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Algebra II Equations Containing Radicals

  1. Jun 16, 2012 #1
    More fun with radicals!

    ...As for the problem, I tend to get problems with this same form (but with slight differences such as two separate binomials underneath individual square roots) wrong. I'm not quite sure what I'm doing wrong, so I was hoping for some help. :D

    1. The problem statement, all variables and given/known data
    Solve. If an equation has no real solution, say so.

    [tex] \sqrt {y} + \sqrt {y + 5} = 5 [/tex]

    Answer: [itex] 4 [/itex]


    2. Relevant equations

    --

    3. The attempt at a solution

    The first thing I have been taught to do is to isolate the radicals on one side. That's already done.

    Next, square both sides to eliminate radicals. I have a feeling that something I'm doing here is incorrect (?)

    [itex] (\sqrt {y} + \sqrt {y + 5})^2 = 5^2 [/itex]
    [itex] y +y + 5 = 25 [/itex]

    Solve for y.

    [itex] 2y + 5 = 25 [/itex]
    [itex] 2y = 20 [/itex]
    [itex] y = 10 [/itex]

    But, since you often have the possibility of extraneous roots; you have to test it out in the original formula. When I do that, I get 'no solution.'

    [itex] \sqrt {10} + \sqrt {10 + 5} \stackrel{?}{=} 5 [/itex]
    [itex] \sqrt {10} + \sqrt {15} \stackrel {?} {=} 5 [/itex]

    Approximate values from here on:

    [itex] 3.16 + 3.87 \stackrel {?} {\approx} 5 [/itex]
    [itex] 7.03 \neq 5 [/itex]

    So, how'd I mess it up this time? (*joking*) Thanks in advance for your help!
     
  2. jcsd
  3. Jun 16, 2012 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    This is an error. [itex](a+ b)^2= a^2+ 2ab+ b^2[/itex], NOT "[itex]a^2+ b^2[/itex]".
    [itex](\sqrt{y}+ \sqrt{y+ 5})^2= (\sqrt{y})^2+ 2\sqrt{y(y+ 5)}+ (\sqrt{y+5})^2= y+ 2\sqrt{y^2+ 5y}+ y+ 5= 25[/itex]
    Now, "isolate" that remaining square root and square again:
    [itex]\sqrt{y^2+ 5y}= 10-y[/itex]
    [itex](y^2+ 5y)= (10+ y)^2= 100+ 20y+ y^2[/itex]

    Solve that for y.

     
  4. Jun 16, 2012 #3

    Mark44

    Staff: Mentor

    When you have an equation with two radicals, it's often more efficient to move them to the opposite sides of the equation.

    ## \sqrt{y} = 5 - \sqrt{y + 5}##
    Now square both sides:
    y = (5 - ##\sqrt{y + 5}##)2 = 25 - 10##\sqrt{y + 5}## + y + 5
    => 0 = 30 - 10##\sqrt{y + 5}##

    This equation is easy to solve for y.
     
  5. Jun 16, 2012 #4
    Here's another method to consider. Multiply both sides of the equation by

    [itex]\sqrt{y+5}[/itex] - [itex]\sqrt{}y[/itex]

    You get

    [itex]\sqrt{y+5}[/itex] - [itex]\sqrt{}y[/itex] = 1

    If you subtract this equation from the original equation, you get

    2[itex]\sqrt{}y[/itex] = 4
     
  6. Jun 18, 2012 #5
    Okay, thanks to everyone's replies I think I see several different ways of going about solving the problem. I have a couple other problems that I happened to get wrong with the same form, so I'm going to try this and see if I can solve them on my own now, but first, I'd like to make sure that I have it down right in the first place.

    Is this the correct form? (I used Mark44's suggestion for this one):

    [itex] 0 = 30 - 10\sqrt {y + 5} [/itex]
    [itex] -30 = - 10 \sqrt {y + 5} [/itex]
    [itex] 3 = \sqrt {y + 5} [/itex]
    [itex] 3^2 = (\sqrt {y + 5})^2 [/itex]
    [itex] 9 = y + 5 [/itex]
    [itex] 4 = y [/itex]

    And checking it for extraneous roots:

    [itex] \sqrt {4} + \sqrt {4 + 5} = 5 [/itex]
    [itex] 2 + \sqrt {9} = 5 [/itex]
    [itex] 2 + 3 = 5 [/itex]

    @HallsofIvy: Thank you for literally spelling it out for me. I had easier problems earlier in my assignment with the (a + b)^2 form, and it took me a few tries to recognize it, but this one was really sneaky. Instead of thinking of it as, [itex] (a + b)^2 [/itex] or [itex] (\sqrt {y} + \sqrt {y + 5})^2 [/itex] as you put it, I was actually thinking of it as: [itex] (\sqrt {y})^2 + (\sqrt {y + 5})^2 [/itex]. I'll have to watch out for this!!

    @Chestermiller: Wow. I wouldn't have even thought of that using a conjugate to multiply and then taking that equation and subtracting from the original. Heh. And I almost wanted to correct you and say it was [itex] \sqrt {y} - \sqrt {y + 5} [/itex], but thank goodness I caught the sign change that comes from subtracting. Thank you for another method. :D
     
    Last edited: Jun 18, 2012
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