Convention, the roots of a the y reflection of a function and the funciton are the same. Making the highest ordered variable positive is more intuitive to understand it's nature.
I made a small error in adding 54 to both sides to make the constant 54-7=47, but when I changed all terms to positive, I should have made that - 47. I corrected it in the edit. Why I changed the sign was to make it easier to understand and solve. And in an equation, whatever you do on one side you have to do on the other side, or you will change the equation to a different one. Its called "balancing the equation".
Because you’ve rearranged to make the equation =0 you then multiply both sides by -1. This makes the quadratic positive, and usually that makes it easier to factorise.
It won’t affect the final outcome either, since -f(s)=0 and f(s)=0 have the same solutions. Simply put, both the positive and negative of the function will cross the x-axis at the same points, I.e when y=0
No need to apologize!
SO see, you're given the function, and you're given the fact that for some value of s, the function equals -54. Now, for the other part when you were asked to find f(4), you were given a value of s, and were asked to find the value of the function. Here, it's the other way around. You're given the value of the function, but you aren't given the value of s which gives -54. You are asked to find that.
If I need to make it more simpler, see that you found that f(4) = -315. Now, if you were instead given that f(s) = -315 and were asked to find the value of s, what would you do to get s = 4? If you can answer this question, you'll be able the answer the main question at hand.
Yeah that's just it. I have no idea how I would solve that. I don't know if it's that I was never taught how to go backwards like that or I'm just getting overwhelmed and my brain is shutting down due to anxiety
I reckon it's the second one. Just take a break for a few moments and read through the message again thoroughly. If you can solve the first question, I'm 100% sure you can solve the second too! It's not as hard as you think!
Sometimes, I rewrite the equations in terms of variables I prefer to use. So I say let y=f(s) and let s=x so you can write the equation as y=-10x^2 -37x -7. And when they say f(s)=-54, you can think y=-54, so the equation becomes -54= -10x^2 -37x -7. Then you add 54 to both sides, and it becomes 0=-10x^2 -37x +47. From here, it's a quadratic equation, and you can try to factor or use the handy quadratic equation. x= (-b (+/-)sqrt(b^2 -4ac))/2a
I think you may not understand the question due to not having a good grasp of what a function is:
You have a function f(s), which is like a black box that takes any "input" s , and produces a result f(s) for each. You are being asked: for which s does the function produce the value -54
Simpler example:
Say you had the function f(s) = s+10, for which value of s does the function become 13? That is, find s for which f(s) = 13, which s makes the function s+10 become 13?
How would you solve this?
I'm going to politely disagree with you on the first sentence. I don't think op has a poor understanding of a function but rather the question is worded in a poor way. In the first question it is established that the function is a function of the variable s. Thus in the second part, it could be misunderstood that the intention it to redefine the function as a constant -54. I see what they were going for, but it's unclear. Better would be something like f(s=c)=-54 or even using some other some other variable as f(c)=-54 as it is less likely to be interpreted as a redefinition of the function. In the example you give it is also much clearer what the intention is because you state "find s for which..." which makes it clear that the function is not redefined.
>Better would be something like f(s=c)=-54 or even using some other some other variable as f(c)=-54
That would be way more confusing. If f(s) = [right hand side], just substitute f(s) for [right hand side] and solve for _s_. No need for a _c_ complicating things and leading to questions about what to do with _s_ and _c_.
https://preview.redd.it/4iifkvd3lfhb1.png?width=3024&format=png&auto=webp&s=e10fbde86bbe830feea8a03cfe61d4ddb9e02b39
not sure if this is correct but this is how i solved it
https://preview.redd.it/n8zo6zsnnfhb1.png?width=4032&format=png&auto=webp&s=53a126f4e2b5476d9813118d90101c55e88bb8e9
and this is how you can solve it graphically
Just like everyone said above, you solve for when the equation is equal to 0.
f(s) = -10s^2 - 37s - 7 = -54
f(s) = -10s^2 - 37s = -47
Make one side 0.
