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

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.

• Use \sqrt{6} for $\sqrt6$ – gen-z ready to perish Mar 16 at 1:25

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}

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 $$[x]\in\{2,3,4,...,11,12\}$$.