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Solve the following quadratic equation using square roots. Round to the nearest hundredth if necessary: (x+27)^2 /−6=−3

.(1 point)
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11 answers

To solve the quadratic equation \(\frac{(x+27)^2}{-6} = -3\), we will first eliminate the fraction by multiplying both sides by -6:

\[ (x+27)^2 = -3 \times -6 \]

This simplifies to:

\[ (x+27)^2 = 18 \]

Next, we take the square root of both sides. Remember to consider both the positive and negative square roots:

\[ x + 27 = \sqrt{18} \quad \text{or} \quad x + 27 = -\sqrt{18} \]

Simplifying \(\sqrt{18}\):

\[ \sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2} \]

So we have:

\[ x + 27 = 3\sqrt{2} \quad \text{or} \quad x + 27 = -3\sqrt{2} \]

Now we will isolate \(x\):

  1. For \(x + 27 = 3\sqrt{2}\):

\[ x = 3\sqrt{2} - 27 \]

  1. For \(x + 27 = -3\sqrt{2}\):

\[ x = -3\sqrt{2} - 27 \]

Now calculating the numerical values for both solutions. First, we find \(3\sqrt{2}\):

\[ 3\sqrt{2} \approx 3 \times 1.414 \approx 4.242 \]

Now substituting this in:

  1. For \(x = 3\sqrt{2} - 27\):

\[ x \approx 4.242 - 27 \approx -22.758 \]

  1. For \(x = -3\sqrt{2} - 27\):

Calculating \(-3\sqrt{2}\):

\[ -3\sqrt{2} \approx -4.242 \]

Then,

\[ x \approx -4.242 - 27 \approx -31.242 \]

Finally, rounding to the nearest hundredth gives us:

  • The first solution: \(x \approx -22.76\)
  • The second solution: \(x \approx -31.24\)

Thus, the solutions to the quadratic equation are:

\[ \boxed{-22.76} \quad \text{and} \quad \boxed{-31.24} \]

Which of the following equations has the solution set x={−95,3}
?(1 point)
Responses

(x+3)(9x−5)=0
left parenthesis x plus 3 right parenthesis left parenthesis 9 x minus 5 right parenthesis equals 0

(x−3)(5x+9)=0
left parenthesis x minus 3 right parenthesis left parenthesis 5 x plus 9 right parenthesis equals 0

(x−3)(9x+5)=0
left parenthesis x minus 3 right parenthesis left parenthesis 9 x plus 5 right parenthesis equals 0

−3x(5x+9)=0
negative 3 x left parenthesis 5 x plus 9 right parenthesis equals 0

To determine which of the given equations has the solution set \( x = {-95, 3} \), we need to check each equation by finding their roots.

1. Equation: \((x + 3)(9x - 5) = 0\)

To find the roots, set each factor equal to zero:

  • \(x + 3 = 0 \Rightarrow x = -3\)
  • \(9x - 5 = 0 \Rightarrow 9x = 5 \Rightarrow x = \frac{5}{9}\)

Solution Set: \({-3, \frac{5}{9}}\)

2. Equation: \((x - 3)(5x + 9) = 0\)

Set each factor equal to zero:

  • \(x - 3 = 0 \Rightarrow x = 3\)
  • \(5x + 9 = 0 \Rightarrow 5x = -9 \Rightarrow x = -\frac{9}{5}\)

Solution Set: \({3, -\frac{9}{5}}\)

3. Equation: \((x - 3)(9x + 5) = 0\)

Set each factor equal to zero:

  • \(x - 3 = 0 \Rightarrow x = 3\)
  • \(9x + 5 = 0 \Rightarrow 9x = -5 \Rightarrow x = -\frac{5}{9}\)

Solution Set: \({3, -\frac{5}{9}}\)

4. Equation: \(-3x(5x + 9) = 0\)

Set each factor equal to zero:

  • \(-3x = 0 \Rightarrow x = 0\)
  • \(5x + 9 = 0 \Rightarrow 5x = -9 \Rightarrow x = -\frac{9}{5}\)

Solution Set: \({0, -\frac{9}{5}}\)

Conclusion

None of the equations yield the specific solution set \( x = {-95, 3} \). Thus, based on the options provided, none of these equations produce that solution set. If there has been a misunderstanding regarding the options or if additional options exist, please clarify.

