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horsegirl27

  • one year ago

Please help me out, will medal!! I'm really confused. It's creating a radical equation. I do not understand radical equations at all.

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  1. horsegirl27
    • one year ago
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    http://prntscr.com/7exg84

  2. horsegirl27
    • one year ago
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    Could someone explain radical equations, then help me make 2 for this question? And what does extraneous mean?

  3. horsegirl27
    • one year ago
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    @phi @ParthKohli @mathmate

  4. anonymous
    • one year ago
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    1) Isolate the radical symbol on one side of the equation 2) Square both sides of the equation to eliminate the radical symbol 3) Solve the equation that comes out after the squaring process 4) Check your answers with the original equation to avoid extraneous values

  5. horsegirl27
    • one year ago
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    How do I isolate it?

  6. phi
    • one year ago
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    I think I would start by picking a number and squaring it. for example 5 and 5*5= 25 and write \[ \sqrt{25} = 5 \]

  7. horsegirl27
    • one year ago
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    Okay.

  8. phi
    • one year ago
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    they want \[ a \sqrt{x+b} + c = d\] so far we have \[ \sqrt{25} = 5\] we could write 25 as the sum of two numbers can you pick a pair of numbers that add up to 25 ?

  9. horsegirl27
    • one year ago
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    16 and 9?

  10. phi
    • one year ago
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    ok. now let x be one of the numbers (which we pretend we don't know) and b be the other number. example: let x be 16 then we can write \[ \sqrt{x+9}= 5 \] (and we know x=16 would make this true) now we want to multiply by "a" \[ a\sqrt{x+9}= a\cdot 5 \] we can pick any number we want for a pick a number for a

  11. horsegirl27
    • one year ago
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    3

  12. phi
    • one year ago
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    If it is not clear, we are creating a complicated equation that matches what they want \[ a \sqrt{x+b} + c = d \]

  13. horsegirl27
    • one year ago
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    I understand how we're replacing numbers for the letters

  14. phi
    • one year ago
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    so if we pick a to be 3, then \[ a\sqrt{x+9}= a\cdot 5 \\ 3\sqrt{x+9}= 3\cdot 5 \\ 3\sqrt{x+9}=15\]

  15. phi
    • one year ago
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    almost done. they want us to add a number "c" to the left side (and to keep things equal) we do the same to the right side \[ 3\sqrt{x+9} + c=15+c \]

  16. horsegirl27
    • one year ago
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    okay, should we use any number for c?

  17. phi
    • one year ago
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    yes, we are making it up. It can be anything (but 0 would be cheating)

  18. horsegirl27
    • one year ago
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    Okay let's use 7.

  19. phi
    • one year ago
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    \[ 3\sqrt{x+9} + c=15+c \\ 3\sqrt{x+9} + 7=15+7 \\ 3\sqrt{x+9} + 7=22 \]

  20. phi
    • one year ago
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    I don't know if this has an extraneous solution but I do know x=16 works because that is how we started... which means when you solve it , you better get x=16 as an answer

  21. horsegirl27
    • one year ago
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    Do you know what extraneous means?

  22. phi
    • one year ago
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    Extraneous means "extra and wrong" to solve this equation we have to undo all the steps we used to make it. to do that, we square the radical , and we get a complicated equation that has more than 1 solution (one of them we do not want)

  23. phi
    • one year ago
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    Let's try to solve your equation \[ 3\sqrt{x+9} + 7=22 \] First add -7 to both sides

  24. horsegirl27
    • one year ago
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    that will take out +7 and change 22 to 15

  25. horsegirl27
    • one year ago
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    \[3\sqrt{x+9}=15 \]

  26. phi
    • one year ago
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    now divide both sides by 3

  27. horsegirl27
    • one year ago
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    how do I do that? I know 15 will become 5, but what about the left side?

  28. phi
    • one year ago
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    I guess you have to know how fractions work. With just numbers: \[ \frac{3 \cdot 2 }{3} = \frac{3}{3} \cdot 2 = 2 \] If the 2 were something else (such as \( \sqrt{x+9} \) it still works the same way

  29. horsegirl27
    • one year ago
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    ohhh ok

  30. phi
    • one year ago
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    so you get \[ \frac{3 \sqrt{x+9}}{3} = \frac{15}{3} \]

  31. phi
    • one year ago
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    which simplifies to ?

  32. horsegirl27
    • one year ago
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    Wait, after simplifying, if I divide 3 by 3, will that be changing it to 1?

