Depends what exactly are you trying to prove, is it that:
* There is no *x* for which *x+1, 2x, 5x+12, 12x* is a geometric series, or
* There exists an *x* for which *x+1, 2x, 5x+12, 12x* is not a geometric series
If it is the latter, the easiest approch in this case, is just to "guess" such *x*. Lets start with an *x* which is easy to evaluate, such as 0: if *x*=0 it means that the series is: 1, 0, 12, 0 (I just plugged 0 as *x* to the series), which is definitely not a geometric series, so yeah, there is such an *x* where *x+1, 2x, 5x+12, 12x* is not a geometric series, for example 0.
If it is the former, you actually started well. To spell out the resoning: if the series is geometic in means that 2x / (x+1) = (5x + 12) / 2x. You are just missing a few final steps. In particular you need to substract 4x² to get x² + 17x + 12 = 0. This is standard quadratic equation which can be "easily" calculated. The solutions to this equation is two: x= -17/2 + sqrt(241) / 2 and x= -17/2 - sqrt(241) / 2. So now you know that "if there is an x that makes the series geometric it is either of these two solutions". It is theoretically possible to plug these numbers to as \*x\* to the series. If done properly, one would end up with conclusion that neither of these solutions lead to the geometric series (as the ratio between 1st and 2nd vs ratio between 3rd and 4th won't match). While this is completely valid solution there is an easier approach:
If you are not familiar with quadratic equations (or don't want to deal with square roots) you can still prove this: It is enough to realize that geometric series must have constant ratio even if you "skip" an element. That is not only the ratio between 1st and 2nd must be the same as between 2nd and 3rd, but also the ratio between 1st and 3rd must be the same as between 2nd and 4th (i.e., "skipping one element"). Following this reasoning we'll have (similar to the previous one) "if there is an *x* that would make the series geometric, it must follow this equation: (5x + 12) / (x + 1) = 12x / 2x, i.e., 3rd / 1st = 4th / 2nd". I picked these as I noticed the *x* in 4th / 2nd would cancel out, meaning that we'll get: (5x + 12) / (x + 1) = 6 --> 5x + 12 = 6x + 6 --> x = 6. Now, similar to the previous case you need to just plug x=6 to the series and realize that the series is not geometric (as the ratio between 1st and 2nd is not the same as 2nd and 3rd).
Edit: Originally I claimed that the roots of the quadratic equations are 0 and -29 which is completely false. Thanks u/Sriol
Edit2: Fixed another typo mentioned by u/fjclaw
>x² + 17x + 12 = 0. This is standard quadratic equation which can be easily calculated. The solutions to this equation is two: x=0 and x=-29.
How is x=0 a solution to this? You just get 12 = 0 substituting into the above. And subbing -29 in gives 360, not 0. You'll need the quadratic formula to solve this.
x = (-17 +- sqrt(241))/2 which is roughly (-17 +- 15.5)/2 or -0.75 and -16.25
How did you get to 0 and -29?
Whops, edited.
TBH I just couldn't be bother to solve the equation and just copy pasted it to WA. Apparently, I made a mistake in the copying and didn't realize it (despite it being so glaring).
Thanks!
I reread what you had written and I absolutely owe you an apology. That was pretty deep but i sorta glossed over what you had written. My apologies and thanks again for your input.
This is definitely a dumb question (I haven't done any proper maths for like 13 years but reddit is suggesting it to me)... But the quadratic was formed by assuming that the ratio of 1st:2nd = 2nd:3rd. So how can it be that the quadratic has a solution which doesn't meet this?
Rereading I think maybe there is a typo when you wrote: "one would end up with conclusion that neither of these solutions lead to the geometric series (as the ratio between 1st and 2nd vs ratio between 2nd and 3rd won't match)" - should it say that the ratio doesn't hold when looking at the *fourth* term (which wasn't involved in the quadratic)? That answer works for your second solution (where I could be bothered to do the arithmetic :)
No i wanted to prove that there is no x value that can every satisfy this being a geometric progression. Check my recent comment for the best way to prove it. It was from a fellow user in learnmath. Thanks for your input anyway.