Joana solved the equation 4x2−4x−3=0
. She followed the steps to solve a quadratic equation by grouping. However, in Step 3 she noticed that the two groupings did not share a common binomial, as shown in her work below. She concluded that the quadratic equation 4x2−4x−3=0
has no solutions. Is Joana’s conclusion correct? Review Joana’s work to answer the question.

Step 1: 4x2+2x+(−6x)−3=0

Step 2: (4x2+2x)+(−6x−3)=0

Step 3: x(4x+2)+(−3(2x+1))=0

(1 point)
Responses

Yes, because the two groupings do not share a common binomial.
Yes, because the two groupings do not share a common binomial.

No, because she did not factor out the GCF from the first grouping.
No, because she did not factor out the GCF from the first grouping.

No, because she can still continue the steps even if the two groupings have no shared binomial.
No, because she can still continue the steps even if the two groupings have no shared binomial.

Yes, because there are no factors of −12
that add up to −4
.

To evaluate Joana's conclusion about the equation \(4x^2 - 4x - 3 = 0\), let's analyze her steps and the nature of quadratic equations.

Joana's Work Review

  1. Step 1: She rewrote the equation as \(4x^2 + 2x - 6x - 3 = 0\).
  2. Step 2: She grouped the terms as \((4x^2 + 2x) + (-6x - 3) = 0\).
  3. Step 3: She factored within each grouping to get \(x(4x + 2) - 3(2x + 1) = 0\).

At this point, Joana identified that the grouped terms do not share a common binomial factor. However, despite this, her conclusion that the quadratic equation has no solutions is not necessarily correct.

Quadratic Equation Analysis

A quadratic equation can always be solved using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In this case:

  • \(a = 4\)
  • \(b = -4\)
  • \(c = -3\)

Calculating the discriminant (\(b^2 - 4ac\)): \[ (-4)^2 - 4(4)(-3) = 16 + 48 = 64 \]

Since the discriminant is positive, this means there are two real solutions. Therefore, the equation does indeed have solutions.

Conclusion

Given Joana's reasoning and work, the correct response is:

No, because she can still continue the steps even if the two groupings have no shared binomial.

This is the correct answer, as the lack of a common binomial does not imply that the quadratic has no solutions.

Each of the following quadratic equations can be solved by factoring. Which equation has only one distinct solution?(1 point)
Responses

x^2−5x+4=0
x squared minus 5 x plus 4 equals 0

x2−4=0
x squared minus 4 equals 0

x2+5x+4=0
x squared plus 5 x plus 4 equals 0

x2+4x+4=0

To determine which of the given quadratic equations has only one distinct solution, we can look for equations where the discriminant (\(D\)) is equal to zero. The discriminant is calculated using the formula:

\[ D = b^2 - 4ac \]

A quadratic equation has one distinct solution when \(D = 0\).

Let’s evaluate each equation:

  1. Equation: \(x^2 - 5x + 4 = 0\)

    • Here, \(a = 1\), \(b = -5\), \(c = 4\)
    • Discriminant: \(D = (-5)^2 - 4(1)(4) = 25 - 16 = 9\) (two distinct solutions)

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

    • This is a difference of squares:
    • Factor: \((x - 2)(x + 2) = 0\)
    • Solutions: \(x = 2\) and \(x = -2\) (two distinct solutions)

  3. Equation: \(x^2 + 5x + 4 = 0\)

    • Here, \(a = 1\), \(b = 5\), \(c = 4\)
    • Discriminant: \(D = (5)^2 - 4(1)(4) = 25 - 16 = 9\) (two distinct solutions)

  4. Equation: \(x^2 + 4x + 4 = 0\)

    • Here, \(a = 1\), \(b = 4\), \(c = 4\)
    • Discriminant: \(D = (4)^2 - 4(1)(4) = 16 - 16 = 0\)
    • Since \(D = 0\), this equation has exactly one distinct solution.

Thus, the equation that has only one distinct solution is:

\[ \boxed{x^2 + 4x + 4 = 0} \]

Solve the following quadratic equation using the perfect square trinomial pattern: 14x2+2x+4=0
.(1 point)
x=

To solve the quadratic equation \(14x^2 + 2x + 4 = 0\) using the perfect square trinomial pattern, we first want to rewrite it in a form that makes factoring easier.

