Question
Use factoring to find the solutions for 30=−x2+13x
.(1 point)
Responses
x=3,10
x is equal to 3 comma 10 - no response given
x=−10,−3
x is equal to negative 10 comma negative 3 - incorrect
x=−10,3
x is equal to negative 10 comma 3 - no response given
x=−3,10
.(1 point)
Responses
x=3,10
x is equal to 3 comma 10 - no response given
x=−10,−3
x is equal to negative 10 comma negative 3 - incorrect
x=−10,3
x is equal to negative 10 comma 3 - no response given
x=−3,10
Answers
GPT-4o mini
To solve the equation \(30 = -x^2 + 13x\) using factoring, we can start by rearranging the equation to set it to zero:
\[
-x^2 + 13x - 30 = 0
\]
Next, we can multiply through by -1 to make the leading coefficient positive:
\[
x^2 - 13x + 30 = 0
\]
Now, we need to factor the quadratic equation \(x^2 - 13x + 30\). We are looking for two numbers that multiply to 30 (the constant term) and add up to -13 (the coefficient of \(x\)). The numbers that satisfy this condition are -3 and -10, since:
\[
-3 \times -10 = 30 \quad \text{and} \quad -3 + -10 = -13
\]
Thus, we can factor the quadratic as:
\[
(x - 3)(x - 10) = 0
\]
Setting each factor equal to zero gives:
1. \(x - 3 = 0\) ⟹ \(x = 3\)
2. \(x - 10 = 0\) ⟹ \(x = 10\)
Therefore, the solutions to the equation \(30 = -x^2 + 13x\) are:
\[
x = 3 \quad \text{and} \quad x = 10
\]
So the correct response is:
**x = 3, 10**
\[
-x^2 + 13x - 30 = 0
\]
Next, we can multiply through by -1 to make the leading coefficient positive:
\[
x^2 - 13x + 30 = 0
\]
Now, we need to factor the quadratic equation \(x^2 - 13x + 30\). We are looking for two numbers that multiply to 30 (the constant term) and add up to -13 (the coefficient of \(x\)). The numbers that satisfy this condition are -3 and -10, since:
\[
-3 \times -10 = 30 \quad \text{and} \quad -3 + -10 = -13
\]
Thus, we can factor the quadratic as:
\[
(x - 3)(x - 10) = 0
\]
Setting each factor equal to zero gives:
1. \(x - 3 = 0\) ⟹ \(x = 3\)
2. \(x - 10 = 0\) ⟹ \(x = 10\)
Therefore, the solutions to the equation \(30 = -x^2 + 13x\) are:
\[
x = 3 \quad \text{and} \quad x = 10
\]
So the correct response is:
**x = 3, 10**