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Question
If `(2 "x" + 7)/3 <= (5 "x" +1)/4` , find the smallest value of x, when:
(i) x ∈ R
(ii) x ∈ Z
Solution
`2/"x"^2 - 5/"x" + 2 = 0`
2 - 5x + 2x2 = 0
2x2 - 5x + 2 = 0
`"x"^2 - 5/2 "x" + 1 = 0`
`"x"^2 - 2"x" - 1/2 "x" + 1 = 0`
`"x" ("x" - 2) - 1/2 ("x" - 2) = 0`
`("x" - 2) ("x" - 1/2) = 0`
(x - 2) = 0 , `("x" - 1/2)` = 0
x = 2 , x = `1/2`
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