square root system of equations
I have a system of equations as follows that I need to solve for $x$:
$$
sqrt{1 - x^2} + sqrt{4 - x^2} = z\
sqrt{4 - y^2} + sqrt{9 - y^2} = z
$$
Originally, I was trying to put $x$ in terms of $z,$ but I’m at a loss how I would solve this system. Trying to solve it just makes it more complicated, no matter if I multiply by the reciprocal or square both sides.
algebra-precalculus systems-of-equations
|
show 2 more comments
I have a system of equations as follows that I need to solve for $x$:
$$
sqrt{1 - x^2} + sqrt{4 - x^2} = z\
sqrt{4 - y^2} + sqrt{9 - y^2} = z
$$
Originally, I was trying to put $x$ in terms of $z,$ but I’m at a loss how I would solve this system. Trying to solve it just makes it more complicated, no matter if I multiply by the reciprocal or square both sides.
algebra-precalculus systems-of-equations
1
Are you looking for real or integer or complex solutions?
– gt6989b
Dec 4 '18 at 16:11
I am looking for real/integer solutions.
– PLZHELP
Dec 4 '18 at 16:12
You ask to solve for $z$, but the system is already solved for $z$ in terms of both $x$ and $y$ -- so what are you asking for exactly?
– gt6989b
Dec 4 '18 at 16:12
If you want integer solutions, it's clear that all four surds must give integer results; there aren't many possiblities.
– rogerl
Dec 4 '18 at 16:14
I guess I’m looking for x or y. I edited the question.
– PLZHELP
Dec 4 '18 at 16:14
|
show 2 more comments
I have a system of equations as follows that I need to solve for $x$:
$$
sqrt{1 - x^2} + sqrt{4 - x^2} = z\
sqrt{4 - y^2} + sqrt{9 - y^2} = z
$$
Originally, I was trying to put $x$ in terms of $z,$ but I’m at a loss how I would solve this system. Trying to solve it just makes it more complicated, no matter if I multiply by the reciprocal or square both sides.
algebra-precalculus systems-of-equations
I have a system of equations as follows that I need to solve for $x$:
$$
sqrt{1 - x^2} + sqrt{4 - x^2} = z\
sqrt{4 - y^2} + sqrt{9 - y^2} = z
$$
Originally, I was trying to put $x$ in terms of $z,$ but I’m at a loss how I would solve this system. Trying to solve it just makes it more complicated, no matter if I multiply by the reciprocal or square both sides.
algebra-precalculus systems-of-equations
algebra-precalculus systems-of-equations
edited Dec 4 '18 at 16:52
amWhy
192k28225439
192k28225439
asked Dec 4 '18 at 16:09
PLZHELP
162
162
1
Are you looking for real or integer or complex solutions?
– gt6989b
Dec 4 '18 at 16:11
I am looking for real/integer solutions.
– PLZHELP
Dec 4 '18 at 16:12
You ask to solve for $z$, but the system is already solved for $z$ in terms of both $x$ and $y$ -- so what are you asking for exactly?
– gt6989b
Dec 4 '18 at 16:12
If you want integer solutions, it's clear that all four surds must give integer results; there aren't many possiblities.
– rogerl
Dec 4 '18 at 16:14
I guess I’m looking for x or y. I edited the question.
– PLZHELP
Dec 4 '18 at 16:14
|
show 2 more comments
1
Are you looking for real or integer or complex solutions?
– gt6989b
Dec 4 '18 at 16:11
I am looking for real/integer solutions.
– PLZHELP
Dec 4 '18 at 16:12
You ask to solve for $z$, but the system is already solved for $z$ in terms of both $x$ and $y$ -- so what are you asking for exactly?
– gt6989b
Dec 4 '18 at 16:12
If you want integer solutions, it's clear that all four surds must give integer results; there aren't many possiblities.
– rogerl
Dec 4 '18 at 16:14
I guess I’m looking for x or y. I edited the question.
– PLZHELP
Dec 4 '18 at 16:14
1
1
Are you looking for real or integer or complex solutions?
– gt6989b
Dec 4 '18 at 16:11
Are you looking for real or integer or complex solutions?
– gt6989b
Dec 4 '18 at 16:11
I am looking for real/integer solutions.
– PLZHELP
Dec 4 '18 at 16:12
I am looking for real/integer solutions.
– PLZHELP
Dec 4 '18 at 16:12
You ask to solve for $z$, but the system is already solved for $z$ in terms of both $x$ and $y$ -- so what are you asking for exactly?
– gt6989b
Dec 4 '18 at 16:12
You ask to solve for $z$, but the system is already solved for $z$ in terms of both $x$ and $y$ -- so what are you asking for exactly?
