Let [t] denote the greatest integer $\le$ t. Then the equation in x,
[x]2 + 2[x+2] - 7 = 0 has :
Solution
${[x]^2} + 2[x + 2] - 7 = 0$
<br><br>$\Rightarrow$ ${[x]^2} + 2[x] + 4 - 7 = 0$
<br><br>Using the property [x + n] = [x] + n ; n $\in$ I
<br><br>$\Rightarrow$ ${[x]^2} + 2[x] - 3 = 0$<br><br>let [x] = y<br><br>${y^2} + 3y - y - 3 = 0$<br><br>$\Rightarrow$ $(y - 1)(y + 3) = 0$<br><br>$[x] = 1\,or\,[x] = - 3$<br><br>$\therefore$ $x \in \left[ {1,2} \right)\,\& \, \in \left[ { - 3, - 2} \right)$
About this question
Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane
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