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.
