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# What values of x satisfy 3 <= x^2 + 7x - 8 <= 9

asicannot2 | Student | eNoter

Posted July 6, 2013 at 5:53 AM via web

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What values of x satisfy 3 <= x^2 + 7x - 8 <= 9

Tagged with inequality, math

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted July 6, 2013 at 6:15 AM (Answer #1)

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The values of x that satisfy `3 <= x^2 + 7x - 8 <= 9` have to be determined.

`3 <= x^2 + 7x - 8 <= 9`

=> `3 <= x^2 + 7x - 8` and `x^2 + 7x - 8 <= 9`

=> `0 <= x^2 + 7x - 11` and `x^2 + 7x - 17 <= 0`

=> `0<= (x-(sqrt(93)+7)/2)(x+(sqrt(93)+7)/2)` and `(x-(3*sqrt(13)+7)/2)(x+(3*sqrt(13)+7)/2) <= 0`

`0<= (x-(sqrt(93)+7)/2)(x+(sqrt(93)+7)/2)`

=> `(x-(sqrt(93)+7)/2)>= 0` and `(x+(sqrt(93)+7)/2) >= 0` or

`(x-(sqrt(93)+7)/2)<= 0` and `(x+(sqrt(93)+7)/2) <= 0`

=> `x >= (sqrt(93)+7)/2` and `x >= -(sqrt(93)+7)/2` or

`x <= (sqrt(93)+7)/2` and `x <= -(sqrt(93)+7)/2`

This gives `x in (-oo, -(sqrt(93)+7)/2]U[(sqrt(93)+7)/2, oo)` ...(1)

`(x-(3*sqrt(13)+7)/2)(x+(3*sqrt(13)+7)/2) <= 0`

=> `(x-(3*sqrt(13)+7)/2) <= 0` and `(x+(3*sqrt(13)+7)/2) >= 0` or

`(x-(3*sqrt(13)+7)/2) >= 0` and `(x+(3*sqrt(13)+7)/2) <= 0`

`x >= (3*sqrt(13)+7)/2` and `x <= -(3*sqrt(13)+7)/2` or `x <= (3*sqrt(13)+7)/2` and `x >= -(3*sqrt(13)+7)/2`

This is true for `x in [-(3*sqrt(13)+7)/2, (3*sqrt(13)+7)/2]`

`(sqrt(93)+7)/2 ~~ 8.32` and `(3*sqrt(13)+7)/2 ~~ 8.90`. This allows us to determine the subset of x that satisfies both the inequalities.

The solution of the inequality is `[-(3*sqrt(13)+7)/2, -(sqrt(93)+7)/2]U[(sqrt(93)+7)/2, (3*sqrt(13)+7)/2]`

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