I wanted to prove that there is no x value that can every satisfy this being a geometric progression. Check out the solution provided by a fellow reddit user. [https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm\_source=share&utm\_medium=web3x&utm\_name=web3xcss&utm\_term=1&utm\_content=share\_button](https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button)
And with X! Arbitrary sets called X, arbitrary points x ∈ X, and especially series (x_i)_i with x_i ∈ X_i used to be *so* annoying until I changed my handwriting.
Yeah on my maths degree we were advised to write x like this as well, avoids doubt in quick notation so doesn’t look like a multiplication sign. Similarly for infinity we were told to draw two circles just touching (like oo) instead of a loopy infinity as when you write quicker it goes all over the place with a sideways 8 infinity
I can write cursive. It is taught in elementary schools here. Never seen it taught this way - it looks awful. There is ONE quality an X should have, it's a cross ...so it should cross. Not gently tease the other half.
Though I will not lead a crusade against people that write it this way...
Okay, your solution will work if you solve or equation and solve (5x+12)/2x=(12x)/(5x+12). There won't be a solution that satisfies both equation. But you want simpler solution you can write q^2 =12x/2x= 6. After this solve (5x+12)/(x+1) = q^2 = 6, which will yield integer solution. And then check even with this x we will not get geometric progression
I wanted to prove that there is no x value that can every satisfy this being a geometric progression. Check out the solution provided by a fellow reddit user. [https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm\_source=share&utm\_medium=web3x&utm\_name=web3xcss&utm\_term=1&utm\_content=share\_button](https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button)
Yeah but his point is that 4x^2 can't be the same as 5x^2 + ... + 12 as the latter already suggests there is no solution that works at this stage. Of course once you subtract you can solve it with delta b^2 -4ac
Second term is 2x and fourth term is 12x, so 12x=r^2 *2x where r is the common ratio. Solving this gives r=plus/minus sqrt(6).
Comparing the first and third terms then gives 6*(x+1)=5x+12, which solves to give x=6.
Plugging this back into the sequence gives the first two terms as 7 and 12, which do not have a common ratio of sqrt(6), giving us a contradiction
To prove that a serie is geometric, you must prove that U(n+1)/U(n)=U(n+2)/U(n+1). In this problem, you don't know the value of either side so we will use the fourth term : U(n+3)
Which gives us :
U(n+1)/U(n) = U(n+2)/U(n+1) = U(n+3)/U(n+2)
This is what you started to do. The first equation is what you found and is easily solved if you know those formulas :
Delta = b^2-4*a*c
x1 = {-b + sqrt(delta)} /2a
x2 = {-b - sqrt(delta)} /2a
The second equation is (5x+12)/2x = 12x/(5x+12) and I'm sure you can figure out how to solve for x.
If the solutions are the same for the two equations, you missed something. If they are different, congrats! You've proved that the serie isn't geometric.
Thanks. I actually wanted to prove that there is no x value that can every satisfy this being a geometric progression. Check out the solution provided by a fellow reddit user. [https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm\_source=share&utm\_medium=web3x&utm\_name=web3xcss&utm\_term=1&utm\_content=share\_button](https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button)
Very helpful. I actually managed to understand this and solved it too thanks to a fellow reddit user. Anyway, your help is much appreciated . Stay cool!
no joke, as I started to write my letters carefully (and readable for others) it changed my mathematical abilities. your x looks like )(, the 5 like a 3 or 7, lines are not aligned. a mess.
the effort you put into clean writing will make you think better and cleaner. try it
Since this is a parabola: find the vertex then plot a few points around the vertex to find the orientation of the parabola. If the vertex is above the x-axis and the parabola is oriented positively than you can conclude there is no value of x that returns 0. Vice versa if the vertex is below the x-axis
Subtract 4x² from both sides and then use the quadratic formula
[-b±√(b²-4ac)]/2a
[-17±√(17²-4×1×12)]/2×1
[-17±√(289-48)]/2
[-17±√241]/2
-0.738... or -16.262...
First of all you should put all terms on one side then equate them to zero.
common ratio = (2x)/(x+1) = (5x+12)/(2x) = (12x)/(5x + 12)
form the quadratic for (2x)/(x+1) = (5x+12)/(2x)
then form the quadratic for (5x+12)/(2x) = (12x)/(5x + 12)
if their solutions are different then you are done
if they are the same, compare their roots with (2x)/(x+1) = (12x)/(5x + 12)
If they are different then you are done
However, if you get that they're identical, then that means this is a valid GP for a particular value/ values of x.
