***Please show how to work these out*****?

Question

Solve the following equations:

(a) sq rt of x - 2 = 1
(b) sq rt of x^3=27
(c) ^3 sq rt of x^2= 9

Answer 1

a) sqrt x - 2 = 1
(sqrt x - 2)^2 = (1)^2
x - 2 = 1
x = 3

b) sqrt x^3 = 27
(sqrt x^3)^2 = (27)^2
x^3 = 729
^3sqrt x^3 = ^3sqrt 729
x = 9

c) ^3 sqrt x^2 = 9
(^3sqrt x^2)^3 = (9)^3
(sqrt x^2)^2 = (729)^2
x^2 = 531441
x = 729

Answer 2

b. Square both sides of the equation

[x^(3/2)]² = (27)²

x³ = 729.

Now take the cube root of both sides:

x = (729)^(1/3) = 9.

To check, multiply 9 x 9 x 9.

9 x 9 = 81
81 x 9 = 729

So the answer is correct.

Let's ask ourselves a question here. Can we use the negative value of x³, since we are extracting a square root, and ordinarily we can use the positive and negative value of the root? If we do, then that implies that x = -9. Therefore x³ = (-9)³ = -729. Can we extract the square root of a negative number using real numbers? No! The result would be an imaginary number. In fact, the result would be 27i, where i is the square root of -1. Only if we are specifically permitted to use imaginary roots can we use these values.

Answer 3

(a) sq rt of x - 2 = 1
sqrt(x-2) =1
x-2= 1
so x= 3
(b) sq rt of x^3=27
sqrt( x^3) = 27
so
x^3 = 27^2
x= cuberoot( 27^2) = 9 .

Answer 4

i'll just do a for you

(x-2)^2 the is sqrt =1
therefor
x-2=√1
so x-2=1
x=1+2
x=3

Answer 5

^3 sqrt x^2 become x^(2/3) remember that 3 sqrt x become x^1/3
x^(2/3)=9
x=9^(3/2)
x= (sqrt 9)^3
x= 3^3
just simplify first