MA 1128: Lecture 18 – 6/19/13

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

Next Slide

x for the positive square root of x, and

x for the negative square root of x.

(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,

Next Slide

2 and 2

4 and 4

42

xx

xx

x

We’ll usually write the two equations as one using .

2

4

42

x

x

x

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

Next Slide

ii

i

2441)4)(1(4

and

1

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 real-world 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 = 22) x = 1,1 = 13) x = i,i = i4) x = 3i,3i = 3i

Next Slide

Square Roots and Quadratic Equations

We can solve many quadratic equations by factoring.

For example,

Next Slide

3 ,3

0)3)(3(

092

x

xx

x

We can also solve this equation using square roots.

3

9

9

092

2

x

x

x

x

(Cont.)

Even if the solutions don’t come out as whole numbers,

it’s still pretty easy to use square roots.

Next Slide

solutions) te(approxima 236.2

solutions)exact (the 5

5

052

2

x

x

x

x

solutions)complex (but two 2

solutions) real (no 4

4

042

2

ix

x

x

x

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 = 3x – 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 = 0,42) x 2 = 3 x = 2 3 x = 5,13) 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.

Next Slide

solutions.exact are 53 and 53

53

53

5)3( 2

xx

x

x

x

The calculator will give us approximate solutions

236.5 and 764.0236.2353

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 = 3x2 – 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. What do we need to add to the complete the square in x2 + 6x = 2?

2. What do we need to add to the complete the square in x2 – 4x = 3?

3. Rewrite x2 + 8x – 3 = 0 in the form (x + a)2 = d.

4. 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.

Next Slide

te)(approxima 4495.4,4495.0

(exact) 62

62

6)2(

4244

]]42 2, is 4 of half [[ 24

024

2

2

22

2

x

x

x

x

xx

xx

xx

1 and 7

43 and 43

43

43

163

16)3(

9796

]]93 3, is 6 of half [[ 76

076

2

2

22

2

x

x

x

x

x

x

xx

xx

xx

Example

Even if the answers are really messy, the process is not much harder.

Next Slide382.0 ,618.2

2

53

2

5

2

3

4

5

2

3

4

5

2

3

4

5

2

3

4

91

4

93

]] 4

9

2

3- and ,

2

3- is 3- of Half [[ 13

013

2

2

22

2

x

x

x

xx

xx

xx

Practice Problems

1. Find the exact solutions to x2 + 12x – 3 = 0.

2. Find the approximate solutions rounded to four decimal places.

3. 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