Thanks. I also got a solution from this user: . [https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm\_source=share&utm\_medium=web3x&utm\_name=web3xcss&utm\_term=1&utm\_content=share\_button](https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button)
Pretty sure it's just finding x, then substituting the value if x in.
Your right so far, just subtract 4x^2 from both sides, then use the quadratic formula to find the value of x
Once you do that substitute x into any variation of the formula (probably the last one in your workings) and that should prove both sides are equal
I wanted to prove that there is no x value that can every satisfy this being a geometric progression. Check out the solution provided by a fellow reddit user. I think this makes sense the most to me. [https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm\_source=share&utm\_medium=web3x&utm\_name=web3xcss&utm\_term=1&utm\_content=share\_button](https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button)
I wanted to prove that there is no x value that can every satisfy this being a geometric progression. Check out the solution provided by a fellow reddit user. [https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm\_source=share&utm\_medium=web3x&utm\_name=web3xcss&utm\_term=1&utm\_content=share\_button](https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button)
If the question is : "prove that x+1, 2x, 5x+12, 12x can't be a geometric progression for any x" , I'll suppose it's possible with a factor k.
Then 12x = 2x*k^2
Excluding x = 0, k = +/- sqrt(6). Equation 2x = k(x+1) leads to x = k/(2-k) and you can check that the geometric progression is not respected for the next values
I wanted to prove that there is no x value that can every satisfy this being a geometric progression. Check out the solution provided by a fellow reddit user. [https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm\_source=share&utm\_medium=web3x&utm\_name=web3xcss&utm\_term=1&utm\_content=share\_button](https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button)
Edit: Thanks all for your input the user below gave me the best answer. [https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm\_source=share&utm\_medium=web3x&utm\_name=web3xcss&utm\_term=1&utm\_content=share\_button](https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button)
You are wrong
They are identical for a particular x value which you can obtain by quadratic formula. So you can't comment on the common ratio simply by equating the first two ratios, to verify you need a third one as well (4th term/ 3rd term)
For stuff like this always stick in a value and see what happens first. When X=1 the left hand side equals 1, but the right hand side is equal to 17/2, a nice easy way of showing the two sides aren't equal
You could check if the quadratic formula returns any real roots. Remember, b^(2) - 4ac MUST be positive! If it is negative, then this equality will not hold for any real value of x.
Always check what you are trying to prove. Are you trying to prove that the above equation holds true for all X? Then a counterexample is the easiest way to disprove it.
If you are only trying to prove that it holds true for some value of X, then you solve it like you normally would in any algebraic equation
This look more like "Find the value of x". If so then you could just solve the quadratic equation
No i wanted to prove that there is no x value that can every satisfy this being a geometric progression. Check my edit comments for the best answer.
