Solve the quadratic equation 5x2+75x=0

Question

by factoring out the GCF.(1 point)
Responses

The solutions are x=0
and x=5
.
The solutions are x equals 0 and x equals 5 .

The solutions are x=−15
and x=0
.
The solutions are x equals negative 15 and x equals 0 .

The solutions are x=0
and x=15
.
The solutions are x equals 0 and x equals 15 .

The solutions are x=−15
and x=−5
.

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 \(5x\):

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

3. Set each factor equal to zero:

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

4. Solve for \(x\):

   • From \(5x = 0\):

     \[ x = 0 \]

   • From \(x + 15 = 0\):

     \[ x = -15 \]

Thus, the solutions to the equation are:

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

Therefore, the correct response is:

The solutions are \(x = -15\) and \(x = 0\).

Answer 2

Which of the following tables shows the correct steps to factor out the GCF and solve the quadratic equation 7x2=56x
?(1 point)
Responses

Put the quadratic in standard form 7x2+56x=0
Factor out the GCF 7x(x+8)=0
Set each factor equal to zero 7x=0
and x+8=0
The solutions are x=−8
and x=0
Put the quadratic in standard form 7 x squared plus 56 x equals 0 Factor out the GCF 7 x left parenthesis x plus 8 right parenthesis equals 0 Set each factor equal to zero 7 x equals 0 and x plus 8 equals 0 The solutions are x equals negative 8 and x equals 0

Put the quadratic in standard form 7x2−56x=0
Factor out the GCF 7x(x−8)=0
Set each factor equal to zero 7x=0
and x−8=0
The solutions are x=−7
and x=8
Put the quadratic in standard form 7 x squared minus 56 x equals 0 Factor out the GCF 7 x left parenthesis x minus 8 right parenthesis equals 0 Set each factor equal to zero 7 x equals 0 and x minus 8 equals 0 The solutions are x equals negative 7 and x equals 8

Put the quadratic in standard form 7x2−56x=0
Factor out the GCF 7x(x−8)=0
Set each factor equal to zero 7x=0
and x−8=0
The solutions are x=0
and x=8
Put the quadratic in standard form 7 x squared minus 56 x equals 0 Factor out the GCF 7 x left parenthesis x minus 8 right parenthesis equals 0 Set each factor equal to zero 7 x equals 0 and x minus 8 equals 0 The solutions are x equals 0 and x equals 8

Put the quadratic in standard form x−8=0
Factor out the GCF No GCF
Set each factor equal to zero x−8=0
The solutions are x=8

Answer 3

To factor out the GCF and solve the quadratic equation \( 7x^2 = 56x \), we will follow the appropriate steps:

1. Put the quadratic in standard form: We need to bring all terms to one side of the equation: \[ 7x^2 - 56x = 0 \]

2. Factor out the GCF: The GCF of \( 7x^2 \) and \( 56x \) is \( 7x \). Thus: \[ 7x(x - 8) = 0 \]

3. Set each factor equal to zero: \[ 7x = 0 \quad \text{and} \quad x - 8 = 0 \]

4. Solve for \( x \):

   • From \( 7x = 0 \): \[ x = 0 \]
   • From \( x - 8 = 0 \): \[ x = 8 \]

Therefore, the solutions are \( x = 0 \) and \( x = 8 \).

The correct response from your options would be:

Put the quadratic in standard form: \( 7x^2 - 56x = 0 \)
Factor out the GCF: \( 7x(x - 8) = 0 \)
Set each factor equal to zero: \( 7x = 0 \) and \( x - 8 = 0 \)
The solutions are: \( x = 0 \) and \( x = 8 \)

So you would select the option that corresponds to these steps.

Answer 4

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 pathh(x)=−x2+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=−x2+15x
. How far will the firework travel before it reaches the ground?(1 point)
Responses

15 feet
15 feet

−15
feet
negative 15 feet

30 feet
30 feet

7.5 feet

Answer 5

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

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

First, we can rewrite the equation to set it in a more standard form:

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

Next, we can factor out the common term:

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

This gives us two factors:

1. \( x = 0 \)
2. \( -x + 15 = 0 \) which simplifies to \( x = 15 \)

The two solutions represent the points at which the firework is at ground level. The first solution \( x = 0 \) indicates the starting point (when the firework is launched), and the second solution \( x = 15 \) indicates the point where it returns to the ground.

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

So the correct answer is:

15 feet.

Answer 6

Hyung-eun is solving the quadratic equation x2−4x−12=0
by grouping. Her work is shown below. At which step does Hyung-eun first make an error?

x2−4x−12=0

Step1: x2+(−6x)+2x−12=0

Step 2: (x2+(−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

(1 point)
Responses

Step 2
Step 2

Step 1
Step 1

Step 4
Step 4

Step 3

Answer 7

Let's analyze Hyung-eun's work step by step to identify where she made an error.

Given Equation:

\[ x^2 - 4x - 12 = 0 \]

Step 1:

She rewrites the equation as: \[ x^2 + (-6x) + 2x - 12 = 0 \] Check: The equation should have the same terms as the original.

• \(-6x + 2x = -4x\) which is correct, so this step is correct.

Step 2:

Then she groups as: \[ (x^2 + (-6x)) + (2x - 12) = 0 \] Check: This step also looks okay.

Step 3:

She factors: \[ x(x + (-6)) + 2(x - 6) = 0 \] This should be: \[ x(x - 6) + 2(x - 6) = 0 \] Error Check: She factored \(2x - 12\) incorrectly. The correct factorization of \(2x - 12\) is \(2(x - 6)\), but she stated it incorrectly as \(2(x + (-6))\).

Thus, the error occurs in Step 3.

Conclusion:

Hyung-eun's first error occurs in Step 3 where she incorrectly factored \(2x - 12\).

The correct response is: Step 3.

Answer 8

Which of the following quadratic equations can be solved by grouping?(1 point)
Responses

x2+10x+21=0
x squared plus 10 x plus 21 equals 0

x2−4x−8=0
x squared minus 4 x minus 8 equals 0

x2+8x−22=0
x squared plus 8 x minus 22 equals 0

x2−12x+18=0

Answer 9

To determine which quadratic equations can be solved by grouping, we look for equations that can be expressed in a form where we can factor the middle term into two terms that can be grouped effectively.

Let's examine each quadratic equation:

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

   • This can be factored directly as \((x + 3)(x + 7) = 0\).
   • Grouping is not necessary here, as it factors cleanly.

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

   • The middle coefficient is negative, and there are no two numbers that add to \(-4\) and multiply to \(-8\).
   • This does not lend itself well to grouping.

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

   • We look for two numbers that add to \(8\) and multiply to \(-22\). These numbers are \(10\) and \(-2\) because \(10 + (-2) = 8\) and \(10 \times (-2) = -20\).
   • This can be rewritten as \(x^2 + 10x - 2x - 22 = 0\) and then factored by grouping: \((x^2 + 10x) + (-2x - 22) = 0\).

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

   • We look for two numbers that add to \(-12\) and multiply to \(18\). These numbers are \(-6\) and \(-6\) (not applicable for grouping since it’s not a distinct pair).
   • It factors but can also be solved directly by the quadratic formula without needing to group.

Conclusion:

The equation that can be solved effectively by grouping is:

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

This is because we can express the middle term in a way that allows for grouping.