  33. phi
    • one year ago
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    on the left , the 3/3 = 1 on the right the 15/3 = 5

  34. horsegirl27
    • one year ago
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    ok, I knew the right side

  35. phi
    • one year ago
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    yes, 3/3 =1 and 1 times anything is the anything

  36. horsegirl27
    • one year ago
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    So it will be \[1\sqrt{x+3}=5\] or is that wrong?

  37. phi
    • one year ago
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    btw, notice \[ \frac{15}{3} = \frac{3 \cdot 5}{3} = \frac{3}{3} \cdot 5 =1\cdot 5= 5 \]

  38. phi
    • one year ago
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    You can only divide one thing. if you have \[ \frac{3\cdot 3 \cdot 3}{3} = \frac{3}{3} \cdot 3 \cdot 3 \] notice you don't divide the other 3's. so in our problem the \(\sqrt{x+9} \) is not divided by 3 if that makes sense?

  39. horsegirl27
    • one year ago
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    Ohhhh yes it finally makes sense now!

  40. phi
    • one year ago
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    another way to keep it straight is to think of dividing as multiplying by the reciprocal in other words \[ \frac{3 \cdot 3 }{3} \text{ means } \frac{1}{3} \cdot 3 \cdot 3 \] (change divide by 3 into multiply by 1/3) it should be more clear you can do \[ \left( \frac{1}{3} \cdot 3 \right) \cdot 3 \]

  41. horsegirl27
    • one year ago
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    ok

  42. phi
    • one year ago
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    and definitely not \[ \left( \frac{1}{3} \cdot 3 \right) \cdot\left( \frac{1}{3} \cdot 3 \right) \]

  43. phi
    • one year ago
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    but back to the problem. what do you have so far?

  44. horsegirl27
    • one year ago
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    um I think \[1\sqrt{x+9}=5\]

  45. phi
    • one year ago
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    yes, but generally we don't bother to show the multiply by 1

  46. horsegirl27
    • one year ago
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    ok

  47. phi
    • one year ago
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    in other words we know 1 times \(\sqrt{x+9}\) is \(\sqrt{x+9}\) so we just write \[ \sqrt{x+9}=5 \]

  48. horsegirl27
    • one year ago
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    ok, that makes sense

  49. phi
    • one year ago
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    we could write (goofy) things like \[ \sqrt{x+9}= 1\cdot 5+ 0 \] but we get into the habit of writing as few symbols as possible

  50. horsegirl27
    • one year ago
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    Yeah, we don't need to write all of that

  51. phi
    • one year ago
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    to "get rid" of the radical operator, square both sides what do we get?

  52. horsegirl27
    • one year ago
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    well, 9 squared is 81, and 5 squared is 25

  53. phi
    • one year ago
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    you have \[ \sqrt{x+9}=5 \] "square both sides" means do \[ \left(\sqrt{x+9}\right)^2 =5^2\]

  54. horsegirl27
    • one year ago
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    ohhhhhhhhhhhhhhhhhhhhhhhh

  55. phi
    • one year ago
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    square means "multiply by itself one time" \[ \left(\sqrt{x+9}\right)\left(\sqrt{x+9}\right) =5 \cdot 5 \]

  56. horsegirl27
    • one year ago
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    that makes sense now lol

  57. phi
    • one year ago
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    by definition of square root \[ \sqrt{z} \cdot \sqrt{z} = \sqrt{z\cdot z}= z \] it does not matter how "complicated" z is (in our case (x+9) )

  58. horsegirl27
    • one year ago
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    ok

  59. phi
    • one year ago
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    or, just remember, squaring a square root undoes the square root

  60. phi
    • one year ago
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    examples: \[ \left( \sqrt{3x}\right)^2 = 3x \\ \left( \sqrt{x^2+y^2}\right)^2 = x^2 + y^2 \]

  61. horsegirl27
    • one year ago
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    that sort of makes sense

  62. phi
    • one year ago
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    It is a definition

  63. horsegirl27
    • one year ago
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    Okay, reading it a few times, I understand

  64. horsegirl27
    • one year ago
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    You square a square root so it's no longer a square root.

  65. phi
    • one year ago
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    yes

  66. horsegirl27
    • one year ago
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    ok, and then if I was squaring \[\sqrt{x+9}\] it will become just x+9, right?

  67. phi
    • one year ago
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    yes and because we have an equation we also square the right side to keep things equal

  68. horsegirl27
    • one year ago
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    ok, and what's that going to be? 25, since it wasn't in a square root?