– gt6989b
Dec 4 '18 at 16:12
If you want integer solutions, it's clear that all four surds must give integer results; there aren't many possiblities.
– rogerl
Dec 4 '18 at 16:14
If you want integer solutions, it's clear that all four surds must give integer results; there aren't many possiblities.
– rogerl
Dec 4 '18 at 16:14
I guess I’m looking for x or y. I edited the question.
– PLZHELP
Dec 4 '18 at 16:14
I guess I’m looking for x or y. I edited the question.
– PLZHELP
Dec 4 '18 at 16:14
|
show 2 more comments
2 Answers
2
active
oldest
votes
Without loss of generality, we can choose to deal only with positive $x,y,z$ values. Notice that $yle2$ from the first radical in the second equation. Then $y$ varies between $0$ and $2$, so $z$ varies between $sqrt 5$ and $5$. Similarly, in the first equation $xin[0,1]$ so $zin [sqrt 3,3]$. Combining the two ranges for $z$ means a possible solution can be found if $zin [sqrt 5,3]$.
Let's focus now on the first equation, and do the substitution $u^2=1-x^2$, with $uge 0$. In fact $uin [0,1]$. We can then rewite the first equation as $$u+sqrt{3+u}=z$$
Leaving just the square root on the left, and squaring the equation, you get $$3+u=(z-u)^2$$
This is a simple quadratic in $u$, which you should be able to solve. Choose the solution that is in the $[0,1]$ interval.
Wouldn’t of thought of that, thanks
– PLZHELP
Dec 4 '18 at 17:04
What is $u$ representing
– PLZHELP
Dec 4 '18 at 17:06
Nothing. I just wanted a different name for the variable. My first idea was $alpha$, but then I decided that it's too much typing
– Andrei
Dec 4 '18 at 18:37
add a comment |
Both the $x$ and $y$ equations have the form
$$sqrt{a^2-h^2} + sqrt{b^2 - h^2} = c$$
For everything to be real, we need $sqrt{|a^2-b^2|} le c le a + b$.
Assume this is satisfied. For positive solution of $h$, this equation has a geometric interpretation. It is the equation for the height $h$ of a triangle of side $a,b,c$ with respect to side $c$. We can use
Heron's formula to figure out the area $Delta$ of the triangle and then
$$begin{align}h = frac{2Delta}{c}
&= frac{1}{2c}sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}\
&= frac{1}{2c}sqrt{4a^2b^2 - (c^2-a^2-b^2)^2}end{align}$$
Apply this to the equations at hand. For positive $x$ and $y$ solutions, we obtain
$$begin{align}
sqrt{1 - x^2} + sqrt{4-x^2} = z &implies
x = frac{1}{2z}sqrt{16 - (z^2-5)^2}\
sqrt{4 - y^2} + sqrt{9-y^2} = z &implies
y = frac{1}{2z}sqrt{144 - (z^2 - 13)^2}
end{align}$$
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%2f3025753%2fsquare-root-system-of-equations%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Without loss of generality, we can choose to deal only with positive $x,y,z$ values. Notice that $yle2$ from the first radical in the second equation. Then $y$ varies between $0$ and $2$, so $z$ varies between $sqrt 5$ and $5$. Similarly, in the first equation $xin[0,1]$ so $zin [sqrt 3,3]$. Combining the two ranges for $z$ means a possible solution can be found if $zin [sqrt 5,3]$.
Let's focus now on the first equation, and do the substitution $u^2=1-x^2$, with $uge 0$. In fact $uin [0,1]$. We can then rewite the first equation as $$u+sqrt{3+u}=z$$
Leaving just the square root on the left, and squaring the equation, you get $$3+u=(z-u)^2$$
This is a simple quadratic in $u$, which you should be able to solve. Choose the solution that is in the $[0,1]$ interval.
Wouldn’t of thought of that, thanks
– PLZHELP
Dec 4 '18 at 17:04
What is $u$ representing
– PLZHELP
Dec 4 '18 at 17:06
Nothing. I just wanted a different name for the variable. My first idea was $alpha$, but then I decided that it's too much typing
– Andrei
Dec 4 '18 at 18:37
add a comment |
Without loss of generality, we can choose to deal only with positive $x,y,z$ values. Notice that $yle2$ from the first radical in the second equation. Then $y$ varies between $0$ and $2$, so $z$ varies between $sqrt 5$ and $5$. Similarly, in the first equation $xin[0,1]$ so $zin [sqrt 3,3]$. Combining the two ranges for $z$ means a possible solution can be found if $zin [sqrt 5,3]$.
