Solve the quadratic equation 5x2+75x=0 by factoring out the GCF.(1 point) Responses The solutions are x=0 and x=15 . The solutions are x equals 0 and x equals 15 . 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=−5 . The solutions are x equals negative 15 and x equals negative 5 . The solutions are x=−15 and x=0 .

To solve the quadratic equation \(5x^2 + 75x = 0\) by factoring out the GCF (Greatest Common Factor), we can start by identifying the GCF of the terms in the equation.

The GCF of \(5x^2\) and \(75x\) is \(5x\). We can factor this out:

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

Now, we set each factor equal to zero:

1. \(5x = 0\)

   • Dividing by 5 gives \(x = 0\).

2. \(x + 15 = 0\)

   • Subtracting 15 gives \(x = -15\).

Thus, the solutions to the equation are:

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

The correct response is:

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

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)=−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 7.5 feet 7.5 feet −15 feet negative 15 feet 15 feet 15 feet 30 feet

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

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

We can factor this equation. First, we can factor out a negative and then factor by grouping:

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

Now, we set each factor equal to zero:

1. \( -x = 0 \)

   • Thus, \( x = 0 \).

2. \( x - 15 = 0 \)

   • Thus, \( x = 15 \).

The solutions to the equation are \(x = 0\) and \(x = 15\).

The \(x = 0\) solution corresponds to the point where the firework is launched, and the \(x = 15\) solution corresponds to the point where the firework hits the ground.

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

The correct response is:

15 feet.

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 4 Step 4 Step 1 Step 1 Step 3 Step 3 Skip to navigation page 16 of 16

Let's analyze Hyung-eun's work step by step to identify where she first makes an error in solving the quadratic equation \(x^2 - 4x - 12 = 0\) by grouping.

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

   • Evaluation: This step is correct, but it doesn't accurately represent how to split \(-4x\) into two terms whose product equals \(-12\) and sum equals \(-4\). The terms should be \(-6x\) and \(2x\) as given, but they need to multiply to \(-12\) and add to \(-4\).

2. Step 2: She groups the terms: \((x^2 + (-6x)) + (2x - 12) = 0\).

   • Evaluation: This is a valid grouping, although as noted before, the choice of terms is inappropriate.

3. Step 3: She factors each group: \(x(x + (-6)) + 2(x - 6) = 0\) and writes \(x(x - 6) + 2(x - 6) = 0\).

   • Evaluation: This step is incorrect due to the sign with \(2(x - 6)\); she should have factored \(2\) out of \(2x - 12\), which concludes as \(2(x - 6)\) but represents a manipulation error from Step 1.

4. Step 4: She states \((x + 6)(x - 2) = 0\).

   • Evaluation: This factorization is incorrect based on her previous work.

5. Step 5: She sets each factor to zero.

   • Evaluation: As she derived the previous factorization incorrectly, this conclusion is also faulty.

Summary: The first step where she introduced an incorrect way to split the equation was in Step 1.

The final answer is:

Step 1

are you sure?

Let’s carefully go through the steps of Hyung-eun’s work once again to find the first error she made in solving the quadratic equation \(x^2 - 4x - 12 = 0\) by grouping.

Hyung-eun's Steps:

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

   • Evaluation: Here, she rewrites \(-4x\) as \(-6x + 2x\). This part of the work is valid as \(-6 + 2 = -4\).

2. Step 2: Group the terms:
   \[ (x^2 + (-6x)) + (2x - 12) = 0 \]

   • Evaluation: This grouping is correct.

3. Step 3: Factor each group:
   \[ x(x + (-6)) + 2(x - 6) = 0 \]
   and then:
   \[ x(x - 6) + 2(x - 6) = 0 \]

   • Evaluation: In this step, she has grouped the coefficients correctly, but during factoring \(2(x - 6)\), the correct factorization should have been \(2(x - 6)\) instead of keeping \(x + (-6)\).

4. Step 4: She combines the factored groups:
   \[ (x + 6)(x - 2) = 0 \]

   • Evaluation: At this stage, she made a mistake. The correct factored form based on what she had in Step 3 should be \( (x - 6)(x + 2) = 0 \) if we derived it correctly.

5. Step 5: Set each factor to zero:
   \[ x + 6 = 0 \quad \text{or} \quad x - 2 = 0 \]
   Which leads to:
   \[ x = -6 \quad \text{or} \quad x = 2 \]

   • Evaluation: These solutions are incorrect because her factorization in Step 4 was wrong.

Conclusion

The first error Hyung-eun made was in Step 4, where she incorrectly factored the expression \((x - 6)(x + 2)\) instead of arriving at the correct product.

So I apologize for the earlier mistake in identifying the step; the correct answer is indeed Step 4. Thank you for your patience!

Question Which of the following quadratic equations can be solved by grouping?(1 point) Responses x2−4x−8=0 x squared minus 4 x minus 8 equals 0 x2+10x+21=0 x squared plus 10 x plus 21 equals 0 x2+8x−22=0 x squared plus 8 x minus 22 equals 0 x2−12x+18=0

To determine which of the given quadratic equations can be solved by grouping, we need to check if we can rewrite them in a way that allows us to express them as a product of two binomials. Typically, we want to find two numbers that multiply to the constant term (the term without \(x\)) and add up to the coefficient of the linear term (the \(x\) term).

Let’s analyze each equation:

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

   • Here, we need two numbers that multiply to \(-8\) and add to \(-4\). The numbers \(-6\) and \(2\) fit this, so we can group and factor it:
     \[(x - 6)(x + 2) = 0\]
   • This can be solved by grouping.

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

   • We need two numbers that multiply to \(21\) and add to \(10\). The numbers \(3\) and \(7\) work:
     \[(x + 3)(x + 7) = 0\]
   • This can also be solved by grouping.

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

   • We need two numbers that multiply to \(-22\) and add to \(8\). The numbers \(11\) and \(-2\) fit:
     \[(x + 11)(x - 2) = 0\]
   • This can be solved by grouping.

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

   • We need two numbers that multiply to \(18\) and add to \(-12\). The numbers that fit do not exist as integers; hence this cannot be factored easily by grouping.

Conclusion

The equations that can be solved by grouping are:

• \(x^2 - 4x - 8 = 0\)
• \(x^2 + 10x + 21 = 0\)
• \(x^2 + 8x - 22 = 0\)

Based on these findings, all except \(x^2 - 12x + 18 = 0\) can be solved by grouping. If you need one specific answer to choose from the options, the best candidates are the first three.

ONLY ONE

Apologies for the oversight! Given that you need to identify only one quadratic equation that can be solved by grouping, let's review the options again:

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

   • Factors to: \((x - 6)(x + 2) = 0\)
   • Can be solved by grouping.

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

   • Factors to: \((x + 3)(x + 7) = 0\)
   • Can be solved by grouping.

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

   • Factors to: \((x + 11)(x - 2) = 0\)
   • Can be solved by grouping.

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

   • Does not factor nicely into integers that fit the criteria for grouping.

Among the above options, if you need to select only one equation, I suggest choosing:

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

This clearly demonstrates a straightforward factoring method using integers, and it is a well-known example that can be grouped effectively.