Question
Part A:
The equation 8x2+2x−3=0
8
𝑥
2
+
2
𝑥
−
3
=
0
can be factored into what product?
(1 point)
Responses
(4x−1)(2x+3)=0
(
4
𝑥
−
1
)
(
2
𝑥
+
3
)
=
0
open paren 4 x minus 1 close paren times open paren 2 x plus 3 close paren is equal to 0
(x+3)(x−1)=0
(
𝑥
+
3
)
(
𝑥
−
1
)
=
0
open paren x plus 3 close paren times open paren x minus 1 close paren is equal to 0
(4x+3)(2x−1)=0
(
4
𝑥
+
3
)
(
2
𝑥
−
1
)
=
0
open paren 4 x plus 3 close paren times open paren 2 x minus 1 close paren is equal to 0
(2x−3)(4x+1)=0
(
2
𝑥
−
3
)
(
4
𝑥
+
1
)
=
0
open paren 2 x minus 3 close paren times open paren 4 x plus 1 close paren is equal to 0
Question 2
Part B:
The two solutions to the equation 8x2+2x−3=0
8
𝑥
2
+
2
𝑥
−
3
=
0
are...
(1 point)
Responses
x=34 and x=−12
𝑥
=
3
4
𝑎
𝑛
𝑑
𝑥
=
−
1
2
x=34 and x=−12
𝑥
=
3
4
𝑎
𝑛
𝑑
𝑥
=
−
1
2
x=−34 and x=12
𝑥
=
−
3
4
𝑎
𝑛
𝑑
𝑥
=
1
2
x=−34 and x=12
𝑥
=
−
3
4
𝑎
𝑛
𝑑
𝑥
=
1
2
x=34 and x=12
𝑥
=
3
4
𝑎
𝑛
𝑑
𝑥
=
1
2
x=34 and x=12
𝑥
=
3
4
𝑎
𝑛
𝑑
𝑥
=
1
2
x=−34 and x=−12
The equation 8x2+2x−3=0
8
𝑥
2
+
2
𝑥
−
3
=
0
can be factored into what product?
(1 point)
Responses
(4x−1)(2x+3)=0
(
4
𝑥
−
1
)
(
2
𝑥
+
3
)
=
0
open paren 4 x minus 1 close paren times open paren 2 x plus 3 close paren is equal to 0
(x+3)(x−1)=0
(
𝑥
+
3
)
(
𝑥
−
1
)
=
0
open paren x plus 3 close paren times open paren x minus 1 close paren is equal to 0
(4x+3)(2x−1)=0
(
4
𝑥
+
3
)
(
2
𝑥
−
1
)
=
0
open paren 4 x plus 3 close paren times open paren 2 x minus 1 close paren is equal to 0
(2x−3)(4x+1)=0
(
2
𝑥
−
3
)
(
4
𝑥
+
1
)
=
0
open paren 2 x minus 3 close paren times open paren 4 x plus 1 close paren is equal to 0
Question 2
Part B:
The two solutions to the equation 8x2+2x−3=0
8
𝑥
2
+
2
𝑥
−
3
=
0
are...
(1 point)
Responses
x=34 and x=−12
𝑥
=
3
4
𝑎
𝑛
𝑑
𝑥
=
−
1
2
x=34 and x=−12
𝑥
=
3
4
𝑎
𝑛
𝑑
𝑥
=
−
1
2
x=−34 and x=12
𝑥
=
−
3
4
𝑎
𝑛
𝑑
𝑥
=
1
2
x=−34 and x=12
𝑥
=
−
3
4
𝑎
𝑛
𝑑
𝑥
=
1
2
x=34 and x=12
𝑥
=
3
4
𝑎
𝑛
𝑑
𝑥
=
1
2
x=34 and x=12
𝑥
=
3
4
𝑎
𝑛
𝑑
𝑥
=
1
2
x=−34 and x=−12
Answers
GPT-4o mini
To solve the equation \( 8x^2 + 2x - 3 = 0 \), we can factor it directly or use the quadratic formula.
**Part A: Factoring the equation**
First, we will attempt to factor the equation.
The equation can be written as:
\[ 8x^2 + 2x - 3 = 0 \]
We need to find two numbers that multiply to \( 8 \times -3 = -24 \) and add to \( 2 \). These numbers are \( 6 \) and \( -4 \).
We then rewrite the middle term:
\[ 8x^2 + 6x - 4x - 3 = 0 \]
Next, we group the terms:
\[ (8x^2 + 6x) + (-4x - 3) = 0 \]
Factoring each group:
\[ 2x(4x + 3) - 1(4x + 3) = 0 \]
This gives us:
\[ (4x + 3)(2x - 1) = 0 \]
Thus, the correct option for Part A is:
**(4x + 3)(2x - 1) = 0**
---
**Part B: Finding the solutions**
Now we can find the solutions for \( (4x + 3)(2x - 1) = 0 \).
Setting each factor to zero:
1. \( 4x + 3 = 0 \)
\[ 4x = -3 \]
\[ x = -\frac{3}{4} \]
2. \( 2x - 1 = 0 \)
\[ 2x = 1 \]
\[ x = \frac{1}{2} \]
Thus, the two solutions to the equation \( 8x^2 + 2x - 3 = 0 \) are:
**\( x = -\frac{3}{4} \) and \( x = \frac{1}{2} \)**
So the correct option for Part B is:
**x=−34 and x=12** (and written as \( x = -\frac{3}{4} \) and \( x = \frac{1}{2} \)).
**Part A: Factoring the equation**
First, we will attempt to factor the equation.
The equation can be written as:
\[ 8x^2 + 2x - 3 = 0 \]
We need to find two numbers that multiply to \( 8 \times -3 = -24 \) and add to \( 2 \). These numbers are \( 6 \) and \( -4 \).
We then rewrite the middle term:
\[ 8x^2 + 6x - 4x - 3 = 0 \]
Next, we group the terms:
\[ (8x^2 + 6x) + (-4x - 3) = 0 \]
Factoring each group:
\[ 2x(4x + 3) - 1(4x + 3) = 0 \]
This gives us:
\[ (4x + 3)(2x - 1) = 0 \]
Thus, the correct option for Part A is:
**(4x + 3)(2x - 1) = 0**
---
**Part B: Finding the solutions**
Now we can find the solutions for \( (4x + 3)(2x - 1) = 0 \).
Setting each factor to zero:
1. \( 4x + 3 = 0 \)
\[ 4x = -3 \]
\[ x = -\frac{3}{4} \]
2. \( 2x - 1 = 0 \)
\[ 2x = 1 \]
\[ x = \frac{1}{2} \]
Thus, the two solutions to the equation \( 8x^2 + 2x - 3 = 0 \) are:
**\( x = -\frac{3}{4} \) and \( x = \frac{1}{2} \)**
So the correct option for Part B is:
**x=−34 and x=12** (and written as \( x = -\frac{3}{4} \) and \( x = \frac{1}{2} \)).