Let's focus now on the first equation, and do the substitution $u^2=1-x^2$, with $uge 0$. In fact $uin [0,1]$. We can then rewite the first equation as $$u+sqrt{3+u}=z$$
Leaving just the square root on the left, and squaring the equation, you get $$3+u=(z-u)^2$$
This is a simple quadratic in $u$, which you should be able to solve. Choose the solution that is in the $[0,1]$ interval.
Wouldn’t of thought of that, thanks
– PLZHELP
Dec 4 '18 at 17:04
What is $u$ representing
– PLZHELP
Dec 4 '18 at 17:06
Nothing. I just wanted a different name for the variable. My first idea was $alpha$, but then I decided that it's too much typing
– Andrei
Dec 4 '18 at 18:37
add a comment |
Without loss of generality, we can choose to deal only with positive $x,y,z$ values. Notice that $yle2$ from the first radical in the second equation. Then $y$ varies between $0$ and $2$, so $z$ varies between $sqrt 5$ and $5$. Similarly, in the first equation $xin[0,1]$ so $zin [sqrt 3,3]$. Combining the two ranges for $z$ means a possible solution can be found if $zin [sqrt 5,3]$.
Let's focus now on the first equation, and do the substitution $u^2=1-x^2$, with $uge 0$. In fact $uin [0,1]$. We can then rewite the first equation as $$u+sqrt{3+u}=z$$
Leaving just the square root on the left, and squaring the equation, you get $$3+u=(z-u)^2$$
This is a simple quadratic in $u$, which you should be able to solve. Choose the solution that is in the $[0,1]$ interval.
Without loss of generality, we can choose to deal only with positive $x,y,z$ values. Notice that $yle2$ from the first radical in the second equation. Then $y$ varies between $0$ and $2$, so $z$ varies between $sqrt 5$ and $5$. Similarly, in the first equation $xin[0,1]$ so $zin [sqrt 3,3]$. Combining the two ranges for $z$ means a possible solution can be found if $zin [sqrt 5,3]$.
Let's focus now on the first equation, and do the substitution $u^2=1-x^2$, with $uge 0$. In fact $uin [0,1]$. We can then rewite the first equation as $$u+sqrt{3+u}=z$$
Leaving just the square root on the left, and squaring the equation, you get $$3+u=(z-u)^2$$
This is a simple quadratic in $u$, which you should be able to solve. Choose the solution that is in the $[0,1]$ interval.
answered Dec 4 '18 at 16:32
Andrei
11.4k21026
11.4k21026
Wouldn’t of thought of that, thanks
– PLZHELP
Dec 4 '18 at 17:04
What is $u$ representing
– PLZHELP
Dec 4 '18 at 17:06
Nothing. I just wanted a different name for the variable. My first idea was $alpha$, but then I decided that it's too much typing
– Andrei
Dec 4 '18 at 18:37
add a comment |
Wouldn’t of thought of that, thanks
– PLZHELP
Dec 4 '18 at 17:04
What is $u$ representing
– PLZHELP
Dec 4 '18 at 17:06
Nothing. I just wanted a different name for the variable. My first idea was $alpha$, but then I decided that it's too much typing
– Andrei
Dec 4 '18 at 18:37
Wouldn’t of thought of that, thanks
– PLZHELP
Dec 4 '18 at 17:04
Wouldn’t of thought of that, thanks
– PLZHELP
Dec 4 '18 at 17:04
What is $u$ representing
– PLZHELP
Dec 4 '18 at 17:06
What is $u$ representing
– PLZHELP
Dec 4 '18 at 17:06
Nothing. I just wanted a different name for the variable. My first idea was $alpha$, but then I decided that it's too much typing
– Andrei
Dec 4 '18 at 18:37
Nothing. I just wanted a different name for the variable. My first idea was $alpha$, but then I decided that it's too much typing
– Andrei
Dec 4 '18 at 18:37
add a comment |
Both the $x$ and $y$ equations have the form
$$sqrt{a^2-h^2} + sqrt{b^2 - h^2} = c$$
For everything to be real, we need $sqrt{|a^2-b^2|} le c le a + b$.