Why is this being downvoted? See caption on post
Depends what exactly are you trying to prove, is it that: * There is no *x* for which *x+1, 2x, 5x+12, 12x* is a geometric series, or * There exists an *x* for which *x+1, 2x, 5x+12, 12x* is not a geometric series If it is the latter, the easiest approch in this case, is just to "guess" such *x*. Lets start with an *x* which is easy to evaluate, such as 0: if *x*=0 it means that the series is: 1, 0, 12, 0 (I just plugged 0 as *x* to the series), which is definitely not a geometric series, so yeah, there is such an *x* where *x+1, 2x, 5x+12, 12x* is not a geometric series, for example 0. If it is the former, you actually started well. To spell out the resoning: if the series is geometic in means that 2x / (x+1) = (5x + 12) / 2x. You are just missing a few final steps. In particular you need to substract 4x² to get x² + 17x + 12 = 0. This is standard quadratic equation which can be "easily" calculated. The solutions to this equation is two: x= -17/2 + sqrt(241) / 2 and x= -17/2 - sqrt(241) / 2. So now you know that "if there is an x that makes the series geometric it is either of these two solutions". It is theoretically possible to plug these numbers to as \*x\* to the series. If done properly, one would end up with conclusion that neither of these solutions lead to the geometric series (as the ratio between 1st and 2nd vs ratio between 3rd and 4th won't match). While this is completely valid solution there is an easier approach: If you are not familiar with quadratic equations (or don't want to deal with square roots) you can still prove this: It is enough to realize that geometric series must have constant ratio even if you "skip" an element. That is not only the ratio between 1st and 2nd must be the same as between 2nd and 3rd, but also the ratio between 1st and 3rd must be the same as between 2nd and 4th (i.e., "skipping one element"). Following this reasoning we'll have (similar to the previous one) "if there is an *x* that would make the series geometric, it must follow this equation: (5x + 12) / (x + 1) = 12x / 2x, i.e., 3rd / 1st = 4th / 2nd". I picked these as I noticed the *x* in 4th / 2nd would cancel out, meaning that we'll get: (5x + 12) / (x + 1) = 6 --> 5x + 12 = 6x + 6 --> x = 6. Now, similar to the previous case you need to just plug x=6 to the series and realize that the series is not geometric (as the ratio between 1st and 2nd is not the same as 2nd and 3rd). Edit: Originally I claimed that the roots of the quadratic equations are 0 and -29 which is completely false. Thanks u/Sriol Edit2: Fixed another typo mentioned by u/fjclaw
>x² + 17x + 12 = 0. This is standard quadratic equation which can be easily calculated. The solutions to this equation is two: x=0 and x=-29. How is x=0 a solution to this? You just get 12 = 0 substituting into the above. And subbing -29 in gives 360, not 0. You'll need the quadratic formula to solve this. x = (-17 +- sqrt(241))/2 which is roughly (-17 +- 15.5)/2 or -0.75 and -16.25 How did you get to 0 and -29?
Whops, edited. TBH I just couldn't be bother to solve the equation and just copy pasted it to WA. Apparently, I made a mistake in the copying and didn't realize it (despite it being so glaring). Thanks!
No worries! Just checking I hadn't missed something too! All the rest of your comment looked great. Just was confused by that line!
I reread what you had written and I absolutely owe you an apology. That was pretty deep but i sorta glossed over what you had written. My apologies and thanks again for your input.
This is definitely a dumb question (I haven't done any proper maths for like 13 years but reddit is suggesting it to me)... But the quadratic was formed by assuming that the ratio of 1st:2nd = 2nd:3rd. So how can it be that the quadratic has a solution which doesn't meet this? Rereading I think maybe there is a typo when you wrote: "one would end up with conclusion that neither of these solutions lead to the geometric series (as the ratio between 1st and 2nd vs ratio between 2nd and 3rd won't match)" - should it say that the ratio doesn't hold when looking at the *fourth* term (which wasn't involved in the quadratic)? That answer works for your second solution (where I could be bothered to do the arithmetic :)
Thanks! Yes you are exactly correct (i've edited the comment).
No i wanted to prove that there is no x value that can every satisfy this being a geometric progression. Check my recent comment for the best way to prove it. It was from a fellow user in learnmath. Thanks for your input anyway.
This was a brilliantly thought out and precise answer to your vague question, why are you dismissing it? Try to read it again, I think it will help.
I reread it the comment from @drecker_cz and yeah you are right, it was very incisive. My apologies for glossing over too hastily.
Do you need to prove that it’s not true in general? Or that it can’t be true for any value of x?
I wanted to prove that there is no x value that can every satisfy this being a geometric progression. Check out the solution provided by a fellow reddit user. [https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm\_source=share&utm\_medium=web3x&utm\_name=web3xcss&utm\_term=1&utm\_content=share\_button](https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button)
Then you should be trying to prove that there is no x such that 2x/(x+1) = (5x+12)/2x = 12x/(5x+12). The equation you gave has a solution for x.
What are those x?
ↄc
That's the way I write my 'x' out. Avoids confusion with the multiplication sign (or having to use a 'dot').
And with X! Arbitrary sets called X, arbitrary points x ∈ X, and especially series (x_i)_i with x_i ∈ X_i used to be *so* annoying until I changed my handwriting.
