anonymous
  • anonymous
Please help with Algebra will award medal
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Task 2 Part 1. Create two radical equations: one that has an extraneous solution, and one that does not have an extraneous solution. Use the equation below as a model. a√x+b+c=d Use a constant in place of each variable a, b, c, and d. You can use positive and negative constants in your equation. Part 2. Show your work in solving the equation. Include the work to check your solution and show that your solution is extraneous. Part 3. Explain why the first equation has an extraneous solution and the second does not.
anonymous
  • anonymous
An extraneous solution is a fake result. You can go through and solve the equation, but when you check the answer it won't work. The easiest thing to do for both equations is to make a = 1 so you don't have to worry about division. Then you're left with \(\sqrt{x+b}+c=d\) Subtract c from both sides to get \(\sqrt{x+b}=d-c\) For the extraneous solution equation, you need d-c to be negative. For the other one d-c will be positive.
anonymous
  • anonymous
so I can choose any number or does It have to be specific

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anonymous
  • anonymous
You can choose any number
anonymous
  • anonymous
ok
anonymous
  • anonymous
so I plug in the numbers ad solve
anonymous
  • anonymous
I couldn't see anything in that file. Can you post a screenshot or use the draw tool?
anonymous
  • anonymous
ok I plugged In numbers a radical sign 18 +24+32=46
anonymous
  • anonymous
so I subtract the 32 from both sides now right
anonymous
  • anonymous
you're still supposed to have a radical sign and an x. The only numbers you're replacing are a,b,c, and d. For example √(x-3)+4=20
anonymous
  • anonymous
For that equation a =1, b=-3, c = 4, and d = 20. Check d - c: d - c = 20 - 4 = 16 This is positive so that's an equation without an extraneous solution
anonymous
  • anonymous
So I don't need to add a number for x
anonymous
  • anonymous
right. x stays
anonymous
  • anonymous
So so I subtract 4 from both sides and then now I have 1\ x+3=16 sorry I don't have a radical sign on my keyboard lol
anonymous
  • anonymous
So I keep simplifying the equation from there
anonymous
  • anonymous
yes after you subtract 4 square both sides
anonymous
  • anonymous
Ok now I have 1/ x + 3 =16
anonymous
  • anonymous
Than I subtracted 3 from both sides and now it's just 1/x=13
anonymous
  • anonymous
no, square both sides first then subtract 3. You can't subtract the 3 while it's under the radical.|dw:1437332991847:dw|
anonymous
  • anonymous
Square the 3 also or no
anonymous
  • anonymous
no, the whole thing under the radical is squared. The √ and ² cancel each other \[\sqrt{x-3}^2=x-3\]
anonymous
  • anonymous
Oh ok so x + 3 =256 so do I subtract the three now or leave it as it is
anonymous
  • anonymous
subtract the 3
anonymous
  • anonymous
x=233
anonymous
  • anonymous
253? Just realized we switched for x-3 to x+3 when we were solving it. Just change the sign in the original equation to +
anonymous
  • anonymous
Ok lol yea it's 253
anonymous
  • anonymous
For the extraneous equation, just swap c and d and go through the same steps √(x-3)+20=4
anonymous
  • anonymous
Ok I just finished that part now how would I explain why the first one was extraneous do I just say because the answer doesn't go with the equation
anonymous
  • anonymous
yes, the answer doesn't check
anonymous
  • anonymous
Ok thank you
anonymous
  • anonymous
you're welcome

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