Assume this is satisfied. For positive solution of $h$, this equation has a geometric interpretation. It is the equation for the height $h$ of a triangle of side $a,b,c$ with respect to side $c$. We can use
Heron's formula to figure out the area $Delta$ of the triangle and then
$$begin{align}h = frac{2Delta}{c}
&= frac{1}{2c}sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}\
&= frac{1}{2c}sqrt{4a^2b^2 - (c^2-a^2-b^2)^2}end{align}$$
Apply this to the equations at hand. For positive $x$ and $y$ solutions, we obtain
$$begin{align}
sqrt{1 - x^2} + sqrt{4-x^2} = z &implies
x = frac{1}{2z}sqrt{16 - (z^2-5)^2}\
sqrt{4 - y^2} + sqrt{9-y^2} = z &implies
y = frac{1}{2z}sqrt{144 - (z^2 - 13)^2}
end{align}$$
add a comment |
Both the $x$ and $y$ equations have the form
$$sqrt{a^2-h^2} + sqrt{b^2 - h^2} = c$$
For everything to be real, we need $sqrt{|a^2-b^2|} le c le a + b$.
Assume this is satisfied. For positive solution of $h$, this equation has a geometric interpretation. It is the equation for the height $h$ of a triangle of side $a,b,c$ with respect to side $c$. We can use
Heron's formula to figure out the area $Delta$ of the triangle and then
$$begin{align}h = frac{2Delta}{c}
&= frac{1}{2c}sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}\
&= frac{1}{2c}sqrt{4a^2b^2 - (c^2-a^2-b^2)^2}end{align}$$
Apply this to the equations at hand. For positive $x$ and $y$ solutions, we obtain
$$begin{align}
sqrt{1 - x^2} + sqrt{4-x^2} = z &implies
x = frac{1}{2z}sqrt{16 - (z^2-5)^2}\
sqrt{4 - y^2} + sqrt{9-y^2} = z &implies
y = frac{1}{2z}sqrt{144 - (z^2 - 13)^2}
end{align}$$
add a comment |
Both the $x$ and $y$ equations have the form
$$sqrt{a^2-h^2} + sqrt{b^2 - h^2} = c$$
For everything to be real, we need $sqrt{|a^2-b^2|} le c le a + b$.
Assume this is satisfied. For positive solution of $h$, this equation has a geometric interpretation. It is the equation for the height $h$ of a triangle of side $a,b,c$ with respect to side $c$. We can use
Heron's formula to figure out the area $Delta$ of the triangle and then
$$begin{align}h = frac{2Delta}{c}
&= frac{1}{2c}sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}\
&= frac{1}{2c}sqrt{4a^2b^2 - (c^2-a^2-b^2)^2}end{align}$$
Apply this to the equations at hand. For positive $x$ and $y$ solutions, we obtain
$$begin{align}
sqrt{1 - x^2} + sqrt{4-x^2} = z &implies
x = frac{1}{2z}sqrt{16 - (z^2-5)^2}\
sqrt{4 - y^2} + sqrt{9-y^2} = z &implies
y = frac{1}{2z}sqrt{144 - (z^2 - 13)^2}
end{align}$$
Both the $x$ and $y$ equations have the form
$$sqrt{a^2-h^2} + sqrt{b^2 - h^2} = c$$
For everything to be real, we need $sqrt{|a^2-b^2|} le c le a + b$.
Assume this is satisfied. For positive solution of $h$, this equation has a geometric interpretation. It is the equation for the height $h$ of a triangle of side $a,b,c$ with respect to side $c$. We can use
Heron's formula to figure out the area $Delta$ of the triangle and then
$$begin{align}h = frac{2Delta}{c}
&= frac{1}{2c}sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}\
&= frac{1}{2c}sqrt{4a^2b^2 - (c^2-a^2-b^2)^2}end{align}$$
Apply this to the equations at hand. For positive $x$ and $y$ solutions, we obtain
$$begin{align}
sqrt{1 - x^2} + sqrt{4-x^2} = z &implies
x = frac{1}{2z}sqrt{16 - (z^2-5)^2}\
sqrt{4 - y^2} + sqrt{9-y^2} = z &implies
y = frac{1}{2z}sqrt{144 - (z^2 - 13)^2}
end{align}$$
answered Dec 4 '18 at 17:31
achille hui
95.6k5130257
95.6k5130257
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3025753%2fsquare-root-system-of-equations%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
1
Are you looking for real or integer or complex solutions?
– gt6989b
Dec 4 '18 at 16:11
I am looking for real/integer solutions.
– PLZHELP
Dec 4 '18 at 16:12
You ask to solve for $z$, but the system is already solved for $z$ in terms of both $x$ and $y$ -- so what are you asking for exactly?
– gt6989b
Dec 4 '18 at 16:12
If you want integer solutions, it's clear that all four surds must give integer results; there aren't many possiblities.
– rogerl
Dec 4 '18 at 16:14
I guess I’m looking for x or y. I edited the question.
– PLZHELP
Dec 4 '18 at 16:14