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)

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- Mar 24th 2007, 09:16 PMjuiicycouturehow 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) - Mar 24th 2007, 09:26 PMJhevon
you mean the x-values as in the roots right?

Quote:

a) y=6x^5 + 192

6x^5 + 192 = 0

=> 6x^5 = -192

=> x^5 = -192/6 = -32

=> x = -32^(1/5) = -2

Quote:

b) 48-3x^2

48-3x^2 = 0

=> 3x^2 = 48

=> x^2 = 48/3 = 16

=> x = +/- sqrt(16) = +/- 4

Quote:

c) 6x^-1 + 5

6x^-1 + 5 = 0

=> 6x^-1 = -5

=> x^-1 = -5/6

=> x = -6/5 ............i just took the inverse

Quote:

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

(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