Question

(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

Answer 1

Is that three different problems?

Answer 2

yes

Answer 3

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

Answer 4

\[(x-1)^{2/3}-13=3\]
x-4 = √(9x)
\[\sqrt{x+6}-\sqrt{x-5}=1\]

Answer 5

@tbates, yes, thats what it looks like

Answer 6

Ok. so the first one add 13 to both sides first

Answer 7

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.

Answer 8

(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

Answer 9

oh i see

Answer 10

To get rid of a square root raise both sides to the 2nd power. (√x is the same as x^(1/2)

Answer 11

(x-4)=√9x
(x-4)^2 = 9x
now do you know how to raise x-4 to the second power?

Answer 12

i gotta do -4(-4) so its gonna be x^2+16=9x?

Answer 13

almost but you also will have a -8x on the left

Answer 14

(x-4)^2 = (x-4)(x-4) and you have to distribute
x^2 - 4x - 4x + 16
x^2 - 8x + 16 = 9x

Answer 15

do i bring the 9x to the lefts side, or the 8x to the right side

Answer 16

I would move the 9x to the left side by subtracting it.

Answer 17

So you will end up with x^2 - 17x + 16 = 0
Now do you know the quadratic formula?

Answer 18

i cant remember

Answer 19

whats the formula for it

Answer 20

[-b+-√(b^2-4ac)]/2a

Answer 21

http://www.scientificpages.net/math/images/Quadratic_Formula.gif

Answer 22

oh i remember this, let me try and solve it real quick and ill write the answers

Answer 23

Excellent!

Answer 24

ok so far i got 17+-*225/32*

Answer 25

(17+-√225)/2
√225 = 15
(17+15)/2 = 16
(17-15)/2 = 1
The answer is x = 1 and 16

Answer 26

how did u get the 2, on the first line

Answer 27

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)

Answer 28

oh!! i multiplied the c on the bottom, oh ok