Mixing
How to solve for x when function can't be inverted
$begingroup$
The sum of the first three members of a geometric progression is equal to 42.
Those same members are, correspondingly, the first, the second and the sixth member of an arithmetic progression. The task is to find out the values of those three elements. (Solution: 2, 8, 32 or 14, 14, 14). After a few lines of solving, I got an equation that describes the difference in the arithmetic progression as a function of x:
$$f(x)=(14-2x)(1+frac{14-x}{14-2x}+frac{(14-x)^2}{(14-2x)^2})$$
$$y=42$$
I'm 100% sure I overcomplicated this, but now I'd really like to know how to get the intersection points of this weird function with a line (the sum). After plotting the equation in Desmos, I get two solutions; 0 and 6. Those are indeed the appropriate differences from which I can obtain the three numbers, but I have no idea how to get to them.
Any help/question reformatting is appreciated!
functions arithmetic-progressions geometric-progressions
asked Jan 27 at 17:17
Andrija RadicaAndrija Radica
83
$endgroup$
add a comment |
$begingroup$
The sum of the first three members of a geometric progression is equal to 42.
Those same members are, correspondingly, the first, the second and the sixth member of an arithmetic progression. The task is to find out the values of those three elements. (Solution: 2, 8, 32 or 14, 14, 14). After a few lines of solving, I got an equation that describes the difference in the arithmetic progression as a function of x:
$$f(x)=(14-2x)(1+frac{14-x}{14-2x}+frac{(14-x)^2}{(14-2x)^2})$$
$$y=42$$
I'm 100% sure I overcomplicated this, but now I'd really like to know how to get the intersection points of this weird function with a line (the sum). After plotting the equation in Desmos, I get two solutions; 0 and 6. Those are indeed the appropriate differences from which I can obtain the three numbers, but I have no idea how to get to them.
Any help/question reformatting is appreciated!
functions arithmetic-progressions geometric-progressions
asked Jan 27 at 17:17
Andrija RadicaAndrija Radica
83
$endgroup$
add a comment |
$begingroup$
The sum of the first three members of a geometric progression is equal to 42.
Those same members are, correspondingly, the first, the second and the sixth member of an arithmetic progression. The task is to find out the values of those three elements. (Solution: 2, 8, 32 or 14, 14, 14). After a few lines of solving, I got an equation that describes the difference in the arithmetic progression as a function of x:
$$f(x)=(14-2x)(1+frac{14-x}{14-2x}+frac{(14-x)^2}{(14-2x)^2})$$
$$y=42$$
I'm 100% sure I overcomplicated this, but now I'd really like to know how to get the intersection points of this weird function with a line (the sum). After plotting the equation in Desmos, I get two solutions; 0 and 6. Those are indeed the appropriate differences from which I can obtain the three numbers, but I have no idea how to get to them.
Any help/question reformatting is appreciated!
functions arithmetic-progressions geometric-progressions
asked Jan 27 at 17:17
Andrija RadicaAndrija Radica
83
$endgroup$
The sum of the first three members of a geometric progression is equal to 42.
Those same members are, correspondingly, the first, the second and the sixth member of an arithmetic progression. The task is to find out the values of those three elements. (Solution: 2, 8, 32 or 14, 14, 14). After a few lines of solving, I got an equation that describes the difference in the arithmetic progression as a function of x:
$$f(x)=(14-2x)(1+frac{14-x}{14-2x}+frac{(14-x)^2}{(14-2x)^2})$$
$$y=42$$
I'm 100% sure I overcomplicated this, but now I'd really like to know how to get the intersection points of this weird function with a line (the sum). After plotting the equation in Desmos, I get two solutions; 0 and 6. Those are indeed the appropriate differences from which I can obtain the three numbers, but I have no idea how to get to them.
Any help/question reformatting is appreciated!
functions arithmetic-progressions geometric-progressions
functions arithmetic-progressions geometric-progressions
asked Jan 27 at 17:17
Andrija RadicaAndrija Radica
83
asked Jan 27 at 17:17
Andrija RadicaAndrija Radica
83
asked Jan 27 at 17:17
Andrija RadicaAndrija Radica
83
asked Jan 27 at 17:17
Andrija RadicaAndrija Radica
83
asked Jan 27 at 17:17
Andrija RadicaAndrija Radica
83
83
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let your numbers be $x,rx,r^2x$. Since $x+rx+r^2x=42$, $xneq 0$. Also,
$$r^2x-x=(r^2-1)x=5(rx-x)=5x(r-1) implies r^2-1=5(r-1) implies (r-1)(r-4)=0.$$
Therefore, $r=1$ or $r=4$. Now use $x+rx+r^2x=42$ to find the sequence.
answered Jan 27 at 17:35
Math LoverMath Lover
14.1k31437
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3089833%2fhow-to-solve-for-x-when-function-cant-be-inverted%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let your numbers be $x,rx,r^2x$. Since $x+rx+r^2x=42$, $xneq 0$. Also,
$$r^2x-x=(r^2-1)x=5(rx-x)=5x(r-1) implies r^2-1=5(r-1) implies (r-1)(r-4)=0.$$
Therefore, $r=1$ or $r=4$. Now use $x+rx+r^2x=42$ to find the sequence.
answered Jan 27 at 17:35
Math LoverMath Lover
14.1k31437
$endgroup$
add a comment |
$begingroup$
Let your numbers be $x,rx,r^2x$. Since $x+rx+r^2x=42$, $xneq 0$. Also,
$$r^2x-x=(r^2-1)x=5(rx-x)=5x(r-1) implies r^2-1=5(r-1) implies (r-1)(r-4)=0.$$
Therefore, $r=1$ or $r=4$. Now use $x+rx+r^2x=42$ to find the sequence.
answered Jan 27 at 17:35
Math LoverMath Lover
14.1k31437
$endgroup$
add a comment |
$begingroup$
Let your numbers be $x,rx,r^2x$. Since $x+rx+r^2x=42$, $xneq 0$. Also,
$$r^2x-x=(r^2-1)x=5(rx-x)=5x(r-1) implies r^2-1=5(r-1) implies (r-1)(r-4)=0.$$
Therefore, $r=1$ or $r=4$. Now use $x+rx+r^2x=42$ to find the sequence.
answered Jan 27 at 17:35
Math LoverMath Lover
14.1k31437
$endgroup$
Let your numbers be $x,rx,r^2x$. Since $x+rx+r^2x=42$, $xneq 0$. Also,
$$r^2x-x=(r^2-1)x=5(rx-x)=5x(r-1) implies r^2-1=5(r-1) implies (r-1)(r-4)=0.$$
Therefore, $r=1$ or $r=4$. Now use $x+rx+r^2x=42$ to find the sequence.
answered Jan 27 at 17:35
Math LoverMath Lover
14.1k31437
answered Jan 27 at 17:35
Math LoverMath Lover
14.1k31437
answered Jan 27 at 17:35
Math LoverMath Lover
14.1k31437
answered Jan 27 at 17:35
Math LoverMath Lover
14.1k31437
14.1k31437
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3089833%2fhow-to-solve-for-x-when-function-cant-be-inverted%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
