Equation is equivalent with: $\lfloor\sqrt {4x+2}\rfloor=9$ ...and now is easy

:: HAW DID YOU REDUCED ?????????

See Solved problem (iii)

roza2010 wrote: See Solved problem (iii) That is for integers only. Is x supposed to be integer in the original problem?

Can someone explain this solution ?

roza2010 wrote: Equation is equivalent with: $\lfloor\sqrt {4x+2}\rfloor=9$ ...and now is easy ... easy, but wrong. Choose for example $x=\frac{79}4$ and you have $\lfloor\sqrt {4x+2}\rfloor=9$ but $\lfloor\sqrt x+\sqrt{x+1}\rfloor$ $+\lfloor\sqrt {4x+2}\rfloor=17\ne 18$ There is not equivalence between the two formulas (just implication) What is easy to show is that $\sqrt x+\sqrt {x+1}+1>\sqrt{4x+2}>\sqrt x+\sqrt {x+1}>0$ And so $\lfloor\sqrt x+\sqrt {x+1}\rfloor+1$ $\ge\lfloor\sqrt{4x+2}\rfloor$ $\ge\lfloor\sqrt x+\sqrt {x+1}\rfloor$ And so equation is equivalent to $\lfloor\sqrt{4x+2}\rfloor=9$ $\boxed{and}$ $\lfloor\sqrt x+\sqrt {x+1}\rfloor=9$ Which gives the answer : $\boxed{x\in\left[\frac{1600}{81},\frac{49}2\right)}$

you need to define the whole part of x, then the space goes to the whole part of 4x + 2.