Lesson 10: Quadratic Equations Digging Deeper solutions

Algebra 1

© 2009 Duke University Talent Identification Program

Page 1 of 6

Quadratic-type Equations

The equations below are not necessarily quadratic equations, but they can

each be solved as if they were. For example, suppose we want to solve the

equation ( ) ( )21 3 1 2 0a a+ + + + = . We could solve this equation in two

different ways.

Method 1:

Use substitution to rewrite ( ) ( )21 3 1 2 0a a+ + + + = as a quadratic

equation.

Let 1x a= + . Then the equation ( ) ( )21 3 1 2 0a a+ + + + = can be rewritten

as ( ) ( )23 2 0x x+ + = . And we can solve this by factoring.

( ) ( )( ) ( )

23 2 0

2 1 0

2 0 or 1 0

2 or 1

x x

x x

x x

x x

+ + =

+ + =

+ = + =

= − = −

So, we know that x is -2 or -1. But we were trying to find the value of a.

So we substitute again and get:

2 or 1

1 2 or 1 1

3 or 2

x x

a a

a a

= − = −

+ = − + = −

= − = −

Checking each value gives:

( ) ( )

( ) ( )

( ) ( )

?2

?2

?2

?

?

1 3 1 2 0

3 1 3 3 1 2 0

2 3 2 2 0

4 6 2 0

0 0

a a+ + + + =

− + + − + + =

− + − + =

− + =

= �

( ) ( )

( ) ( )

( ) ( )

?2

?2

?2

?

?

1 3 1 2 0

2 1 3 2 1 2 0

1 3 1 2 0

1 3 2 0

0 0

a a+ + + + =

− + + − + + =

− + − + =

− + =

= �

Lesson 10: Quadratic Equations Digging Deeper solutions

Algebra 1

© 2009 Duke University Talent Identification Program

Page 2 of 6

Method 2:

Factor ( ) ( )21 3 1 2 0a a+ + + + = as if it were a quadratic equation.

The factors of ( )21a + are ( )1a + and ( )1a + . The factors of 2 are 2 and 1.

Write these as binomials and use FOIL to verify that they give

( ) ( )21 3 1 2 0a a+ + + + = .

( ) ( )( ) ( )

21 3 1 2

1 2 1 1

a a

a a

+ + + +

+ + + +

The product of the First terms is ( ) ( )1 1a a+ + , which is ( )21a + .

The product of the Outer terms is ( )1 1a⋅ + .

The product of the Inner terms is ( )2 1a⋅ + .

The product of the Last terms is 2 1 2⋅ = .

Combining these terms, we get

( ) ( ) ( )( ) ( )

2

2

1 1 1 2 1 2

1 3 1 2

a a a

a a

+ + + + + +

= + + + +

Since we have factored it correctly, we can use the Zero Product property

to solve the equation.

( ) ( )( ) ( )

( ) ( )

21 3 1 2 0

1 2 1 1 0

1 2 0 or 1 1 0

3 0 or 2 0

3 or 2

a a

a a

a a

a a

a a

+ + + + =

+ + + + =

+ + = + + =

+ = + =

= − = −

From Method 1, we know that these values check.

Lesson 10: Quadratic Equations Digging Deeper solutions

Algebra 1

© 2009 Duke University Talent Identification Program

Page 3 of 6

Solve the following quadratic-type equations.

1. 4 23 28x x− =

( ) ( )

4 2

4 2

2 2

2 2

2 2

3 28

3 28 0

7 4 0

7 0 or 4 0

7 or 4

7 or 4

7 or 2

x x

x x

x x

x x

x x

x x

x x i

− =

− − =

− + =

− = + =

= = −

= ± = ± −

= ± = ±

2. ( ) ( )22 24 2 4 24 0x x x x+ − + − =

( ) ( )( ) ( )( ) ( )

22 2

2 2

2 2

2 2

4 2 4 24 0

4 6 4 4 0

4 6 0 or 4 4 0

4 6 0 or 4 4 0

x x x x

x x x x

x x x x

x x x x

+ − + − =

+ − + + =

+ − = + + =

+ − = + + =

( ) ( )( )

