square root system of equations












3















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







share|cite|improve this question




















  • 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
















3















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







share|cite|improve this question




















  • 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














3












3








3


2






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







share|cite|improve this question
















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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














  • 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










2 Answers
2






active

oldest

votes


















2














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.






share|cite|improve this answer





















  • 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



















0














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



$$begin{align}
sqrt{1 - x^2} + sqrt{4-x^2} = z &implies
x = frac{1}{2z}sqrt{16 - (z^2-5)^2}\
sqrt{4 - y^2} + sqrt{9-y^2} = z &implies
y = frac{1}{2z}sqrt{144 - (z^2 - 13)^2}
end{align}$$






share|cite|improve this answer





















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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    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.






    share|cite|improve this answer





















    • 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
















    2














    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.






    share|cite|improve this answer





















    • 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














    2












    2








    2






    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.






    share|cite|improve this answer












    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.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    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


















    • 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











    0














    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



    $$begin{align}
    sqrt{1 - x^2} + sqrt{4-x^2} = z &implies
    x = frac{1}{2z}sqrt{16 - (z^2-5)^2}\
    sqrt{4 - y^2} + sqrt{9-y^2} = z &implies
    y = frac{1}{2z}sqrt{144 - (z^2 - 13)^2}
    end{align}$$






    share|cite|improve this answer


























      0














      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



      $$begin{align}
      sqrt{1 - x^2} + sqrt{4-x^2} = z &implies
      x = frac{1}{2z}sqrt{16 - (z^2-5)^2}\
      sqrt{4 - y^2} + sqrt{9-y^2} = z &implies
      y = frac{1}{2z}sqrt{144 - (z^2 - 13)^2}
      end{align}$$






      share|cite|improve this answer
























        0












        0








        0






        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



        $$begin{align}
        sqrt{1 - x^2} + sqrt{4-x^2} = z &implies
        x = frac{1}{2z}sqrt{16 - (z^2-5)^2}\
        sqrt{4 - y^2} + sqrt{9-y^2} = z &implies
        y = frac{1}{2z}sqrt{144 - (z^2 - 13)^2}
        end{align}$$






        share|cite|improve this answer












        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



        $$begin{align}
        sqrt{1 - x^2} + sqrt{4-x^2} = z &implies
        x = frac{1}{2z}sqrt{16 - (z^2-5)^2}\
        sqrt{4 - y^2} + sqrt{9-y^2} = z &implies
        y = frac{1}{2z}sqrt{144 - (z^2 - 13)^2}
        end{align}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 4 '18 at 17:31









        achille hui

        95.6k5130257




        95.6k5130257






























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            I want to find a topological embedding $f : X rightarrow Y$ and $g: Y rightarrow X$, yet $X$ is not...

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            0 0 $begingroup$ This is related to a problem of showing that none of the intervals $(0,1], (0,1), [0,1]$ are homeomorphic to another, since it's in the same problem block. I've tried using $f(x) = 1 - x , f(x) = 1/(x-1)$ between varying pairs of those intervals to no avail. Do I need a function as complicated as the topologist's sine to show this? general-topology connectedness share | cite | improve this question edited Dec 10 '18 at 3:46 user1101010 775 1 7 30 asked Sep 24 '13 at 23:15
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