NIMCET 2018 Question Paper with solutions

$\sqrt{{\left(224\right)}_{\mathit{r}}}$ = ${\left(13\right)}_{\mathit{r}}$
On squaring both sides,
${\left(224\right)}_{\mathit{r}}$ = ${\left(13\right)}_{\mathit{r}}^2$
Now, converting to decimal number
$2{\mathit{r}}^2$ + 2r + 4 = ${\left(1\mathit{r}+3\right)}^2$
$2{\mathit{r}}^2$ + 2r + 4 = ${\left(\mathit{r}+3\right)}^2$
$2{\mathit{r}}^2$ + 2r + 4 = ${\mathit{r}}^2$ + 9 + 6r
${\mathit{r}}^2$ - 4r - 5 = 0
Root of the above equation is 5, -1
Therefore, radix = 5