Asked by adaleta
Solve the following quadratic equation using square roots. Round to the nearest hundredth if necessary: (x+27)^2/−6=−3
why is -23 the wrong answer?
Correct answers:
−22.76, −31.24 And why is this the correct answer
why is -23 the wrong answer?
Correct answers:
−22.76, −31.24 And why is this the correct answer
Answers
Answered by
GPT-4o mini
To solve the quadratic equation \((x + 27)^2 / -6 = -3\), let's start by isolating the squared term:
1. Multiply both sides by -6 to eliminate the fraction:
\[
(x + 27)^2 = -3 \times -6
\]
\[
(x + 27)^2 = 18
\]
2. Now take the square root of both sides:
\[
x + 27 = \pm \sqrt{18}
\]
Simplifying \(\sqrt{18}\):
\[
\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}
\]
(Approximately \(3 \times 1.414 \approx 4.242\))
3. Now, we have:
\[
x + 27 = 3\sqrt{2} \quad \text{or} \quad x + 27 = -3\sqrt{2}
\]
4. Solve for \(x\) by subtracting 27 from both sides in both cases:
\[
x = -27 + 3\sqrt{2} \quad \text{or} \quad x = -27 - 3\sqrt{2}
\]
5. Calculate \(x\):
For \(x = -27 + 3\sqrt{2}\):
\[
x \approx -27 + 4.242 \approx -22.76
\]
For \(x = -27 - 3\sqrt{2}\):
\[
x \approx -27 - 4.242 \approx -31.24
\]
So the correct answers are approximately \(-22.76\) and \(-31.24\).
### Why is \(-23\) the wrong answer?
- \(-23\) is incorrect because it does not satisfy the original equation. When substituting \(-23\) back in:
\[
(x + 27)^2 / -6 = -3 \\
(-23 + 27)^2 / -6 = -3 \\
(4)^2 / -6 \neq -3 \quad \text{(This simplifies to } \frac{16}{-6} = -\frac{8}{3}, \text{ which is not } -3)
\]
Thus, \(-23\) is not a valid solution of the equation.
In conclusion, the correct solutions, rounded to two decimal places, are indeed approximately \(-22.76\) and \(-31.24\).
1. Multiply both sides by -6 to eliminate the fraction:
\[
(x + 27)^2 = -3 \times -6
\]
\[
(x + 27)^2 = 18
\]
2. Now take the square root of both sides:
\[
x + 27 = \pm \sqrt{18}
\]
Simplifying \(\sqrt{18}\):
\[
\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}
\]
(Approximately \(3 \times 1.414 \approx 4.242\))
3. Now, we have:
\[
x + 27 = 3\sqrt{2} \quad \text{or} \quad x + 27 = -3\sqrt{2}
\]
4. Solve for \(x\) by subtracting 27 from both sides in both cases:
\[
x = -27 + 3\sqrt{2} \quad \text{or} \quad x = -27 - 3\sqrt{2}
\]
5. Calculate \(x\):
For \(x = -27 + 3\sqrt{2}\):
\[
x \approx -27 + 4.242 \approx -22.76
\]
For \(x = -27 - 3\sqrt{2}\):
\[
x \approx -27 - 4.242 \approx -31.24
\]
So the correct answers are approximately \(-22.76\) and \(-31.24\).
### Why is \(-23\) the wrong answer?
- \(-23\) is incorrect because it does not satisfy the original equation. When substituting \(-23\) back in:
\[
(x + 27)^2 / -6 = -3 \\
(-23 + 27)^2 / -6 = -3 \\
(4)^2 / -6 \neq -3 \quad \text{(This simplifies to } \frac{16}{-6} = -\frac{8}{3}, \text{ which is not } -3)
\]
Thus, \(-23\) is not a valid solution of the equation.
In conclusion, the correct solutions, rounded to two decimal places, are indeed approximately \(-22.76\) and \(-31.24\).
Answered by
adaleta
Joana solved the equation 4x^2−4x−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 4x^2−4x−3=0
has no solutions. Is Joana’s conclusion correct? Review Joana’s work to answer the question.
Step 1: 4x^2+2x+(−6x)−3=0
Step 2: (4x^2+2x)+(−6x−3)=0
Step 3: x(4x+2)+(−3(2x+1))=0
where is the shared binomial.
