(x-1)^2/3-13=3 x-4=9x *the 9x is a radical *x+6*-*x-5*=1 the stuff inside the stars is in its own radical sign

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(x-1)^2/3-13=3 x-4=9x *the 9x is a radical *x+6*-*x-5*=1 the stuff inside the stars is in its own radical sign

Mathematics
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Is that three different problems?
yes
if I understand the problem the approach is: multiply both sides by 3: (x-1)^2 -39=9 add 39 to both sides: (x-1)^2 = 48 x=2* (3)^1/2 +1

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\[(x-1)^{2/3}-13=3\] x-4 = √(9x) \[\sqrt{x+6}-\sqrt{x-5}=1\]
@tbates, yes, thats what it looks like
Ok. so the first one add 13 to both sides first
tbates when u get done will u help me with this one question? i got the answer i just need to know if its right.
(x−1)^(2/3) = 16 now raise both sides to the 3/2 power x-1 = 16^(3/2) x-1 = 64 so x = 65
oh i see
To get rid of a square root raise both sides to the 2nd power. (√x is the same as x^(1/2)
(x-4)=√9x (x-4)^2 = 9x now do you know how to raise x-4 to the second power?
i gotta do -4(-4) so its gonna be x^2+16=9x?
almost but you also will have a -8x on the left
(x-4)^2 = (x-4)(x-4) and you have to distribute x^2 - 4x - 4x + 16 x^2 - 8x + 16 = 9x
do i bring the 9x to the lefts side, or the 8x to the right side
I would move the 9x to the left side by subtracting it.
So you will end up with x^2 - 17x + 16 = 0 Now do you know the quadratic formula?
i cant remember
whats the formula for it
[-b+-√(b^2-4ac)]/2a
http://www.scientificpages.net/math/images/Quadratic_Formula.gif
oh i remember this, let me try and solve it real quick and ill write the answers
Excellent!
ok so far i got 17+-*225/32*
(17+-√225)/2 √225 = 15 (17+15)/2 = 16 (17-15)/2 = 1 The answer is x = 1 and 16
how did u get the 2, on the first line
in the denominator of the quadratic formula you have 2a and in the problem a = 1 (the number in front of the x^2 term)
oh!! i multiplied the c on the bottom, oh ok

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