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lolab
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Erika's Equations | Math Solutions & Help
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In summary, Dan is providing guidance on how to solve a system of equations given three equations and three unknowns. He suggests solving for one variable and plugging it into the other equations. He also mentions a specific example and advises to post any difficulties for further assistance. However, the conversation abruptly shifts to discussing a hypothetical scenario, where it is impossible for two different values of f(2) to exist. The conversation then provides an explanation of the given parabola and equations, and poses a question to evaluate the possibility of a, b, and c values to satisfy the equations.
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The link is bad.lolab said:
-Dan
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lolab
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Do you understand how they got the first two equations? Can you get part a)?
As to the second part you have three equations in three unknowns. I'd solve the first equation for c, then plug that into the other two equations. Then solve one of them for b and plug that into the next. Then solve the last equation for a.
See what you can do with this and if you are still having problems post what you've got and we can take a look at it.
-Dan
As to the second part you have three equations in three unknowns. I'd solve the first equation for c, then plug that into the other two equations. Then solve one of them for b and plug that into the next. Then solve the last equation for a.
See what you can do with this and if you are still having problems post what you've got and we can take a look at it.
-Dan
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HallsofIvy
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This is immediately impossible. f(2) cannot have two different values!
We are given $y= f(x)= ax^2+ bx+ c$. The fact that (1, 5) is on its graph (a parabola) means that $y= 5= a(1)^2+ b(1)+ c= a+ b+ c$.
The fact that (2, 10) is on the graph means that $y= 10= a(2)^2+ b(2)+ c= 4a+ 2b+ c$.
If (2, 19) were also on the graph we would have $y= 19= a(2)^2+ b(2)+ c= 4a+ 2b+ c$.
So we have both $10= 4a+ 2b+ c$ and $19= 4a+ 2b+ c$.
What do you get if you subtract the first equation from the second? Is there ANY value of a, b, and c which will make that true?
We are given $y= f(x)= ax^2+ bx+ c$. The fact that (1, 5) is on its graph (a parabola) means that $y= 5= a(1)^2+ b(1)+ c= a+ b+ c$.
The fact that (2, 10) is on the graph means that $y= 10= a(2)^2+ b(2)+ c= 4a+ 2b+ c$.
If (2, 19) were also on the graph we would have $y= 19= a(2)^2+ b(2)+ c= 4a+ 2b+ c$.
So we have both $10= 4a+ 2b+ c$ and $19= 4a+ 2b+ c$.
What do you get if you subtract the first equation from the second? Is there ANY value of a, b, and c which will make that true?
FAQ: Erika's Equations | Math Solutions & Help
What is Erika's Equations?
Erika's Equations is a website that provides math solutions and help for various mathematical problems. It was created by a mathematician named Erika to assist students and professionals in understanding and solving complex equations.
Who can benefit from Erika's Equations?
Erika's Equations can benefit anyone who needs help with solving math problems, from students struggling with homework to professionals looking for quick solutions to complex equations. The website is designed to be user-friendly and accessible to all levels of math proficiency.
How does Erika's Equations work?
Erika's Equations works by providing step-by-step solutions to various math problems. Users can input their equations or select from a list of common problems, and the website will provide a detailed explanation of how to solve it. The solutions also include graphs and diagrams to aid in understanding.
Is Erika's Equations free to use?
Yes, Erika's Equations is completely free to use. The website is funded by advertisements and donations, so users do not have to pay anything to access the math solutions and help provided.
Can I request a specific equation to be solved on Erika's Equations?
Yes, Erika's Equations has a feature where users can submit a request for a specific equation to be solved. The website's team of mathematicians will work on solving the requested equation and add it to the list of solutions for others to benefit from.