. 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 4x^2−4x−3=0
has no solutions. Is Joana’s conclusion correct? Review Joana’s work to answer the question.
Step 1: 4x^2+2x+(−6x)−3=0
Step 2: (4x^2+2x)+(−6x−3)=0
Step 3: x(4x+2)+(−3(2x+1))=0
where is the shared binomial.
Answered by
GPT-4o mini
To determine whether Joana's conclusion is correct, let's review her work step-by-step and identify where the issues may have arisen.
### Joana's Steps:
**Step 1:**
Joana rewrote the equation as:
\[
4x^2 + 2x - 6x - 3 = 0
\]
**Step 2:**
She grouped the terms as follows:
\[
(4x^2 + 2x) + (-6x - 3) = 0
\]
**Step 3:**
She factored out from the groups:
\[
x(4x + 2) - 3(2x + 1) = 0
\]
### Analysis:
In **Step 3**, Joana attempted to factor out \(x\) from the first group and \(-3\) from the second group, resulting in the following expression:
\[
x(4x + 2) - 3(2x + 1) = 0
\]
The issue arises that \(4x + 2\) and \(2x + 1\) are not the same binomial, which leads to Joana's incorrect conclusion that there is no common binomial shared between the two groups.
### Identifying the Common Binomial:
If we analyze the terms she worked with, we can see that:
1. From the first grouping \(4x^2 + 2x\), the binomial factor is \(2(2x + 1)\).
2. From the second grouping \(-6x - 3\), the factors with the same terms can also break down to \(-3(2x + 1)\).
Thus during her factoring, both expressions can actually be expressed in terms of \(2x + 1\):
- \(4x^2 + 2x\) can be factored as \(2(2x + 1)\).
- \(-6x - 3\) can be factored as \(-3(2x + 1)\).
### Step 4 (Correct Factorization):
So a correct grouping should look like this:
\[
2(2x + 1) - 3(2x + 1) = 0
\]
Factoring out the common binomial \(2x + 1\):
\[
(2x + 1)(2 - 3) = 0 \rightarrow (2x + 1)(-1) = 0
\]
### Solving for \(x\):
Setting the factor equal to zero gives:
\[
2x + 1 = 0 \\
2x = -1 \\
x = -\frac{1}{2}
\]
### Conclusion:
Joana's original conclusion that the quadratic equation \(4x^2 - 4x - 3 = 0\) has no solutions is incorrect. The quadratic equation actually has a solution, \(x = -\frac{1}{2}\).
So, the **correct conclusion is**: the quadratic equation has solutions, and specifically \(x = -\frac{1}{2}\) is one of the solutions.
### Joana's Steps:
**Step 1:**
Joana rewrote the equation as:
\[
4x^2 + 2x - 6x - 3 = 0
\]
**Step 2:**
She grouped the terms as follows:
\[
(4x^2 + 2x) + (-6x - 3) = 0
\]
**Step 3:**
She factored out from the groups:
\[
x(4x + 2) - 3(2x + 1) = 0
\]
### Analysis:
In **Step 3**, Joana attempted to factor out \(x\) from the first group and \(-3\) from the second group, resulting in the following expression:
\[
x(4x + 2) - 3(2x + 1) = 0
\]
The issue arises that \(4x + 2\) and \(2x + 1\) are not the same binomial, which leads to Joana's incorrect conclusion that there is no common binomial shared between the two groups.
### Identifying the Common Binomial:
If we analyze the terms she worked with, we can see that:
1. From the first grouping \(4x^2 + 2x\), the binomial factor is \(2(2x + 1)\).
2. From the second grouping \(-6x - 3\), the factors with the same terms can also break down to \(-3(2x + 1)\).
Thus during her factoring, both expressions can actually be expressed in terms of \(2x + 1\):
- \(4x^2 + 2x\) can be factored as \(2(2x + 1)\).
- \(-6x - 3\) can be factored as \(-3(2x + 1)\).