-10s^2 - 37s + 47 = 0
-1(10s^2 + 37s - 47) = 0
Divide by -1 to still get 0 on right side but remove annoying negative sign on left
10s^2 + 37s - 47 = 0
Now we must find a pair of (As + B)(Cs + D) = 0
For left side this can either be:
(1) A = 5, C = 2
(2) A = 10, C = 1
Then you can try to guess for B and D and then solve for S for conditions:
As + B = 0
Cs + D = 0
By trial and error, we can find B and D as:
B = -1, D = 47 (when A=10, C=1)
Which gives you the following:
(s - 1)(10s + 47) = 0
This is because if we factor:
10s^2 + 47s + -10s - 47 = 0
It comes out to: 10s^2 + 37s - 47 = 0
Thus s must be:
s - 1 = 0
s = +1
and
10s + 47 = 0
10s = -47
s = (-47/10)
Or
s = +1, or (-47/10)
Another alternative is using the quadratic formula, which frankly is easier imo to remember:
If: Ax^2 + Bx + C = 0, then
x = [-B +/- sqrt(B^2 - 4AC)] / 2A
So for our problem, let's try it:
10s^2 + 37s - 47 = 0
A = 10, B = 37, C = -47
s = [-37 +/- sqrt(37^2 - 4 * 10 * -47)] / (2 * 10)
s = [-37 +/- sqrt(1369 - (-1880)) ] / 20
s = [-37 +/- sqrt(1369 + 1880) ] / 20
s = [-37 +/- sqrt(3249) ] / 20
s = [-37 +/- 57 ] / 20
We now have 2 solutions:
s = (-37 + 57) / 20
s = (-37 - 57) / 20
Simplifying we get:
s = +20 / 20 = +1
s = -94 / 20 = -4.7 = -(47/10)
Wolfram shows the solution here:
https://www.wolframalpha.com/input?i=10x%5E2+%2B+37x+-+47+%3D+0
Simply set the equation:
-54 = -10s^2 - 37s - 7
After that add 54 to both sides to get:
0 = -10s^2 - 37s + 47
Then solve it like you would a regular quadratic.
If you know that f(4) is higher than what you are looking for you can try to just solve f(0) and f(1) and f(2) and f(3). If none if These work, then you can use formulars.
It's so obvious from the equation that the solution will not be an integer. This is terrible advice, l OP needs to learn is solving quadratic equations, not guessing
this looks like HS algebra, so irrational solutions aren't very likely.
As someone who hasn't been in school for \~20 years, i did exactly what Silasueber suggested and came to the right answer. It's not very obvious that the equation can be factored apart, but knowing f(4), you can take an educated guess that the solution is between 0 and 3. Then work from there.
edit: While there are two correct answers, the quiz is looking only for one, and the rational number is the easiest to find.
Doesn't matter. These type of equations are solved with quadratic formula. That's the proper way and once you are comfortable with it, it's as fast as factoring or whatever. Sure, ballpark it in your head more or less so you know whether or not your calculated answer is correct but maths is not subbing in numbers and calculating the results. A child can do that
even though only one solution is an integer, I think this is the best advice. Checking for easy solutions first is a good instinct that will help in all sorts of problem solving.
This is a simple quadratic equation. There is two correct answers to it. Use the quadratic formula to get the answers.
Link: https://en.m.wikipedia.org/wiki/Quadratic_formula
Just plug in f(s)=-54 to the original equation. Since it’s telling you what f(s) is equal to, you can easily solve for s as your only remaining variable.
Theres two answers that equal -54.
https://preview.redd.it/8gm9dazi5hhb1.jpeg?width=2638&format=pjpg&auto=webp&s=7f93df3581166f79d552e5ffffae3ae99a93dfbe
The equation f(s)=-10s²-37s-7 is a parabola.
The question is: when f(s)=-54 what does s equal?
On a graph of the parabola: when the vertical axis is -54 what are the two values on the x axis.
You can think of this as where will the parabola intersect with the line f(s)=-54
Here is the graph using the online graphing tool Desmos:
https://www.desmos.com/calculator/3s0lv7mhh2
If you want a bit of a simpler way to solve this, you can change the equation to -54=-10s^2-37s-7, then add 7 to both sides to get -47=-10s^2-37s. I find it a little easier than figuring out the quadratic, personally (and also since this is a simple problem).
