Question

Find integral value of $x$, such that value of $x^2+19x+92$ is a perfect square of an

Answer 1

$x^2+19x+92$ is perfect square $=y^2$

$\Rightarrow$ $x^2+19x+92-y^2=0$ ----- ( 1 )

$\Rightarrow$ $x=\frac{-19\pm \sqrt{(19{)}^2-4(19-y^2)}}{2}$

$\Rightarrow$ $x=\frac{-19\pm \sqrt{361-4(92-y^2)}}{2}$

$\therefore$ $x$ will attain integral values, if $361-4(92-y^2)$ is a perfect square of an odd integer.

$\Rightarrow$ If $361-4(92-y^2)=(2n+1{)}^2$

For some integer $n$.

$\Rightarrow$ $4y^2-7=(2n+1{)}^2$

$\Rightarrow$ $4y^2-(2n+1{)}^2=7$

$\Rightarrow$ $(2y-2n-1)(2y+2n+1)=7$

$\Rightarrow$ $2y+2n+1=7$ and $2y-2n-1=1$

$\Rightarrow$ $y+n=3$ ----- ( 2 ) and $y-n=1$ --- ( 3 )

By solving equation ( 2 ) and ( 3 ) we get,

$\Rightarrow$ $y=2$ and $n=1$

Putting $y=2$ in equation ( 1 ), we get

$x^2+19x+88=0$

$\Rightarrow$ $x^2+11x+8x+88=0$

$\Rightarrow$ $x(x+11)+8(x+11)=0$

$\Rightarrow$ $(x+11)(x+8)=0$

$\Rightarrow$ $x=-11$ and $x=-8$

$\therefore$ $x^2+19x+92=0$ is a perfect square for $x=-8$ and $x=-11$