SOLUTION: Please Help, I am confused about how to work with third roots.
I am supposed to multiply then simplify by factoring.
(3rd root)of y^7 * (3rd root) of 16y^8
Use Rational exp
Algebra ->
Polynomials-and-rational-expressions
-> SOLUTION: Please Help, I am confused about how to work with third roots.
I am supposed to multiply then simplify by factoring.
(3rd root)of y^7 * (3rd root) of 16y^8
Use Rational exp
Log On
Question 303313: Please Help, I am confused about how to work with third roots.
I am supposed to multiply then simplify by factoring.
(3rd root)of y^7 * (3rd root) of 16y^8
Use Rational exponents to simplify the third root of x^6
Find the 16th root of (-8)^16
Rewrite the 4th root of 23 with a rational exponent.
Simplify the 3rd root of -1/64
Simplify by taking roots of the numerator and denominator
(3rd root) of 1000x^14/y^3
Solve
The square root of 5x+29 = x+3
I am completely lost when working with roots. Any help would be greatly appreciated! Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! multiply then simplify by factoring.
Use this as a cube root symbol
:
(3rd root)of y^7 * (3rd root) of 16y^8
:
We can combine these under single radical
:
Add exponents when you multiply
:
Factor to reveal perfect cubes
:
Extract those perfect cubes
:
:
Use Rational exponents to simplify the third root of x^6
to raise on power to another power, multiply exponents = =
:
:
Find the 16th root of (-8)^16 = = = -8
:
:
Rewrite the 4th root of 23 with a rational exponent.
:
:
Simplify the 3rd root of -1/64 = =
:
:
Simplify by taking roots of the numerator and denominator
(3rd root) of 1000x^14/y^3
Extract cube root 1/y^3
Extract the cube root of 1000
Factor inside the radical
Extract
:
;
Solve
The square root of 5x+29 = x+3
square both sides:
5x + 29 = (x+3)^2
FOIL the right side
5x + 29 = x^2 + 6x + 9
combine like terms on the right
0 = x^2 + 6x - 5x + 9 - 29
A quadratic equation
x^2 + x - 20 = 0
Factors to
(x+5)(x-4) = 0
two solutions
x = -5
and
x = +4
:
Check both solution in original equation
x = -5 not a solution
x = 4 ; a good solution (x=4)