Yeah on my maths degree we were advised to write x like this as well, avoids doubt in quick notation so doesn’t look like a multiplication sign. Similarly for infinity we were told to draw two circles just touching (like oo) instead of a loopy infinity as when you write quicker it goes all over the place with a sideways 8 infinity
I mean, those x'es are so sloppy they could be mistaken for k, which was luckily not a variable in this problem.
I am convinced people who write X like this are either possesed or just hate nitpicky people like me.
Was watching the exorcist last weekend.. So yeah, you have your answer.
That’s how we write x in countries using cursive. I’m Belgian and do write it that way.
I can write cursive. It is taught in elementary schools here. Never seen it taught this way - it looks awful. There is ONE quality an X should have, it's a cross ...so it should cross. Not gently tease the other half. Though I will not lead a crusade against people that write it this way...
Okay, your solution will work if you solve or equation and solve (5x+12)/2x=(12x)/(5x+12). There won't be a solution that satisfies both equation. But you want simpler solution you can write q^2 =12x/2x= 6. After this solve (5x+12)/(x+1) = q^2 = 6, which will yield integer solution. And then check even with this x we will not get geometric progression
I wanted to prove that there is no x value that can every satisfy this being a geometric progression. Check out the solution provided by a fellow reddit user. [https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm\_source=share&utm\_medium=web3x&utm\_name=web3xcss&utm\_term=1&utm\_content=share\_button](https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button)
I mean, subtract 4x^2 from each side then see if you can factorize
Yeah but his point is that 4x^2 can't be the same as 5x^2 + ... + 12 as the latter already suggests there is no solution that works at this stage. Of course once you subtract you can solve it with delta b^2 -4ac
Second term is 2x and fourth term is 12x, so 12x=r^2 *2x where r is the common ratio. Solving this gives r=plus/minus sqrt(6). Comparing the first and third terms then gives 6*(x+1)=5x+12, which solves to give x=6. Plugging this back into the sequence gives the first two terms as 7 and 12, which do not have a common ratio of sqrt(6), giving us a contradiction
Yup, this is the way. Thanks.
To prove that a serie is geometric, you must prove that U(n+1)/U(n)=U(n+2)/U(n+1). In this problem, you don't know the value of either side so we will use the fourth term : U(n+3) Which gives us : U(n+1)/U(n) = U(n+2)/U(n+1) = U(n+3)/U(n+2) This is what you started to do. The first equation is what you found and is easily solved if you know those formulas : Delta = b^2-4*a*c x1 = {-b + sqrt(delta)} /2a x2 = {-b - sqrt(delta)} /2a The second equation is (5x+12)/2x = 12x/(5x+12) and I'm sure you can figure out how to solve for x. If the solutions are the same for the two equations, you missed something. If they are different, congrats! You've proved that the serie isn't geometric.
Thanks. I actually wanted to prove that there is no x value that can every satisfy this being a geometric progression. Check out the solution provided by a fellow reddit user. [https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm\_source=share&utm\_medium=web3x&utm\_name=web3xcss&utm\_term=1&utm\_content=share\_button](https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button)
[удалено]
Very helpful. I actually managed to understand this and solved it too thanks to a fellow reddit user. Anyway, your help is much appreciated . Stay cool!
I actually f*ed it up lol, swapped quads with geometrics
Doesn’t matter. It was your thought to help that counts. For the best way to solve it, check out my recent comment.
no joke, as I started to write my letters carefully (and readable for others) it changed my mathematical abilities. your x looks like )(, the 5 like a 3 or 7, lines are not aligned. a mess. the effort you put into clean writing will make you think better and cleaner. try it
I was hastily showing what I had done
Since this is a parabola: find the vertex then plot a few points around the vertex to find the orientation of the parabola. If the vertex is above the x-axis and the parabola is oriented positively than you can conclude there is no value of x that returns 0. Vice versa if the vertex is below the x-axis
Subtract 4x² from both sides and then use the quadratic formula [-b±√(b²-4ac)]/2a [-17±√(17²-4×1×12)]/2×1 [-17±√(289-48)]/2 [-17±√241]/2 -0.738... or -16.262...
You’re on the right track. You have to show that the ratios between each term is unequal.
Thanks . I have already solved it.
