Question

The number of integral values of ${}^{\prime }{x}^{\prime }$ for which $x^2+19x+92$ is a perfect square is:

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $0$

The above quadratic equation can be expressed as : $x^2+19x+92={(x+\frac{19}{2})}^2+92-\frac{361}{4}$

$={(x+\frac{19}{2})}^2+\frac{7}{4}$

From the above expression, we can infer that no integer value of x will make it a perfect square because we have a fraction in the expression.