If x is an integer and $(5 x - 1) < {(x + 1)}^{2} < (7 x - 3)$ , then the value of x is

Open in App

Solution

There are two parts $(5 x - 1) < {(x + 1)}^{2} and {(x + 1)}^{2} < (7 x - 3) Taking the first part, we have (5 x - 1) < {(x + 1)}^{2} \Rightarrow 5 x - 1 < x^{2} + 2 x + 1 \Rightarrow x^{2} - 3 x + 2 > 0 \Rightarrow (x - 2) (x - 1) > 0 \Rightarrow x < 1 or x > 2 . . . . . [1] Taking the second part, we have {(x + 1)}^{2} < (7 x - 3) \Rightarrow x^{2} + 2 x + 1 < 7 x - 3 \Rightarrow x^{2} - 5 x + 4 < 0 \Rightarrow (x - 1) (x - 4) < 0 \Rightarrow 1 < x < 4 . . . . . [2] From [1] and [2], we get 2 < x < 4 ∴ x = 3$

Suggest Corrections

Similar questions

If 17(2-x)-5(x+12) divided by 1-7x =8 .Then value of x is .......

\frac{17 (2 - x) - 5 (x + 12)}{1 - 7 x} = 8

x

x

5 x - 1 < (x + 1)^{2} < 7 x - 3

(5 x - 1) < {(x + 1)}^{2} < (7 x - 3) .

Given

$\frac{(x^{2} - 5 x + 6)}{(x^{2} - 7 x + 12)} = \frac{(x^{2} - x - 6)}{(x^{2} + 7 x + 10)}$ , then the value of $x$ is

Join BYJU'S Learning Program