For what value of $\displaystyle x+\frac { 1 }{ 4 } \sqrt { x } +{ a }^{ 2 }$ will be perfect square -
$\displaystyle \pm { 1 }/{ 18 }$
$\displaystyle \pm { 1 }/{ 8 }$
$\displaystyle \pm { 1 }/{ 5 }$
$\displaystyle { 1 }/4$
If $\displaystyle x+\frac { 1 }{ 4 } \sqrt { x } +{ a }^{ 2 }$ is a perfect squarethen $\displaystyle \frac { 1 }{ 4 } \sqrt { x } =2\times \sqrt { x } \times \left( \pm a \right) $$\displaystyle \therefore \quad a=\pm \frac { 1 }{ 8 } $
$10$
$4$
$15$
$5$