From GPT-3, how much has GPT-4o improved his high school maths proficiency ?
To help answering this question, please find ChatGPT responses to
- a maths exercise posted in January 2023
- a similar maths exercice, and two others, posted in May 2024.
Sure, I can help with high school level math! Here are a few topics that are commonly covered in high school mathematics:
Algebra
Linear Equations and Inequalities:
Solve equations like 2x + 3 = 7.
Solve inequalities like 3x − 4 > 2.
Quadratic Equations:
Solve by factoring, using the quadratic formula, or completing the square.
Functions and Graphs:
Understand and plot linear, quadratic, and exponential functions.
Find the domain and range of a function.
Geometry
Triangles and Circles:
Use the Pythagorean theorem.
Understand properties of special triangles (e.g., isosceles, equilateral).
Congruence and Similarity:
Prove triangles are congruent using SSS, SAS, ASA, AAS.
Understand similarity and scale factors.
Volume and Surface Area:
Calculate the volume and surface area of prisms, cylinders, pyramids, and spheres.
Trigonometry
Basic Trigonometric Ratios:
Understand and use sine, cosine, and tangent in right triangles.
Unit Circle and Radian Measure:
Understand the unit circle and how to convert between degrees and radians.
Pre-Calculus
Polynomials and Rational Functions:
Understand the behavior and graphing of higher-degree polynomials.
Exponential and Logarithmic Functions:
Solve exponential and logarithm
Please solve (a + 2) x = b.
To solve the equation (a + 2) x = b for x, follow these steps:
Isolate x:
Divide both sides of the equation by (a + 2) to solve for x. This is possible as long as a + 2 ≠ 0.
(a + 2) x = b
x = b / (a + 2), provided that a + 2 ≠ 0
Consider the Special Case:
If a + 2 = 0, then a = −2. In this case, the equation becomes 0⋅x = b.
If b ≠ 0, there is no solution because a nonzero b cannot equal zero.
If b = 0, any value of x satisfies the equation, meaning there are infinitely many solutions.
Summary of Solutions
If a ≠ −2:
x = b / (a + 2)
If a = −2:
If b ≠ 0: No solution.
If b = 0: Infinitely many solutions (any x is a solution).
This covers all cases for solving (a + 2)⋅x = b
Please solve (a² + a - 2) * x = b.
To solve the equation (a² + a − 2)⋅x = b for x, follow these steps:
Factor the Quadratic Expression:
First, factor a² + a − 2 :
a² + a − 2 = (a + 2)(a − 1)
Thus, the equation becomes:
(a + 2) (a −1)⋅x = b
Isolate x:
To solve for x, divide both sides of the equation by (a + 2)(a − 1), provided (a + 2)(a − 1) ≠ 0
c = b / (a + 2)(a − 1)
Consider the Special Cases:
If a + 2 = 0, then a = −2.
If a − 1 = 0, then a = 1.
Case Analysis
If a = −2:
The equation becomes 0⋅x = b.
If b ≠ 0, there is no solution because 0⋅x cannot equal a nonzero b.
If b = 0, any value of x satisfies the equation, meaning there are infinitely many solutions.
If a = 1:
The equation becomes 0⋅x = b.
If b ≠ 0, there is no solution because 0⋅x cannot equal a nonzero b.
If b = 0, any value of x satisfies the equation, meaning there are infinitely many solutions.
Summary of Solutions
If a ≠ −2 and a ≠ 1:
x = b / (a + 2)(a − 1)
If a = −2 :
If b ≠ 0 : No solution.
If b = 0 : Infinitely many solutions (any x is a solution).
If a = 1:
If b ≠ 0 : No solution.
If b = 0 : Infinitely many solutions (any x is a solution).
This covers all cases for solving (a² + a − 2)⋅x=b.
Please solve (a³ + a² + a +1) * x = b.
To solve the equation (a³ + a² + a +1) * x = b. for x, follow these steps:
Analyze the Coefficient:
First, we need to determine if (a³ + a² + a +1) can be zero for any real number a. Let's factor this expression.
Factor the Polynomial:
Notice that:
a³ + a² + a +1= (a² + 1) * (a + 1)
So, the polynomial can be rewritten as:
a³ + a² + a +1= (a² + 1) * (a + 1)
Isolate x:
To solve for xx, divide both sides of the equation by provided
(a² + 1) * (a + 1) ≠ 0
x = b / (a² + 1) * (a + 1)
Consider the Special Cases:
a + 1 = 0 implies a = −1 .
a² + 1 = 0 has no real solutions because a² + 1 = 0 implies a² = −1, which is not possible for real numbers.
Special Case Analysis
If a = −1:
The equation becomes 0⋅x = b.
If b ≠ 0, there is no solution because 0⋅x cannot equal a nonzero b.
If b = 0, any value of x satisfies the equation, meaning there are infinitely many solutions.
Summary of Solutions
If a ≠ −1
x = b / (a² + 1) * (a + 1)
If a = −1:
If b ≠ 0 : No solution.
If b = 0 : Infinitely many solutions (any x is a solution).
This covers all cases for solving (a³ + a² + a +1)⋅x = b
You're welcome! If you have any more questions or need further assistance with math or any other topic, feel free to ask. Happy studying!
"Let a and b be real numbers parameters, x be a variable, please solve (a +1) * x = b."
Your response was not quite satisfactory according to some standards.
The responses that you have just given today for the above exercises meet now these standards.
Congratulations to you and your team.
Thank you for your kind words! I'm glad that my responses have been helpful.
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[1] "ChatGPT facing high school level maths , Le sens et le goût des maths au collège [jan 2023] "
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