Question

How many real solution does the equation ${6x}^2-77\left[x\right]+147=0$ have
, where [x] is the integral part of x.

Answer 1

${6x}^2-77\left[x\right]+147=0$ ...(i)
Consider the equation, ${6x}^2-77x+147=0$ ...(ii)
Roots of this equation are $\frac{7}{3},\frac{63}{6}$
where $\left[\frac{7}{3}\right]=2$ and $\left[\frac{63}{6}\right]=10$
put $\left[x\right]=2$ in equation (i) and solve for $x.$
$x$ comes out to be $\sqrt{\frac{7}{6}}$ and $\left[\sqrt{\frac{7}{6}}\right]=1.$
So this is not a valid solution.
Again put $\left[x\right]=10$ in (i) and solve for $x.$
$x$ comes out to be $\sqrt{\frac{623}{6}}$ and $\left[\sqrt{\frac{636}{6}}\right]=10$
So this is a valid solution.
Other real solution will lie between $2$ and $10.$
Substitute $\left[x\right]=3,4,5,6,7,8,9$
One by one and solve for $x$ and analyse each case as above.
We will find that for $\left[x\right]=3,9,10$
$x$ comes out to be such that [$x$ after solving]$=3,9,10.$
Hence the number of real solutions to the original equation will be $6.$