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The solutions to 1-2/(x-2)-(x-1)A(-∞,5)∪(1,∞)   B(-∞,5)∪(-2,1)  ...

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melissazamora7 | Student, Grade 11 | Honors

Posted December 11, 2012 at 7:33 PM via web

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The solutions to 1-2/(x-2)-(x-1)

A(-∞,5)∪(1,∞)

 

B(-∞,5)∪(-2,1)

 

C(-5,-2)∪(1,5)

 

D(-2,1)∪(5,∞)

 

Tagged with math, solutions infinite

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted December 12, 2012 at 6:30 AM (Answer #1)

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The request of the problem is not complete since it is not mentioned if it is an equation or an inequality. The given answers indicates that is an inequality, but it could be both ways, larger or smaller. Notice that the options larger or equal or smaller or equal are excluded, since the answers provided indicate only opened intervals, hence, the inequality may be strictly larger or strictly smaller.

Hence, supposing that the inequality you need to solve is `1 - 2/(x-2) - (x-1) > 0`  yields:

`1 - 2/(x-2) - (x-1) > 0 => (x - 2 - 2 - (x-1)(x-2))/(x-2)>0`

`(x - 4 - x^2 + 3x - 2)/(x -2)>0`

`(-x^2 + 4x - 6)/(x -2)>0 => (x^2- 4x+ 6)/(x -2)<0`

You need to find the solutions to the equation `x^2 - 4x + 6 = 0`  such that:

`x_(1,2) = (4+-sqrt(16 - 24))/2`

Notice that `x_(1,2) !in R` , hence, the parabola `y = x^2 - 4x + 6`  opens upward and it does not intersect x axis.

 

Hence, for any `x in R, x^2 - 4x + 6 > 0.`  Since the inequality `(x^2 - 4x + 6)/(x -2)<0` , then the denominator needs to be negative, hence `x in (-oo,2).`

Notice that neither of the provided answers are not suitable for any of the two possible inequalities `1 - 2/(x-2) - (x-1) > 0`  or `1 - 2/(x-2) - (x-1)< 0` .

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