2

2

4 6 0

4 4 4 1 6

2 1

4 16 24

2

4 40

2

4 2 10

2

2 10

x x

x

x

x

x

x

+ − =

− ± − −=

− ± +=

− ±=

− ±=

= − ±

or ( )

2

2

4 4 0

2 0

2 0 with multiplicity 2

2 with multiplicity 2

x x

x

x

x

+ + =

+ =

+ =

= −

Answer: 2 10 or 2x x= − ± = −

Lesson 10: Quadratic Equations Digging Deeper solutions

Algebra 1

© 2009 Duke University Talent Identification Program

Page 4 of 6

3. 4 24 8 0b b− + =

( ) ( ) ( ) ( )( )

4 2

2

2

2

2

2

2

4 8 0

4 4 4 1 8

2 1

4 16 32

2

4 16

2

4 4

2

2 2

b b

b

b

b

ib

b i

− + =

− − ± − −=

± −=

± −=

±=

= ±

Since 2 2 2b i= ± , we can use the Square Root property to find

the value of b.

2

2

2 2

2 2

2 2

b i

b i

b i

= ±

= ± ±

= ± ±

Answer: 2 2b i= ± ±

4. 4 5 0a a− − =

( ) ( )

( ) ( ) ( )2 2 22

4 5 0

5 1 0

5 0 or 1 0

5 or 1

5 or 1

25 or 1

a a

a a

a a

a a

a a

a a

− − =

− + =

− = + =

= = −

= = −

= =

Lesson 10: Quadratic Equations Digging Deeper solutions

Algebra 1

© 2009 Duke University Talent Identification Program

Page 5 of 6

Check: 25a = 1a =

?

?

?

?

?

4 5 0

25 4 25 5 0

25 4 5 5 0

25 20 5 0

0 0

a a− − =

− − =

− ⋅ − =

− − =

= �

?

?

?

?

?

1 4 1 5 0

1 4 1 5 0

1 4 5 0

3 5 0

8 0

− − =

− ⋅ − =

− − =

− − =

− = �

Answer: 25a =

Note: We could have eliminated 1a = as a solution earlier. When

we set each factor equal to zero, we got 1a = − . The principal

square root is never negative, so we know that 1a = − is an extraneous solution.

5. ( )289 3 3 0

3

x x− + =

Notice that ( ) ( )22 29 3 3 3x

x x x= = = .

Then ( )289 3 3 0

3

x x− + = is the same as ( ) ( )2 283 3 3 0

3

x x− + = .

This is an equation in quadratic form. Since it’s difficult to factor

an expression with rational terms, we’ll multiply both sides by the LCD to get integer coefficients.

Lesson 10: Quadratic Equations Digging Deeper solutions

Algebra 1

© 2009 Duke University Talent Identification Program

Page 6 of 6

( )

( ) ( )

( ) ( ) ( )

( ) ( )( )( ) ( )( )

2

2

2

289 3 3 0

3

283 3 3 0

3

283 3 3 3 0 3

3

3 3 28 3 9 0

3 3 1 3 9 0

x x

x x

x x

x x

x x

− + =

− + =

⋅ − + = ⋅

⋅ − + =

⋅ − − =

( )( )

3 3 1 0 or 3 9 0

3 3 1 or 3 9

13 or 3 9

3

1 or 2

x x

x x

x x

x x

⋅ − = − =

⋅ = =

= =

= − =

Check: 1x = − 2x =

( )

( )

?

?1 1

?

?

?

289 3 3 0

3

289 3 3 0

3

1 28 13 0

9 3 3

1 28 270

9 9 9

0 0

x x

− −

− + =

− + =

− ⋅ + =

− + =

= �

( )

( )

?

?2 2

?

?

?

289 3 3 0

3

289 3 3 0

3

2881 9 3 0

3

81 84 3 0

0 0

x x− + =

− + =

− ⋅ + =

− + =

= �

Answer: 1 or 2x x= − =