Question
Y^2=5x^3-x ; Find g(2)
Answer is +-sqrt 38
Answer is +-sqrt 38
Answers
Answer
How did you take the square root from both sides? Where did you put them?
Answer
Oh ok, so this is how I have it written out is this good?
Y^2=5x^3-x
Y^2=5(2^3)-2 Since in Pemdas exponents come first which are in parentheses, it’s 2x2x2=8
Y^2=5(8)-2 Multiply 5x8=40
Square root of Y^2= square root of 40-2 Subtract and answer is 38
Y^2=+-square root of 38
Is this right?
Y^2=5x^3-x
Y^2=5(2^3)-2 Since in Pemdas exponents come first which are in parentheses, it’s 2x2x2=8
Y^2=5(8)-2 Multiply 5x8=40
Square root of Y^2= square root of 40-2 Subtract and answer is 38
Y^2=+-square root of 38
Is this right?
Answered by
GPT-4o mini
To find \( g(2) \) for the equation \( y^2 = 5x^3 - x \), we first need to evaluate the right-hand side when \( x = 2 \).
Start by substituting \( x = 2 \) into the equation:
\[
y^2 = 5(2^3) - 2
\]
Calculating \( 2^3 \):
\[
2^3 = 8
\]
Now substitute back into the equation:
\[
y^2 = 5(8) - 2
\]
Calculating \( 5 \times 8 \):
\[
5 \times 8 = 40
\]
Now subtract \( 2 \):
\[
y^2 = 40 - 2 = 38
\]
Thus, we have:
\[
y^2 = 38
\]
Now, taking the square root of both sides gives:
\[
y = \pm \sqrt{38}
\]
Therefore, \( g(2) = \pm \sqrt{38} \), which matches your provided answer.
Start by substituting \( x = 2 \) into the equation:
\[
y^2 = 5(2^3) - 2
\]
Calculating \( 2^3 \):
\[
2^3 = 8
\]
Now substitute back into the equation:
\[
y^2 = 5(8) - 2
\]
Calculating \( 5 \times 8 \):
\[
5 \times 8 = 40
\]
Now subtract \( 2 \):
\[
y^2 = 40 - 2 = 38
\]
Thus, we have:
\[
y^2 = 38
\]
Now, taking the square root of both sides gives:
\[
y = \pm \sqrt{38}
\]
Therefore, \( g(2) = \pm \sqrt{38} \), which matches your provided answer.
Answered by
GPT-4o mini
When we solve the equation \( y^2 = 38 \), we take the square root of both sides to find the values of \( y \). Here's a step-by-step explanation of that process.
Starting from the equation:
\[
y^2 = 38
\]
To solve for \( y \), we can take the square root of both sides. When we do that, we must remember that taking the square root can yield both a positive and a negative solution. The mathematical expression is written as:
\[
y = \pm \sqrt{38}
\]
This means:
1. We recognize two possible values for \( y \) because the square root operation (in the context of real numbers) produces both a positive and negative value:
- The positive root, \( \sqrt{38} \)
- The negative root, \( -\sqrt{38} \)
So, combining both results gives us:
\[
g(2) = \sqrt{38} \quad \text{or} \quad g(2) = -\sqrt{38}
\]
To summarize, we placed the positive and negative roots on the same equation to represent both possible solutions. Hence, when we evaluate \( g(2) \), it can be either \( \sqrt{38} \) or \( -\sqrt{38} \), which is where \( g(2) = \pm \sqrt{38} \) comes from.
Starting from the equation:
\[
y^2 = 38
\]
To solve for \( y \), we can take the square root of both sides. When we do that, we must remember that taking the square root can yield both a positive and a negative solution. The mathematical expression is written as:
\[
y = \pm \sqrt{38}
\]
This means:
1. We recognize two possible values for \( y \) because the square root operation (in the context of real numbers) produces both a positive and negative value:
- The positive root, \( \sqrt{38} \)
- The negative root, \( -\sqrt{38} \)
So, combining both results gives us:
\[
g(2) = \sqrt{38} \quad \text{or} \quad g(2) = -\sqrt{38}
\]
To summarize, we placed the positive and negative roots on the same equation to represent both possible solutions. Hence, when we evaluate \( g(2) \), it can be either \( \sqrt{38} \) or \( -\sqrt{38} \), which is where \( g(2) = \pm \sqrt{38} \) comes from.
Answered by
GPT-4o mini
You've got the right approach, but there are a few places where the explanation could be clarified or slightly corrected. Here's a revised version of your explanation:
---
1. Start with the given equation:
\[
y^2 = 5x^3 - x
\]
2. Substitute \( x = 2 \):
\[
y^2 = 5(2^3) - 2
\]
3. Calculate \( 2^3 \):
Since in the order of operations (PEMDAS), exponents come first, we have:
\[
2^3 = 8
\]
4. Substitute back into the equation:
\[
y^2 = 5(8) - 2
\]
5. Multiply:
\[
5 \times 8 = 40
\]
6. Substitute:
\[
y^2 = 40 - 2
\]
7. Subtract:
\[
y^2 = 38
\]
8. Now, take the square root of both sides:
\[
y = \pm \sqrt{38}
\]
---
### Summary
- You summarized the calculations nicely! Just ensure that you clarify each step, particularly where you subtract \( 2 \) from \( 40 \) to arrive at \( 38 \), and when you take the square root, you include both the positive and negative results explicitly.
Overall, your structure is good; just focus on clear transitions from one calculation to the next!
---
1. Start with the given equation:
\[
y^2 = 5x^3 - x
\]
2. Substitute \( x = 2 \):
\[
y^2 = 5(2^3) - 2
\]
3. Calculate \( 2^3 \):
Since in the order of operations (PEMDAS), exponents come first, we have:
\[
2^3 = 8
\]
4. Substitute back into the equation:
\[
y^2 = 5(8) - 2
\]
5. Multiply:
\[
5 \times 8 = 40
\]
6. Substitute:
\[
y^2 = 40 - 2
\]
7. Subtract:
\[
y^2 = 38
\]
8. Now, take the square root of both sides:
\[
y = \pm \sqrt{38}
\]
---
### Summary
- You summarized the calculations nicely! Just ensure that you clarify each step, particularly where you subtract \( 2 \) from \( 40 \) to arrive at \( 38 \), and when you take the square root, you include both the positive and negative results explicitly.
Overall, your structure is good; just focus on clear transitions from one calculation to the next!