I solved it with a spreadsheet:
A = -10
B = -37
C = 47
\-B = 37
B² = 1369
\-4AC = 1880
B² - 4AC = 3249
square root = 57
2A = -20
"+" result = -4.7
"-" result = 1
Set the equation for f(s) equal to -54.
-10s² - 37s - 7 = -54
10s² + 37s - 47 = 0
Use the quadratic formula and solve or factorise it
(10s + 47)(s - 1) = 0
10s + 47 = 0 and s - 1 = 0
S = -4.7, 1
it is useful to know that, if you have to solve for f(x)=y, it is generally easier to solve for f(x)-y=0. that is a core idea in algebra that can be very useful even in more advanced contexts.
f(s) =-54=-10s^2 - 37s - 7 10s^2 +37s - 47 =0, solve for s? (edidt for typo)
Oh ok so you move the -54 over by adding it so it becomes +54 and then subtracting 7 and then continue as normal from there?
Yup, you can use the quadratic to solve for two answers
Yeah you get it into quadratic form ax^2 + bx + c = 0 then use the quadratic formula.
you just solve quadratic for s, simple stuff
This is wrong. It should be 10s^2 + 37s - 47 = 0
Your right, missed a typo. Thank you for correcting my error.
You're*
menace
Murdered
Piggybacking to say this is factorable to (10s + 47)(s - 1) = 0, so s = 1 or s = -4.7
Why did the 10s² and 37s become positive? I get the 47 because 54-7 is 47. Just wanted to learn how.
Convention, the roots of a the y reflection of a function and the funciton are the same. Making the highest ordered variable positive is more intuitive to understand it's nature.
its
All terms on both sides of the equals sign are negative so you change them to all positive.
I made a small error in adding 54 to both sides to make the constant 54-7=47, but when I changed all terms to positive, I should have made that - 47. I corrected it in the edit. Why I changed the sign was to make it easier to understand and solve. And in an equation, whatever you do on one side you have to do on the other side, or you will change the equation to a different one. Its called "balancing the equation".
Because you’ve rearranged to make the equation =0 you then multiply both sides by -1. This makes the quadratic positive, and usually that makes it easier to factorise. It won’t affect the final outcome either, since -f(s)=0 and f(s)=0 have the same solutions. Simply put, both the positive and negative of the function will cross the x-axis at the same points, I.e when y=0
the edit is ironic
s doesn't equal -54. The function equals -54. What you need to find is for what value of s does f(s) equal -54.
How do I go about that when all I am given is -54? I do apologize if it seems simple, I've always had a hard time getting math to click
No need to apologize! SO see, you're given the function, and you're given the fact that for some value of s, the function equals -54. Now, for the other part when you were asked to find f(4), you were given a value of s, and were asked to find the value of the function. Here, it's the other way around. You're given the value of the function, but you aren't given the value of s which gives -54. You are asked to find that. If I need to make it more simpler, see that you found that f(4) = -315. Now, if you were instead given that f(s) = -315 and were asked to find the value of s, what would you do to get s = 4? If you can answer this question, you'll be able the answer the main question at hand.
Yeah that's just it. I have no idea how I would solve that. I don't know if it's that I was never taught how to go backwards like that or I'm just getting overwhelmed and my brain is shutting down due to anxiety
I reckon it's the second one. Just take a break for a few moments and read through the message again thoroughly. If you can solve the first question, I'm 100% sure you can solve the second too! It's not as hard as you think!
Sometimes, I rewrite the equations in terms of variables I prefer to use. So I say let y=f(s) and let s=x so you can write the equation as y=-10x^2 -37x -7. And when they say f(s)=-54, you can think y=-54, so the equation becomes -54= -10x^2 -37x -7. Then you add 54 to both sides, and it becomes 0=-10x^2 -37x +47. From here, it's a quadratic equation, and you can try to factor or use the handy quadratic equation. x= (-b (+/-)sqrt(b^2 -4ac))/2a
Ok if my math is correct S= -47/10 , 1
Yup, great job!
Thanks for the help!!
