Solve the quadratic equation 5x^2+75x=0

Question

by factoring out the GCF.
Responses:

The solutions are x=0 and x=5

The solutions are x=−15and x=0

The solutions are x=−15 and x=−5

The solutions are x=0 and x=15

Answer 1

To solve the quadratic equation \( 5x^2 + 75x = 0 \) by factoring out the greatest common factor (GCF), we can follow these steps:

1. Identify the GCF of the terms \( 5x^2 \) and \( 75x \). The GCF is \( 5x \).

2. Factor out the GCF from the equation:

\[ 5x(x + 15) = 0 \]

3. Set each factor equal to zero:

\[ 5x = 0 \quad \text{or} \quad x + 15 = 0 \]

4. Solve for \( x \):

For \( 5x = 0 \): \[ x = 0 \]

For \( x + 15 = 0 \): \[ x = -15 \]

Thus, the solutions to the equation \( 5x^2 + 75x = 0 \) are:

\[ x = 0 \quad \text{and} \quad x = -15 \]

The correct response is: The solutions are \( x = -15 \) and \( x = 0 \).

Answer 2

As Isla prepares to set off fireworks, she wants to ensure the spectators are a safe distance away. When shot in the air, one firework will follow the path h(x)=−x^2+15x , where h(x) is the height of the firework, and x is the horizontal distance it travels in feet. To determine how far the firework will travel before it reaches the ground, Isla sets the height equal to zero, and solves the following quadratic equation: 0=−x^2+15x . How far will the firework travel before it reaches the ground?
Responses

30 feet

7.5 feet

15 feet

−15 feet

Answer 3

To determine how far the firework will travel before it reaches the ground, we need to solve the quadratic equation:

\[ 0 = -x^2 + 15x \]

1. Rearranging the equation gives:

\[ -x^2 + 15x = 0 \]

2. We can factor out the common term \( x \):

\[ x(15 - x) = 0 \]

3. Setting each factor equal to zero gives:

\[ x = 0 \quad \text{or} \quad 15 - x = 0 \]

4. Solving \( 15 - x = 0 \) gives:

\[ x = 15 \]

Thus, the firework will travel 15 feet before it reaches the ground.

The correct response is: 15 feet.

Answer 4

Hyung-eun is solving the quadratic equation x^2−4x−12=0 by grouping. Her work is shown below. At which step does Hyung-eun first make an error?

x^2−4x−12=0



Step1: x^2+(−6x)+2x−12=0



Step 2: (x^2+(−6x))+(2x−12)=0



Step 3: x(x+(−6))+2(x−6)=0



x(x−6)+2(x−6)=0



Step 4: (x+6)(x−2)=0



Step 5: x+6=0 or x−2=0



x=−6 or x=2



Responses:

Step 3

Step 2

Step 4

Step 1

Answer 5

Let's analyze Hyung-eun's steps to identify where the error occurs in her solution of the quadratic equation \( x^2 - 4x - 12 = 0 \) by grouping.

1. The original equation is: \[ x^2 - 4x - 12 = 0 \]

2. Step 1: She rewrites the equation as: \[ x^2 + (-6x) + 2x - 12 = 0 \]

   • This seems wrong since \(-6 + 2 = -4\). This is correct for her to express \(-4x\) as \(-6x + 2x\).

3. Step 2: Hyung-eun groups the terms: \[ (x^2 + (-6x)) + (2x - 12) = 0 \]

   • This is correct grouping from Step 1; the equation remains valid.

4. Step 3: In this step, she factors the groups: \[ x(x + (-6)) + 2(x - 6) = 0 \]

   • Here she should have factored \(2(x - 6)\) since the constant term is -12 but gets \(x - 6\), this simplifies incorrectly since the factor should have been \(2(x + 6)\). This is where the error happens.

5. Step 4: She now writes: \[ (x + 6)(x - 2) = 0 \]

   • This is based on the incorrect grouping in Step 3.

6. Step 5: She finds solutions: \[ x + 6 = 0 \quad \text{or} \quad x - 2 = 0 \]

   • This leads to \(x = -6\) or \(x = 2\), which also stem from the previous mistakes.

Therefore, the first error occurs in Step 3, where she incorrectly factors out from the grouped terms.

The correct response is: Step 3.

Answer 6

Which of the following quadratic equations can be solved by grouping?
Responses

x^2+10x+21=0

x^2+8x−22=0

x^2−4x−8=0

x^2−12x+18=0

Answer 7

To determine which of the given quadratic equations can be solved by grouping, we need to find if the equation can be factored into two binomials in a way that allows simple grouping.

For the quadratic equation \( ax^2 + bx + c = 0 \), we look for two numbers that multiply to \( ac \) (the product of the coefficient of \( x^2 \) and the constant term) and add up to \( b \) (the coefficient of \( x \)). This process allows for the grouping method.

Let's analyze each option:

1. \( x^2 + 10x + 21 = 0 \):

   • Here, \( a = 1 \), \( b = 10 \), and \( c = 21 \).
   • \( ac = 1 \cdot 21 = 21 \).
   • We need two numbers that multiply to \( 21 \) and add to \( 10 \): \( 3 \) and \( 7 \).
   • This can be factored as \( (x + 3)(x + 7) = 0 \).
   • Can be solved by grouping.

2. \( x^2 + 8x - 22 = 0 \):

   • Here, \( a = 1 \), \( b = 8 \), and \( c = -22 \).
   • \( ac = 1 \cdot (-22) = -22 \).
   • We need two numbers that multiply to \( -22 \) and add to \( 8 \): \( 11 \) and \( -2 \) (since \( 11 + (-2) = 9\)).
   • This does NOT work for grouping.

3. \( x^2 - 4x - 8 = 0 \):

   • Here, \( a = 1 \), \( b = -4 \), and \( c = -8 \).
   • \( ac = 1 \cdot (-8) = -8 \).
   • We need two numbers that multiply to \( -8 \) and add to \( -4 \): \( 2 \) and \( -4 \) (it can also become complex).
   • This does NOT seem to effectively group nicely for integers.

4. \( x^2 - 12x + 18 = 0 \):

   • Here, \( a = 1 \), \( b = -12 \), and \( c = 18 \).
   • \( ac = 1 \cdot 18 = 18 \).
   • We need two numbers that multiply to \( 18 \) and add to \( -12 \): \( -6 \) and \( -3 \) (as \( -6 + -3 = -9\)).
   • This does NOT work for grouping.

Conclusion: The only quadratic equation that can be solved by grouping is:

\( x^2 + 10x + 21 = 0 \).