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Solve the Equation getting rid of square roots

  • #1
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Solve the Equation.... getting rid of square roots PLZ

1+ √((1+x)√(x^2-24)) = x
(x*√(x^2-24)) = (x-1)^2 (Square both side)
(-1)+1+(x^√(x^2-24) = (x^2 - 2x +1)-1
x^2(x^2-24) = (x^2 -2x)^2 (Square both sides)
x^4 -24x^2 = (x^2 - 2x)^2
x^4 -24x = x^4 - 4x^3 +4x^2
0 = -4x^3 +4x^2 +4x^2
0 = 4x^2 +4x+ 24

4x^2 -4x -24 = 0


What mistake am I making...

I honestly have tried a bunch of different times, and i had another student check it for me, and i cant catch the mistake i'm making!
 

Answers and Replies

  • #2
280
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edit:

I misread your equation, I think. Is it
[tex]1+\sqrt{(x+1)\sqrt{x^2-24}}=x[/tex]
 
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  • #3
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edit:

I misread your equation, I think. Is it
[tex]1+\sqrt{(x+1)\sqrt{x^2-24}}=x[/tex]
It is, he has the extra set of parentheses to imply this.

Edit: Now that I "solved" this (turns out to be a cubic equation) the answer isn't pretty and the equation might be something else that yields an easier answer).

1+ √((1+x)√(x^2-24)) = x
(x*√(x^2-24)) = (x-1)^2 (Square both side)
The right hand side is fine (you subtracted the 1 and then squared) but the left hand side is incorrect.
 
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  • #4
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It is, he has the extra set of parentheses to imply this.

Edit: Now that I "solved" this (turns out to be a cubic equation) the answer isn't pretty and the equation might be something else that yields an easier answer).



The right hand side is fine (you subtracted the 1 and then squared) but the left hand side is incorrect.
Honestly i feel like a tool.... I just rewrote the question on my board and did it... and this is the way i solved it earlier for my teacher..

unfortunately when i divided 28/4 i got... 6.. not 7.. i couldn't do simple math in my head... x.x

hers my solution i just took a picture off my whiteboard.. c.c

http://postimage.org/image/oku2wqccn/ [Broken]
 
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  • #5
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I find... when i plug in 7 into the right side

1+all-sqrt(1+x)sqrt(x^2-24) = x

it resolves as 7...

therefore LS = RS Since X=7

Is This CorrecT?
 
  • #6
392
17


Yes, 7 is the correct answer! You miswrote your equation in your first post which is why it confused us. You should have wrote:

1+ sqrt(1+x * sqrt[(x^2)-24]) = x

Or better yet:

[itex]\displaystyle 1 + \sqrt{1+x \cdot \sqrt{x^2-24}}=x[/itex]


You had parentheses aroung the 1 + x originally which implied their sum was being multiplied by the inside square root.
 
  • #7
174
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Yes, 7 is the correct answer! You miswrote your equation in your first post which is why it confused us. You should have wrote:

1+ sqrt(1+x * sqrt[(x^2)-24]) = x

Or better yet:

[itex]\displaystyle 1 + \sqrt{1+x \cdot \sqrt{x^2-24}}=x[/itex]


You had parentheses aroung the 1 + x originally which implied their sum was being multiplied by the inside square root.
Yea i'm silly like that.. Sorry, but thank you very much for your help, how do i write the equations like you guys are writing so they are much more visually evident.
 
  • #8
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Yea i'm silly like that.. Sorry, but thank you very much for your help, how do i write the equations like you guys are writing so they are much more visually evident.
You end it with[\tex]. You start it with [tex]. Then you use various commands that format into what we wrote.
\sqrt{} is square root
a^b is a power, but you can also do a^{b+c+d}. \frac{numerator}{denominator} is for fractions. I use this:
http://en.wikipedia.org/wiki/Help:Displaying_a_formula

You can also always quote someone who has done it. You will then see how he created his equations in the quoted text. For example, try quoting me or scurty and seeing the [tex] statement.
 
  • #9
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√Oh okay thanks a lot. Also I'm coming to find there are 2 solutions for this problem but I can't seen to find the other. I remember in trig there was a way to prove for ex.
Cosx + sinx = 0
Divide by cosx to get tanx

But because u got rid of cosx u could prove it wasn't part of solution anyway in lOst on finding the second solution...

at the point

1+√1+x√x^2-24 = x
√1+x√x^2-24 = x-1 (square both sdie)
1 +x√x^2-24 = (x-1)^2 (subtract 1 from both sides)
x√x^2-24 = x-2x
Divide by x here, and i beleive this is where i determine the other solution, but how exactly?

√x^2-24 = X-2
Square both side

x^2-24 = (x-2)^2
x^2-24 = x^2-4x+4
4x = 28
x = 7
so yea this other solution :(
 
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  • #10
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okay so another teacher helped me today at school with the solution.. My original solution is incorrect.

Variables cannot be divided by, you must factor them, or they can be removed by subtraction/addition.

1 +x√x^2-24 = (x-1)^2

so from this point.

x√x^2-24 = x^2 -2x +1 -1
x√x^2-24 - x^2 +2x = 0

x(√x^2-24 - x + 2)= 0

x = 0 or
√x^2-24 -x + 2 = 0
√x^2-24 = x -2
square both sides
x^2 - 24 = (x-2)^2

x^2 -24 = x^2 -4x +4
4x = 28
x = 7

Plug back in,1+√1+x√X^2-24 = x

X=7 is a solution
x= 0 not a solution due to extraneous roots.
 

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