Question

Find the number of integers satisfying the following inequalities
x 2(square) +6x - 7 < 2 ​

Answer 1

Step-by-step explanation:

We have,

${x}^{2} + 6 x - 7 \leqslant 2$

$= > {x}^{2} + 6x - 9 \leqslant 0$

$= > (x - (3 + 3 \sqrt{2} ) )(x - (3 - 3 \sqrt{2})) \leqslant 0$

$= > x \in [ 3 - 3 \sqrt{2} \: \: to \: \: 3 + 3 \sqrt{2}]$

so, the integers are -1, 0, 1, 2, 3, 4, 5, 6, 7