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

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To solve the quadratic equation \(5x^2 + 75x = 0\) by factoring out the greatest common factor (GCF), we first identify the GCF of the terms in the equation.

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

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

Now, we can set each factor equal to zero:

  1. \(5x = 0\)

    • Dividing both sides by 5, we have \(x = 0\).

  2. \(x + 15 = 0\)

    • Subtracting 15 from both sides, we have \(x = -15\).

Thus, the solutions to the equation are \(x = 0\) and \(x = -15\).

So the correct response is: The solutions are x = -15 and x = 0.

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=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 Put the quadratic in standard form x minus 8 equals 0 Factor out the GCF No GCF Set each factor equal to zero x minus 8 equals 0 The solutions are 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=−7 and x=8

To solve the quadratic equation \(7x^2 = 56x\) by factoring out the GCF, we first need to rewrite the equation in standard form. The correct steps are as follows:

  1. Put the quadratic in standard form: \[7x^2 - 56x = 0\]

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

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

  4. Solve for x:

    • From \(7x = 0\), we find \(x = 0\).
    • From \(x - 8 = 0\), we find \(x = 8\).

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

Looking through the provided responses, the correct choice is:

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.

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 −15 feet negative 15 feet 15 feet 15 feet 30 feet 30 feet 7.5 feet 7.5 feet

To determine how far the firework will travel before it reaches the ground, we need to solve the quadratic equation given by the height function \( h(x) = -x^2 + 15x \).

Setting the height equal to zero gives us the equation:

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

We can rearrange this to:

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

Next, we factor out the GCF, which is \( x \):

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

Setting each factor equal to zero gives us:

  1. \( x = 0 \)
  2. \( x - 15 = 0 \) ⟹ \( x = 15 \)

The solutions to the equation are \( x = 0 \) and \( x = 15 \). Since \( x = 0 \) corresponds to the initial launch point of the firework, we are interested in the other solution.

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 3 Step 3 Step 1 Step 1 Step 2 Step 2 Step 4

To analyze Hyung-eun's work on the quadratic equation \(x^2 - 4x - 12 = 0\) and identify at which step she made an error, let's review each step:

Hyung-eun's Steps

  1. Step 1: \(x^2 + (-6x) + 2x - 12 = 0\)

    • The quadratic is rewritten with the terms \( -4x \) split into \( -6x + 2x \). (This is valid.)

  2. Step 2: \((x^2 + (-6x)) + (2x - 12) = 0\)

    • The terms are grouped correctly. (This is valid.)

  3. Step 3: \(x(x + (-6)) + 2(x - 6) = 0\)

    • That part here is incorrect.
    • Correctly factoring \(2x - 12\) should give \(2(x - 6)\). The first part should have been \(x(x - 6)\), and not \(x + (-6)\).

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

    • Hyung-eun incorrectly factored the previous step.

  5. Step 5: \(x + 6 = 0\) or \(x - 2 = 0\)

    • From the incorrect factors, leading to incorrect solutions of \(x = -6\) or \(x = 2\).

Conclusion

The first error occurs in Step 3, where she incorrectly factors \(x(x + (-6))\) instead of the proper form \(x(x - 6)\).

Thus, the correct response is: Step 3.

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 x2−12x+18=0 x squared minus 12 x plus 18 equals 0 x2−4x−8=0 x squared minus 4 x minus 8 equals 0 x2+8x−22=0

To determine which of the given quadratic equations can be solved by grouping, we need to find an equation where we can factor it into the form \((x + p)(x + q) = 0\) or similar, where the product of \(p\) and \(q\) equals the constant term and their sum equals the coefficient of the \(x\) term.

Let's analyze each equation:

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

    • Factors as \((x + 3)(x + 7) = 0\). This can be solved by factoring but not typically by grouping as there are no two pairs being grouped.

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

    • Factors as \((x - 6)^2 - 18 = 0\); can be solved by completing the square or quadratic formula, but not grouping.

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

    • To apply grouping, rewrite as \(x^2 - 4x + 4 - 8 = 0\).
    • This can be transformed into \( (x - 2)^2 - 12 = 0\), but does not lend itself to simple grouping.

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

    • Rearranging or looking for factors, it does not lend itself to simple integer factoring that would fit grouping.

To identify an equation that can be solved through grouping, we generally look for a factorable trinomial where the middle term can be split into two parts.

None of the quadratics listed have coefficients that clearly allow them to be grouped for solutions.

However, commonly quadratic equations of the form \(ax^2 + bx + c = 0\) can be solved if \(b\) can break into two rational parts.

If we focus purely on the methodology of grouping:

We can't derive definitive solutions from what's above simply based on grouping. However, most mathematically typical form is yielding:

  • \(x^2 - 4x - 8 = 0\) is rearrangeable, though it’s not typical for simple grouping compared to others, it can be manipulated to some forms leading to solutions.

So if asked simply to choose which can be attempted with some creative manipulation:

The best option to choose would be(though ideally not simple solving by standard grouping): \(x^2 - 4x - 8 = 0\).

If the intention is to have pure grouping without any special transformations, it’s fair to assume no are entirely suitable for classic grouping approach.

In conclusion, while quad. like \(x^2 + 10x + 21\) is simple volunteer grouping by interests of straight factoring, go with typical exploratory of the \(x^2 + 10x + 21\) if forced choice.

Re-clarifying, please adjust if category strictly fits definition options for grouping as need --- but assess through classic group splitting capacities, none are strongly suited for grouping thus in final determination:=

\(x^2 - 4x - 8 = 0\) as best potential if aiming for even flexibilities at group explorative or transformations.