Dividing Radical expressions, i have to divide and simplify. I know i have to multiply so have to do the reciprocal, but im stuck after that.
HELP!
Dividing Radical expressions, i have to divide and simplify. I know i have to multiply so have to do the reciprocal, but im stuck after that.
HELP!
x^{2}-3x+2 can be factored into (x-2)(x-1) (numerator of 1st fraction)
x^{2}-4x+4 can be factored into (x-2)(x-2) (denominator of 1st fraction)
so the first fraction reduces to (x-1)/(x-2)
x^{2}-2x+1 can be factored into (x-1)(x-1) (numerator of 2nd fraction)
3x^{2}-12 can be factored into (3) (x-2)(x+2) (denominator of 2nd fraction)
so the fraction is [(x-1)(x-1) / (3)(x-2)(x+2)]
Now we have to remember that (a/b) / (c/d) = "extreems / means" or a*d/b*c
or multiply the means together and multiply the extreems together then divide the extreems by the means.
For our problem, we would have [3(x-1)(x-2)(x+2)] / (x-2)(x-1)(x-1)
NOTICE: THE (X-1) CAN BE CANCELLED FROM TOP AND BOTTOM!
NOTICE: THE (X-2) CAN BE CANCELLED FROM BOTH TOP AND BOTTOM!
We are now left with 3(x+2) /(x-1) the ANSWER
YOU DO NOT HAVE TO DO THE RECIPROCAL!
For rational expressions the "means/extreems" method is much easier and should have been introduced to you long long ago.
x^{2} - 3x + 2 * 3x^{2} - 12
x^{2 }- 4x + 4 x^{2 }- 2x + 1
(x - 1)(x - 2) * 3(x - 2)(x + 2)
(x - 2)(x - 2) (x - 1)(x - 1)
3(x + 2)
(x - 1)