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System of two equations with radical expression

  1. Aug 7, 2012 #1
    1. The problem statement, all variables and given/known data

    I am trying to solve the following two systems of equations and at every attempt I get completely stuck.

    x2 + y2 = 1

    -x + √3y = 0

    2. Relevant equations
    not applicable

    3. The attempt at a solution

    the answers have already been given to me but I have no idea how to get the answers: x = (√3)/2, -(√3)/2
    y = 1/2, -1/2
  2. jcsd
  3. Aug 7, 2012 #2


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    Using either equation, you can solve for y in terms of x (or for x in terms of y). Then, substituting that expression for y (or x) into the other equation, you are now left with an equation with only one unknown variable. I think you know how to solve that!
  4. Aug 7, 2012 #3
    Yes I realize in a very general sense how to solve a system of equations. I was trying to use the substitution method but this equation seem so unusual that I cannot figure out how to do it. I thought I had it nearly solved but my answer was nowhere close to what I know is the right answer. It's the radical expression that is throwing me off.
  5. Aug 7, 2012 #4


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    It's not so bad. I would start by solving the 2nd equation for x, and then substitute that into the first. Try it, and post your workings here.

    By the way, in the second equation, is it
    [itex]-x + \sqrt{3} \cdot y = 0[/itex]
    [itex]-x + \sqrt{3y} = 0[/itex]?
  6. Aug 7, 2012 #5


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    What the OP wrote was ##\sqrt{3} \cdot y##, but that's not necessarily what was intended.
  7. Aug 7, 2012 #6
    Show your attempt, maybe we can help you figure out what went wrong.

    Based on the answer he gave, I would think that it has to be the first equation that you listed.
  8. Aug 7, 2012 #7
    Yes it is √3 (multiplied by) y. I couldn't tell whether the variable was underneath the radical sign based on how it was written in the textbook. I plugged in the answers they gave me to make sure. By the way, how do I insert the multiplication dot symbol?
  9. Aug 7, 2012 #8
    If you're using latex, you type \cdot.

    Otherwise you can't, but in order to avoid confusion you can write (√3)y to illustrate the difference.

    But before we can help you figure out the problem, you need to show the work you already did.
  10. Aug 9, 2012 #9
    I was finally able to get the solution by substituting the second equation into the first. I was making things much too complicated and it was actually quite simple to get the solution. But I have challenged myself to find the solution by substituting the first equation into the second one, and once again I'm really stuck. I think I actually did it correctly once, but now have no idea how I did that. When I try to isolate the y in the first equation I get y = √1 - x . However, when I plug that into the second equation then suddenly I have the difficulty of multiplying √3 by (√1 - x). How am I supposed to accomplish that, or am I already on the wrong track?
  11. Aug 9, 2012 #10


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    $$\sqrt{a} \cdot \sqrt{b + c} = \sqrt{a(b+c)}$$
    This is just a rule, but if you think about it, it is consistent with the rules of exponents. Just interpret sqrt(a) as a1/2 etc. and you'll see why the above must be true if the rules of exponents are true.
  12. Aug 9, 2012 #11


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    NO. You get ##y = \sqrt{1 - x^2}## if you do it correctly
  13. Aug 9, 2012 #12

    Ray Vickson

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    No, you get [itex] y = \pm \sqrt{1-x^2}[/itex] if you do it properly.

  14. Aug 9, 2012 #13


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  15. Aug 9, 2012 #14
    Wouldn't y^2 = √1 - x^2 be the same thing as y = (√1) - x?
  16. Aug 9, 2012 #15
    There is no rule that says ##\sqrt{a+b} = \sqrt{a} + \sqrt{b}##.
  17. Aug 9, 2012 #16
    You're right. I didn't really mean to say that. What I think it should be is y = 1 - x.
  18. Aug 9, 2012 #17
    But wasn't it established already that ##y=\pm\sqrt{1-x^2}##, or do you not agree with how it got to that?
  19. Aug 9, 2012 #18
    It started out as y^2 = 1 - x^2. If you try to remove the exponent for y then you should also remove the exponent from x, but then you would also end up with √1.
  20. Aug 9, 2012 #19


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    NO! You can't do that. You started with this:
    [itex]y^2 = 1 - x^2[/itex]
    You take the square root of both sides (to isolate the y) and you get this:
    [itex]y = \pm\sqrt{1 - x^2}[/itex]
    The following are not necessarily equal:
    [itex]\sqrt{1 - x^2} \ne 1 - x[/itex]

    You can't just take the square root of each term underneath the square root if you have an addition or subtraction. Does
    [itex]\sqrt{25 - 9} = 5 - 3[/itex]?
  21. Aug 9, 2012 #20
    So you want to take the square root of each side.
    If ##y^2 = 1 - x^2## then ##\sqrt{y^2} = \sqrt{1 - x^2}##, which is only the same as ##y = 1 - x## if ##\sqrt{a + b} = \sqrt{a} + \sqrt{b}##.
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