  69. phi
    • one year ago
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    yes \[ \left(\sqrt{x+9}\right)^2 =5^2 \\ x+9= 25 \]

  70. horsegirl27
    • one year ago
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    Then I do -9, to get x=16

  71. horsegirl27
    • one year ago
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    -9 on both sides

  72. phi
    • one year ago
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    yes. so that means we have a radical equation \[ 3\sqrt{x+9} + 7=22 \] that does not have an extraneous solution (it has a "true" solution) to make an equation with an extraneous solution, start over with \[ \sqrt{25} = -5 \] notice if we square both sides we get \[ \left( \sqrt{25}\right)^2 = -5 \cdot -5 \\ 25 = 25 \] because -5 , along with +5, is a square root of 25

  73. horsegirl27
    • one year ago
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    ok

  74. horsegirl27
    • one year ago
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    So it doesn't matter if something is negative or positive to be a square root?

  75. phi
    • one year ago
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    a real number has two square roots, and you might see \[ \pm \sqrt{x} \] to remind us but to continue do the same thing as before let 25 be x+9 let a=3, and multiply both sides by 3

  76. phi
    • one year ago
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    \[ \sqrt{25} = -5 \\ \sqrt{x+9 } = -5\\3 \sqrt{x+9 } = -5\cdot 3 \]

  77. phi
    • one year ago
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    and add 7 to both sides to get \[3 \sqrt{x+9 } +7 = -8 \]

  78. horsegirl27
    • one year ago
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    ok

  79. phi
    • one year ago
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    can you solve this radical equation ?

  80. horsegirl27
    • one year ago
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    I think so, let me work on it

  81. horsegirl27
    • one year ago
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    So, first do -7 on both sides and get \[3\sqrt{x+9}=-15\] right? then divide by 3 the same way as before and get \[\sqrt{x+9}=-5\] and then square it to get x+9=25 again. Right?

  82. phi
    • one year ago
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    yes, looks good

  83. horsegirl27
    • one year ago
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    ok, so it's basically the same.

  84. phi
    • one year ago
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    yes, you get x=16 now "put in 16" for x in the original equation. what do you get ?

  85. horsegirl27
    • one year ago
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    the original equation being \[3\sqrt{x+9}+7=15?\]

  86. phi
    • one year ago
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    this one \[ 3 \sqrt{x+9 } +7 = -8 \]

  87. horsegirl27
    • one year ago
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    ok

  88. phi
    • one year ago
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    put in 16 for x on the left side, and simplify the left side. what do you get ?

  89. horsegirl27
    • one year ago
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    I'm not quite sure how I should simplify it

  90. phi
    • one year ago
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    the left side is \[ 3 \sqrt{x+9 } +7 \] replace x with 16: \[ 3 \sqrt{16+9 } +7 = -8 \] order of operations: do the stuff in side the radical first

  91. phi
    • one year ago
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    in other words add 16+9

  92. horsegirl27
    • one year ago
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    ok, that's 25.

  93. phi
    • one year ago
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    yes, you now have \[ 3 \sqrt{25 } +7 \] now do the square root next

  94. horsegirl27
    • one year ago
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    5

  95. phi
    • one year ago
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    ok, so it's now \[ 3 \cdot 5 +7 \] order of operations: multiply before add

  96. horsegirl27
    • one year ago
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    15+7 22

  97. phi
    • one year ago
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    yes, and we just found \[ 3 \sqrt{x+9 } +7 = -8 \] with x= 16 becomes \[ 22 = -8 \]

  98. phi
    • one year ago
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    and that is just goofy. in other words, x=16 is an "extraneous solution" (it is not a correct solution)

  99. horsegirl27
    • one year ago
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    ok

  100. phi
    • one year ago
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    the problem happens because when we take the square root of 25 we always choose the "principle root" +5, and that leads to 22 on the left side if we chose -5 as the square root of 25, we would get -8 = -8 But (for whatever reason, I am not sure why), when evaluating a radical expression we only choose the positive square root, unless there is a \( \pm\) sign in front of the square root.

  101. horsegirl27
    • one year ago
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    ok

  102. phi
    • one year ago
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    You are now done with this question

  103. horsegirl27
    • one year ago
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    http://prntscr.com/7eygkr that's the next part, but we already finished that in checking it and you explained what extraneous means

  104. horsegirl27
    • one year ago
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    thank you so much for helping me!!!!

  105. phi
    • one year ago
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    ok, but be careful. They want the first equation to be the one with the extraneous solution. we did it the other way round. so use \[ \text{ Eq 1 } 3 \sqrt{x+9 } +7 = -8 \\ \text{ Eq 2 } 3 \sqrt{x+9 } +7 = 22 \]

  106. horsegirl27
    • one year ago
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    yup, I noticed that

  107. horsegirl27
    • one year ago
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    thanks again, I finally understand radical equations!!

  108. phi
    • one year ago
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    yw

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