Question

Esma and Hunter were trying to solve the equation:

Answer 1

Answer:

both

Step-by-step explanation:

When would these strategies work?

Esma wants to solve by completing the square. If our equation looks like

a

x

2

+

b

x

+

c

=

k

ax

2

+bx+c=ka, x, squared, plus, b, x, plus, c, equals, k, this strategy would work. If

a

=

1

a=1a, equals, 1, we can square half of

b

bb to find what number completes the square. If

a

>

1

a>1a, is greater than, 1, we need to factor before we complete the square.

Hunter wants to solve using the zero product property. If we have a factored expression that equals zero, this strategy would work.

Whose strategy would work to solve

x

2

+

8

x

=

2

x

−

8

x

2

+8x=2x−8x, squared, plus, 8, x, equals, 2, x, minus, 8?

Hint #22 / 4

Esma's strategy

Esma is correct that adding

1

11 to both sides completes the square. She can factor

x

2

+

6

x

+

9

x

2

+6x+9x, squared, plus, 6, x, plus, 9 and rewrite the equation as

(

x

+

3

)

2

=

1

(x+3)

2

=1left parenthesis, x, plus, 3, right parenthesis, squared, equals, 1. Then she can solve using square roots.

So Esma's strategy would work.

[Show me this strategy worked out.]

Hint #33 / 4

Hunter's strategy

Hunter is correct that he can factor

x

2

+

6

x

+

8

x

2

+6x+8x, squared, plus, 6, x, plus, 8 as

(

x

+

2

)

(

x

+

4

)

(x+2)(x+4)left parenthesis, x, plus, 2, right parenthesis, left parenthesis, x, plus, 4, right parenthesis.

He can then solve

(

x

+

2

)

(

x

+

4

)

=

0

(x+2)(x+4)=0left parenthesis, x, plus, 2, right parenthesis, left parenthesis, x, plus, 4, right parenthesis, equals, 0 using the zero product property since he has a factored expression equal to zero.

So Hunter's strategy would work.

[Show me this strategy worked out.]

Hint #44 / 4

Answer

Both of Esma's and Hunter's strategies would work.