Number of real solutions of the equation $6x^2 -77[x] + 147 =0 $.

Question

How many real solutions of $$6x^2 -77[x] +147=0$$ are there, where $[x]$ is the integral part of $x$?

The answer says 4 solutions but I got none.

As:

$6x^2 + 147 = 77[x]$ LHS= integer Therefore, RHS = integer

Then $6x^2$ just be integer

Then CASE 1: if $x$ is integer then $[x] = x$ By solving we get $x = 21/3$ and $7/2$ But x is integer , therefore no solution

CASE 2: if $x$ is not integer but $6x^2$ is integer I.e. $x^2= 1/6$. so $x = \pm 1/√6$ But non e satisfies the equation so no solution

Hence overall no solution.

I don't understand how there are 4 real solutions.

Answer 1

Hint: Use the fact that $$x-1<[x] \leq x$$

so $$77x-77<6x^2+147\leq 77x$$

and solve these two inequalites. From second we see $x>0$ and $$ 6x^2+144<90x\implies x^2-15x+26<0$$ so $$(x-13)(x-2)<0\implies x \in (2,13)$$

Now you can solve the equation for $[x]\in\{2,3,4,...,11,12\}$.

Answer 2

Let's write $x=k+r$ with $k\in\mathbb{Z}$ and $0\le r\lt1$. Then the quadratic equation becomes $6(k+r)^2-77k+147=0$, so that we have

$$(k+r)^2={77k-147\over6}$$

hence

$$0\le r=\sqrt{77k-147\over6}-k\lt1$$

from which we conclude that the integer $k$ must simultaneously satisfy

$$6k^2-77k+147\le0\qquad\text{and}\qquad0\lt6k^2-65k+153$$

The first of these is satisfied for $3\le k\le10$; the latter is satisfied when $k\le3$ or $k\ge8$. The values for $k=\lfloor x\rfloor$ are thus $3$, $8$, $9$, and $10$, which lead to the following four values of $x$:

$$\begin{align} \sqrt{77\cdot3-147\over6}&=\sqrt{14}\\ \sqrt{77\cdot8-147\over6}&=\sqrt{469\over6}\\ \sqrt{77\cdot9-147\over6}&=\sqrt{91}\\ \sqrt{77\cdot10-147\over6}&=\sqrt{623\over6}\\ \end{align}$$