Thread: how do u find the value of x for these:

i know these are easy questions, but i am just a bit confused:


please indicate which question you are answering:

a) y=6x^5 + 192

b) 48-3x^2

c) 6x^-1 + 5

d) y= (1-x^2)(x^-1)

Originally Posted by juiicycouture

i know these are easy questions, but i am just a bit confused:


please indicate which question you are answering:

a) y=6x^5 + 192

b) 48-3x^2

c) 6x^-1 + 5

d) y= (1-x^2)(x^-1)

you mean the x-values as in the roots right?

a) y=6x^5 + 192

for the roots


6x^5 + 192 = 0
=> 6x^5 = -192
=> x^5 = -192/6 = -32
=> x = -32^(1/5) = -2

b) 48-3x^2

for roots:

48-3x^2 = 0
=> 3x^2 = 48
=> x^2 = 48/3 = 16
=> x = +/- sqrt(16) = +/- 4

c) 6x^-1 + 5

for roots:

6x^-1 + 5 = 0
=> 6x^-1 = -5
=> x^-1 = -5/6
=> x = -6/5 ............i just took the inverse

d) y= (1-x^2)(x^-1)

for roots:

(1-x^2)(x^-1) = 0

=> (1 - x^2) = 0 or (x^-1)=0

since x^-1 is never 0, we must have

(1 - x^2) = 0 ...........this is the difference of 2 squares
=> (1 + x)(1 - x) = 0
=> x = -1 or x = 1