11Simply putting \frac{x-1}{6}\le \sqrt{x} gives bounds on natural x.
\large \dpi{100} $\mathbf{Solve for $x\in \mathbb{R}$ in x = 6\left[\sqrt{x}\;\right]+1}$\\\\ $Where $[\;.\;]=$ Greatest Integer function.
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4 Answers
11What does [.] symbolise?
Simply on solving the question the answer comes out to be 17±6√10
341Let x=y^2
Then we have y^2 = 6 [y]+1. Note that y2 is an integer and y≥1
[y] \le (y+1) \Rightarrow y^2 \le 6 (y+1)+1 \Rightarrow y^2-6y-7 \le 0 \Rightarrow y \le 7
Also, [y] \ge (y-1) \Rightarrow y^2 \ge 6 (y-1)+1 \Rightarrow y^2-6y+5 \ge 0 \Rightarrow y \ge 5
So [y] can be 5,6, or 7.
For [y]=5, we get x=31; [y]=6, gives x=37 and [y]=7 gives x=43 which does not satisfy the eqn.
So 31, 37 are the solutions
1708Thanks Sir (Brilliant Method Sir).
I have solved very Lengthy way.