Step 1: Simplify the equation

We can factor out the greatest common factor, which is 2, from the equation:

\[ 2(7x^2 + x + 2) = 0 \]

Now we can focus on solving:

\[ 7x^2 + x + 2 = 0 \]

Step 2: Rearrange into a perfect square trinomial

To create a perfect square trinomial, we want to express \(7x^2 + x + 2\) in the form \((ax + b)^2\). However, since the leading coefficient isn't 1, we first divide the entire equation by 7 for easier manipulation:

\[ x^2 + \frac{1}{7}x + \frac{2}{7} = 0 \]

Step 3: Completing the square

To complete the square, we'll take the coefficient of \(x\) (which is \(\frac{1}{7}\)), divide it by 2 and square it:

\[ \left(\frac{1}{2 \cdot 7}\right)^2 = \left(\frac{1}{14}\right)^2 = \frac{1}{196} \]

Now, add and subtract this value inside the equation:

\[ x^2 + \frac{1}{7}x + \frac{1}{196} - \frac{1}{196} + \frac{2}{7} = 0 \]

Rearranging:

\[ \left( x + \frac{1}{14} \right)^2 - \frac{1}{196} + \frac{2}{7} = 0 \]

Convert \(\frac{2}{7}\) into a fraction with a denominator of 196:

\[ \frac{2}{7} = \frac{56}{196} \]

So we have:

\[ \left( x + \frac{1}{14} \right)^2 - \frac{1}{196} + \frac{56}{196} = 0 \]

This simplifies to:

\[ \left( x + \frac{1}{14} \right)^2 + \frac{55}{196} = 0 \]

Step 4: Isolate the perfect square

Now,

\[ \left( x + \frac{1}{14} \right)^2 = -\frac{55}{196} \]

Since we have a negative on the right, we realize this indicates no real solutions, but instead, we will calculate for complex solutions:

Step 5: Taking the square root

Taking the square root of both sides:

\[ x + \frac{1}{14} = \pm \sqrt{-\frac{55}{196}} \]

This becomes:

\[ x + \frac{1}{14} = \pm \frac{\sqrt{55}}{14}i \]

Step 6: Solve for \(x\)

Now, we isolate \(x\):

\[ x = -\frac{1}{14} \pm \frac{\sqrt{55}}{14}i \]

Thus, the solutions are:

\[ x = -\frac{1}{14} + \frac{\sqrt{55}}{14}i \quad \text{and} \quad x = -\frac{1}{14} - \frac{\sqrt{55}}{14}i \]

Conclusion

The solutions can be presented as:

\[ x = -\frac{1}{14} \pm \frac{\sqrt{55}}{14}i \]

Where \(x\) has no real solutions since the quadratic expression yields complex roots. If you need the solution in a form to plug into a testing system, just note that it returns as two complex numbers.

Gwen was asked to solve 3x^2−20=7
by factoring using the difference of squares pattern. What is/are the correct solution(s)?(1 point)
Responses

x=9
, x=−9
x equals 9 , x equals negative 9

x=3
x equals 3

x=−3
, x=3
x equals negative 3 , x equals 3

x=−3

To solve the equation \(3x^2 - 20 = 7\) by factoring using the difference of squares pattern, we first need to rearrange the equation. Here’s how to do it step by step:

Step 1: Rearranging the equation

Starting with the equation:

\[ 3x^2 - 20 = 7 \]

We can move \(7\) to the left side:

\[ 3x^2 - 20 - 7 = 0 \]

This simplifies to:

\[ 3x^2 - 27 = 0 \]

Step 2: Factor out the GCF

Next, we can factor out the greatest common factor (GCF), which is \(3\):

\[ 3(x^2 - 9) = 0 \]

Step 3: Recognizing the difference of squares

Now, we can recognize that \(x^2 - 9\) is a difference of squares:

\[ x^2 - 9 = (x - 3)(x + 3) \]

So, we can write:

\[ 3(x - 3)(x + 3) = 0 \]

Step 4: Solve for \(x\)

Setting each factor equal to zero gives us:

  1. \(x - 3 = 0\) \(\Rightarrow\) \(x = 3\)
  2. \(x + 3 = 0\) \(\Rightarrow\) \(x = -3\)

Conclusion

The solutions are:

\[ x = 3 \quad \text{and} \quad x = -3 \]

Thus, the correct response is:

x = -3, x = 3 (or written as \(x = 3\) and \(x = -3\)).