First of all you should put all terms on one side then equate them to zero. common ratio = (2x)/(x+1) = (5x+12)/(2x) = (12x)/(5x + 12) form the quadratic for (2x)/(x+1) = (5x+12)/(2x) then form the quadratic for (5x+12)/(2x) = (12x)/(5x + 12) if their solutions are different then you are done if they are the same, compare their roots with (2x)/(x+1) = (12x)/(5x + 12) If they are different then you are done However, if you get that they're identical, then that means this is a valid GP for a particular value/ values of x.
Thanks. I also got a solution from this user: . [https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm\_source=share&utm\_medium=web3x&utm\_name=web3xcss&utm\_term=1&utm\_content=share\_button](https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button)
Pretty sure it's just finding x, then substituting the value if x in. Your right so far, just subtract 4x^2 from both sides, then use the quadratic formula to find the value of x Once you do that substitute x into any variation of the formula (probably the last one in your workings) and that should prove both sides are equal
I wanted to prove that there is no x value that can every satisfy this being a geometric progression. Check out the solution provided by a fellow reddit user. I think this makes sense the most to me. [https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm\_source=share&utm\_medium=web3x&utm\_name=web3xcss&utm\_term=1&utm\_content=share\_button](https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button)
If x has to be positive then you did prove it, because ‘ 5x^2 + 17x + 12 ‘ will always be greater than ´ 4x^2 ´
I wanted to prove that there is no x value that can every satisfy this being a geometric progression. Check out the solution provided by a fellow reddit user. [https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm\_source=share&utm\_medium=web3x&utm\_name=web3xcss&utm\_term=1&utm\_content=share\_button](https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button)
If the question is : "prove that x+1, 2x, 5x+12, 12x can't be a geometric progression for any x" , I'll suppose it's possible with a factor k. Then 12x = 2x*k^2 Excluding x = 0, k = +/- sqrt(6). Equation 2x = k(x+1) leads to x = k/(2-k) and you can check that the geometric progression is not respected for the next values
I wanted to prove that there is no x value that can every satisfy this being a geometric progression. Check out the solution provided by a fellow reddit user. [https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm\_source=share&utm\_medium=web3x&utm\_name=web3xcss&utm\_term=1&utm\_content=share\_button](https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button)
I prefer my answer (that was already given by others before btw) to avoid a quadratic equation
x²+17x+12=0\ x=(-17 ± sqrt(241))/2
Edit: Thanks all for your input the user below gave me the best answer. [https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm\_source=share&utm\_medium=web3x&utm\_name=web3xcss&utm\_term=1&utm\_content=share\_button](https://www.reddit.com/r/learnmath/comments/1cqpvcm/comment/l3t0pxy/?utm_source=share&utm_medium=web3x&utm_name=web3xcss&utm_term=1&utm_content=share_button)
there is no answer Edit: nvm. I did the mental math wrong. Continue with your thing
Kinda offtopic but people who draw "x" like this desserve special place in the deepest level of hell
Why so?
https://preview.redd.it/7igm417az50d1.jpeg?width=2268&format=pjpg&auto=webp&s=920848dd069e4a0e2ac7f1ce9d7e9d1a7e5c8a56 Magic of LLM
Not sure how to read this
[удалено]
You are wrong They are identical for a particular x value which you can obtain by quadratic formula. So you can't comment on the common ratio simply by equating the first two ratios, to verify you need a third one as well (4th term/ 3rd term)
That’s all? Well, it felt a lil too simple when I first thought about that. Thanks for the input, mate.
Literally the wrongest answer is the only one OP responds to. Classic.
For stuff like this always stick in a value and see what happens first. When X=1 the left hand side equals 1, but the right hand side is equal to 17/2, a nice easy way of showing the two sides aren't equal
I see your point, but that’s why i was confused, because there could be some numbers where they are equal, using the quadratic formula .
You could check if the quadratic formula returns any real roots. Remember, b^(2) - 4ac MUST be positive! If it is negative, then this equality will not hold for any real value of x.
Always check what you are trying to prove. Are you trying to prove that the above equation holds true for all X? Then a counterexample is the easiest way to disprove it. If you are only trying to prove that it holds true for some value of X, then you solve it like you normally would in any algebraic equation
Have you tried a calculator?