x 2  4 x  2 [[ half of 4 is 2, 2 2  4]]
x 2  6 x  7 [[ half of 6 is 3, 32  9]]
x2  4x  4  2  4
x2  6x  9  7  9
( x  2) 2  6
( x  3) 2  16
x2 6
x  3   16
x  3  4
x  3 4
x  3  4 and 3  4
x  2  6 (exact)
x  0.4495,4.4495 (approxima te)
x  7 and  1
Next Slide
Example
Even if the answers are really messy, the process is not much harder.
x 2  3x  1  0
2
-3
9
 -3
x  3x  1 [[ Half of - 3 is , and    ]]
2
4
 2
9
9
x 2  3x   1 
4
4
2
2
3
5

x




2
4

x
3
5

2
4
3
5 3
5 3 5
x 
 

 2.618, 0.382
2
4 2 2
2
Next Slide
Practice Problems
1.
2.
3.
Find the exact solutions to x2 + 12x – 3 = 0.
Find the approximate solutions rounded to four decimal places.
Find the exact solutions to x2 + 6x + 10 = 0.
Answers:
1) x = -6  sqrt(39)
2) x = 0.2450,-12.2450
3) x = -3  i
End