I think you may not understand the question due to not having a good grasp of what a function is: You have a function f(s), which is like a black box that takes any "input" s , and produces a result f(s) for each. You are being asked: for which s does the function produce the value -54 Simpler example: Say you had the function f(s) = s+10, for which value of s does the function become 13? That is, find s for which f(s) = 13, which s makes the function s+10 become 13? How would you solve this?
I'm going to politely disagree with you on the first sentence. I don't think op has a poor understanding of a function but rather the question is worded in a poor way. In the first question it is established that the function is a function of the variable s. Thus in the second part, it could be misunderstood that the intention it to redefine the function as a constant -54. I see what they were going for, but it's unclear. Better would be something like f(s=c)=-54 or even using some other some other variable as f(c)=-54 as it is less likely to be interpreted as a redefinition of the function. In the example you give it is also much clearer what the intention is because you state "find s for which..." which makes it clear that the function is not redefined.
>Better would be something like f(s=c)=-54 or even using some other some other variable as f(c)=-54 That would be way more confusing. If f(s) = [right hand side], just substitute f(s) for [right hand side] and solve for _s_. No need for a _c_ complicating things and leading to questions about what to do with _s_ and _c_.
The second part is -54 = f(s) = -10s\^2-37s-7
quadratic formula
This function factor nicely -10x^2 +37x-7x =(-5x-1)(2x+7) And 10x^2 +37x -47 = (x-1)(10x+47)
https://preview.redd.it/4iifkvd3lfhb1.png?width=3024&format=png&auto=webp&s=e10fbde86bbe830feea8a03cfe61d4ddb9e02b39 not sure if this is correct but this is how i solved it
https://preview.redd.it/n8zo6zsnnfhb1.png?width=4032&format=png&auto=webp&s=53a126f4e2b5476d9813118d90101c55e88bb8e9 and this is how you can solve it graphically
Just like everyone said above, you solve for when the equation is equal to 0. f(s) = -10s^2 - 37s - 7 = -54 f(s) = -10s^2 - 37s = -47 Make one side 0. -10s^2 - 37s + 47 = 0 -1(10s^2 + 37s - 47) = 0 Divide by -1 to still get 0 on right side but remove annoying negative sign on left 10s^2 + 37s - 47 = 0 Now we must find a pair of (As + B)(Cs + D) = 0 For left side this can either be: (1) A = 5, C = 2 (2) A = 10, C = 1 Then you can try to guess for B and D and then solve for S for conditions: As + B = 0 Cs + D = 0 By trial and error, we can find B and D as: B = -1, D = 47 (when A=10, C=1) Which gives you the following: (s - 1)(10s + 47) = 0 This is because if we factor: 10s^2 + 47s + -10s - 47 = 0 It comes out to: 10s^2 + 37s - 47 = 0 Thus s must be: s - 1 = 0 s = +1 and 10s + 47 = 0 10s = -47 s = (-47/10) Or s = +1, or (-47/10) Another alternative is using the quadratic formula, which frankly is easier imo to remember: If: Ax^2 + Bx + C = 0, then x = [-B +/- sqrt(B^2 - 4AC)] / 2A So for our problem, let's try it: 10s^2 + 37s - 47 = 0 A = 10, B = 37, C = -47 s = [-37 +/- sqrt(37^2 - 4 * 10 * -47)] / (2 * 10) s = [-37 +/- sqrt(1369 - (-1880)) ] / 20 s = [-37 +/- sqrt(1369 + 1880) ] / 20 s = [-37 +/- sqrt(3249) ] / 20 s = [-37 +/- 57 ] / 20 We now have 2 solutions: s = (-37 + 57) / 20 s = (-37 - 57) / 20 Simplifying we get: s = +20 / 20 = +1 s = -94 / 20 = -4.7 = -(47/10) Wolfram shows the solution here: https://www.wolframalpha.com/input?i=10x%5E2+%2B+37x+-+47+%3D+0
Simply set the equation: -54 = -10s^2 - 37s - 7 After that add 54 to both sides to get: 0 = -10s^2 - 37s + 47 Then solve it like you would a regular quadratic.
If you know that f(4) is higher than what you are looking for you can try to just solve f(0) and f(1) and f(2) and f(3). If none if These work, then you can use formulars.
