Question
4. Solve the factored quadratic equation (19x−3)(9+x)=0
. Enter the lesser number first.(2 points)
The solution set is {
,
}
Question 2
5. Which of the following quadratic equations is not solvable by grouping?(1 point)
Responses
2x2−2x−10=0
2 x squared minus 2 x minus 10 is equal to 0
x2−12x+35=0
x squared minus 12 x plus 35 is equal to 0
2x2+14x+12=0
2 x squared plus 14 x plus 12 is equal to 0
x2−2x+1=0
x squared minus 2 x plus 1 is equal to 0
Question 3
6. Factor the expression 49x4−4y6
by typing into the blank spaces. (4 points)
(7
+
) (
-2
)
. Enter the lesser number first.(2 points)
The solution set is {
,
}
Question 2
5. Which of the following quadratic equations is not solvable by grouping?(1 point)
Responses
2x2−2x−10=0
2 x squared minus 2 x minus 10 is equal to 0
x2−12x+35=0
x squared minus 12 x plus 35 is equal to 0
2x2+14x+12=0
2 x squared plus 14 x plus 12 is equal to 0
x2−2x+1=0
x squared minus 2 x plus 1 is equal to 0
Question 3
6. Factor the expression 49x4−4y6
by typing into the blank spaces. (4 points)
(7
+
) (
-2
)
Answers
GPT-4o mini
**Question 1: Solve the factored quadratic equation (19x−3)(9+x)=0**
To solve the equation, we can set each factor equal to zero:
1. \(19x - 3 = 0\)
- \(19x = 3\)
- \(x = \frac{3}{19}\)
2. \(9 + x = 0\)
- \(x = -9\)
Thus, the solution set is:
\[
\{ -9, \frac{3}{19} \}
\]
**Question 2: Which of the following quadratic equations is not solvable by grouping?**
We can analyze how to factor each equation:
1. \(2x^2 - 2x - 10 = 0\)
- Can be factored as \(2(x^2 - x - 5) = 0\).
2. \(x^2 - 12x + 35 = 0\)
- Can be factored as \((x - 5)(x - 7) = 0\).
3. \(2x^2 + 14x + 12 = 0\)
- Can be factored as \(2(x^2 + 7x + 6) = 0\) and further as \(2(x + 6)(x + 1) = 0\).
4. \(x^2 - 2x + 1 = 0\)
- Can be factored as \((x - 1)(x - 1) = 0\) or \((x - 1)^2 = 0\).
The equation that is **not solvable by grouping** is:
\[
2x^2 - 2x - 10 = 0
\]
as it requires a different factoring approach.
**Question 3: Factor the expression \(49x^4 - 4y^6\)**
This expression is a difference of squares, which can be factored using the formula \(a^2 - b^2 = (a - b)(a + b)\).
1. \(a^2 = (7x^2)^2\) and \(b^2 = (2y^3)^2\)
Thus, we can rewrite it as:
\[
49x^4 - 4y^6 = (7x^2 - 2y^3)(7x^2 + 2y^3)
\]
So the factored expression is:
\[
(7x^2 - 2y^3)(7x^2 + 2y^3)
\]
To fill in the blank spaces:
1. First blank: \(x^2\)
2. Second blank: \(y^3\)
Final factorization:
\[
(7 \, x^2 - 2y^3)(7 \, x^2 + 2y^3)
\]
To solve the equation, we can set each factor equal to zero:
1. \(19x - 3 = 0\)
- \(19x = 3\)
- \(x = \frac{3}{19}\)
2. \(9 + x = 0\)
- \(x = -9\)
Thus, the solution set is:
\[
\{ -9, \frac{3}{19} \}
\]
**Question 2: Which of the following quadratic equations is not solvable by grouping?**
We can analyze how to factor each equation:
1. \(2x^2 - 2x - 10 = 0\)
- Can be factored as \(2(x^2 - x - 5) = 0\).
2. \(x^2 - 12x + 35 = 0\)
- Can be factored as \((x - 5)(x - 7) = 0\).
3. \(2x^2 + 14x + 12 = 0\)
- Can be factored as \(2(x^2 + 7x + 6) = 0\) and further as \(2(x + 6)(x + 1) = 0\).
4. \(x^2 - 2x + 1 = 0\)
- Can be factored as \((x - 1)(x - 1) = 0\) or \((x - 1)^2 = 0\).
The equation that is **not solvable by grouping** is:
\[
2x^2 - 2x - 10 = 0
\]
as it requires a different factoring approach.
**Question 3: Factor the expression \(49x^4 - 4y^6\)**
This expression is a difference of squares, which can be factored using the formula \(a^2 - b^2 = (a - b)(a + b)\).
1. \(a^2 = (7x^2)^2\) and \(b^2 = (2y^3)^2\)
Thus, we can rewrite it as:
\[
49x^4 - 4y^6 = (7x^2 - 2y^3)(7x^2 + 2y^3)
\]
So the factored expression is:
\[
(7x^2 - 2y^3)(7x^2 + 2y^3)
\]
To fill in the blank spaces:
1. First blank: \(x^2\)
2. Second blank: \(y^3\)
Final factorization:
\[
(7 \, x^2 - 2y^3)(7 \, x^2 + 2y^3)
\]