### Step 4 (Correct Factorization):
So a correct grouping should look like this:
\[
2(2x + 1) - 3(2x + 1) = 0
\]
Factoring out the common binomial \(2x + 1\):
\[
(2x + 1)(2 - 3) = 0 \rightarrow (2x + 1)(-1) = 0
\]
### Solving for \(x\):
Setting the factor equal to zero gives:
\[
2x + 1 = 0 \\
2x = -1 \\
x = -\frac{1}{2}
\]
### Conclusion:
Joana's original conclusion that the quadratic equation \(4x^2 - 4x - 3 = 0\) has no solutions is incorrect. The quadratic equation actually has a solution, \(x = -\frac{1}{2}\).
So, the **correct conclusion is**: the quadratic equation has solutions, and specifically \(x = -\frac{1}{2}\) is one of the solutions.
Answered by
adaleta
but where is the shared binomial???
Answered by
GPT-4o mini
Apologies for any confusion in my previous messages. Let's clarify the identification of the common (or shared) binomial in Joana's work.
### Joana’s Original Work
1. **Step 1:**
\[
4x^2 + 2x - 6x - 3 = 0
\]
2. **Step 2:**
\[
(4x^2 + 2x) + (-6x - 3) = 0
\]
3. **Step 3:**
\[
x(4x + 2) - 3(2x + 1) = 0
\]
### Identifying the Shared Binomial
In Step 3, Joana factored the first grouping as \(x(4x + 2)\) and the second grouping as \(-3(2x + 1)\). However, it looks like she made an error in recognizing the common binomial factor.
Observing the expressions:
- From the first factor, we have \(4x + 2\), which can be factored as \(2(2x + 1)\).
- From the second factor, we have \(-3(2x + 1)\).
### Revealing the Shared Binomial
Now, it is clear that the binomials are:
- The factor from the first grouping, when correctly factored, is \(2(2x + 1)\).
- The factor from the second grouping is \(-3(2x + 1)\).
The shared binomial is indeed **\((2x + 1)\)**.
### Corrected Step 4 with the Shared Binomial
Now that we know there is a shared binomial, the equation can be rewritten as:
\[
(2x + 1)(2 - 3) = 0
\]
Since \(2 - 3 = -1\):
\[
(2x + 1)(-1) = 0
\]
This simplifies to finding:
\[
2x + 1 = 0
\]
### Solve for \(x\):
\[
2x = -1 \\
x = -\frac{1}{2}
\]
### Conclusion
Thus, to answer your question directly, the shared binomial in Joana's work is **\(2x + 1\)**. This means that the quadratic equation actually does have a solution, specifically \(x = -\frac{1}{2}\). Joana’s error was in not recognizing that there was a common binomial to factor out. Thank you for your patience, and I hope this clears up the confusion!
### Joana’s Original Work
1. **Step 1:**
\[
4x^2 + 2x - 6x - 3 = 0
\]
2. **Step 2:**
\[
(4x^2 + 2x) + (-6x - 3) = 0
\]
3. **Step 3:**
\[
x(4x + 2) - 3(2x + 1) = 0
\]
### Identifying the Shared Binomial
In Step 3, Joana factored the first grouping as \(x(4x + 2)\) and the second grouping as \(-3(2x + 1)\). However, it looks like she made an error in recognizing the common binomial factor.
Observing the expressions:
- From the first factor, we have \(4x + 2\), which can be factored as \(2(2x + 1)\).
- From the second factor, we have \(-3(2x + 1)\).
### Revealing the Shared Binomial
Now, it is clear that the binomials are:
- The factor from the first grouping, when correctly factored, is \(2(2x + 1)\).
- The factor from the second grouping is \(-3(2x + 1)\).
The shared binomial is indeed **\((2x + 1)\)**.
### Corrected Step 4 with the Shared Binomial
Now that we know there is a shared binomial, the equation can be rewritten as:
\[
(2x + 1)(2 - 3) = 0
\]
Since \(2 - 3 = -1\):
\[
(2x + 1)(-1) = 0
\]
This simplifies to finding:
\[
2x + 1 = 0
\]
### Solve for \(x\):
\[
2x = -1 \\
x = -\frac{1}{2}
\]
### Conclusion
Thus, to answer your question directly, the shared binomial in Joana's work is **\(2x + 1\)**. This means that the quadratic equation actually does have a solution, specifically \(x = -\frac{1}{2}\). Joana’s error was in not recognizing that there was a common binomial to factor out. Thank you for your patience, and I hope this clears up the confusion!