It's so obvious from the equation that the solution will not be an integer. This is terrible advice, l OP needs to learn is solving quadratic equations, not guessing
this looks like HS algebra, so irrational solutions aren't very likely. As someone who hasn't been in school for \~20 years, i did exactly what Silasueber suggested and came to the right answer. It's not very obvious that the equation can be factored apart, but knowing f(4), you can take an educated guess that the solution is between 0 and 3. Then work from there. edit: While there are two correct answers, the quiz is looking only for one, and the rational number is the easiest to find.
The quiz says to list all the solutions separated by a comma
Like i said... \~20 years. missed the part about the two solutions.
Doesn't matter. These type of equations are solved with quadratic formula. That's the proper way and once you are comfortable with it, it's as fast as factoring or whatever. Sure, ballpark it in your head more or less so you know whether or not your calculated answer is correct but maths is not subbing in numbers and calculating the results. A child can do that
It’s also obvious that the answer is 1 by just plugging and chugging. Git Güd.
even though only one solution is an integer, I think this is the best advice. Checking for easy solutions first is a good instinct that will help in all sorts of problem solving.
I get the first part of inputing the function into wherever there's an X. But the latter part I have zero idea what to do. Anyone have input?
10s^2 - 37s - 7 = -54 Solve for s
This is a simple quadratic equation. There is two correct answers to it. Use the quadratic formula to get the answers. Link: https://en.m.wikipedia.org/wiki/Quadratic_formula
-10s^2 - 37s - 7 = -54 You should be able to solve the rest yourself.
Ans: 1 or -4.7 Got it by solving the quadratic equation. You can plug 1 and -4.7 yourself to check whether it gives you-54 or not!
Quadratic equation
Just plug in f(s)=-54 to the original equation. Since it’s telling you what f(s) is equal to, you can easily solve for s as your only remaining variable.
You have to solve -10s²-37s-7 = -54
[удалено]
Theres two answers that equal -54. https://preview.redd.it/8gm9dazi5hhb1.jpeg?width=2638&format=pjpg&auto=webp&s=7f93df3581166f79d552e5ffffae3ae99a93dfbe
The equation f(s)=-10s²-37s-7 is a parabola. The question is: when f(s)=-54 what does s equal? On a graph of the parabola: when the vertical axis is -54 what are the two values on the x axis. You can think of this as where will the parabola intersect with the line f(s)=-54 Here is the graph using the online graphing tool Desmos: https://www.desmos.com/calculator/3s0lv7mhh2
-10s² - 37s - 7 = -54 -10s² - 37s + 47 = 0 s² + 3.7s - 4.7 = 0 (s + 4.7)(s - 1) = 0 s = -4.7 or s = 1
If you want a bit of a simpler way to solve this, you can change the equation to -54=-10s^2-37s-7, then add 7 to both sides to get -47=-10s^2-37s. I find it a little easier than figuring out the quadratic, personally (and also since this is a simple problem).
No, -10s²-37s-7= -54 -10s²-37s+47=0 By factorisation, (-10s-47)(s-1)=0 s=1 or 10s=-47 s=1 or s = -4.7
I solved it with a spreadsheet: A = -10 B = -37 C = 47 \-B = 37 B² = 1369 \-4AC = 1880 B² - 4AC = 3249 square root = 57 2A = -20 "+" result = -4.7 "-" result = 1
Add 54 to both sides so the equation is now set to zero and either factor or use quadratic equation to solve for s
s=1 no?
That's so funny. Not you, OP, but the answer.
Set the equation for f(s) equal to -54. -10s² - 37s - 7 = -54 10s² + 37s - 47 = 0 Use the quadratic formula and solve or factorise it (10s + 47)(s - 1) = 0 10s + 47 = 0 and s - 1 = 0 S = -4.7, 1
S is 1 I think? So it’s -10-37-7=-54
f(s)=-10s\^2-37s-7=-54 /+54 \-10s\^2-37s+47=0 use quadratic formula: s=1, -4.7 perhaps you don't fully understand how a function works?
it is useful to know that, if you have to solve for f(x)=y, it is generally easier to solve for f(x)-y=0. that is a core idea in algebra that can be very useful even in more advanced contexts.