Let $x_1$ and $x_2$ be the real root of the equation $x^2-kx+(k^2+7k+15)=0$, if the maximum value of $(x_1^2+x_2^2)=\dfrac{18}{x}$, then find the value of $x$?
My attempt is as follows:-
$$x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2$$ $$x_1^2+x_2^2=k^2-2(k^2+7k+15)$$ $$x_1^2+x_2^2=-k^2-14k-30$$
Max value of $x_1^2+x_2^2=-\dfrac{D}{4a}$
$$(x_1^2+x_2^2)_{max}=\dfrac{196-120}{4}$$ $$(x_1^2+x_2^2)_{max}=19$$
hence $76=\dfrac{18}{x}$, $x=\dfrac{18}{19}$. But actual answer is $1$.
First, it seems like you're using the symbol $x$ to denote $2$ different things (the variable in the quadratic equation, and the number we need in $18/x$). Avoid doing that because it causes confusion.
Next, the problem with your solution is that you have a condition on $k$. The discriminant $D=-3k^2-28k-60$ has to be non-negative for the quadratic to have real roots. Solving $D\geq 0$ gives $k\in[-6,-10/3]$. So, you have to find the maximum of $f(k)=-k^2-14k-30$ on $[-6,-10/3]$. Instead, you found the maximum over $\mathbb{R}$ which is obtained when $k=-7$, and that's not in the interval. That's why you got a wrong answer. You'll find that $f$ is decreasing in $[-6,-10/3]$, so the maximum is $f(-6)=18$, and the answer follows.