Find the set of $'a'$ for which the equation has real roots $x^{2}+2ax+\dfrac{1}{16}=-a+\sqrt{a^{2}+x-\dfrac{1}{16}}$
- $a \in \left(-\infty, \dfrac{1}{4}\right] \cup \left[\dfrac{3}{4}, \infty\right)$
- $ a \in (-\infty, 1]$
- $a \in $\right[\dfrac{1}{4},\dfrac{3}{4}\right]$
- $a \in $\right[\dfrac{1}{4},\dfrac{1}{2}\right]$
For the equation to have real roots, the discriminant of the quadratic must be non-negative and the square root term must be defined. Solving the inequality leads to the interval (-infinity, 1/4] union [3/4, infinity).
Rearrange the equation to get x squared plus 2 a x plus a squared equals the quantity a plus the square root of the quantity a squared plus x minus the fraction with numerator 1 and denominator 16. The left side is the perfect square of the quantity x plus a. Squaring both sides of the original rearranged form, substituting the perfect square, and simplifying eventually yields a relationship that tests the boundary conditions for a. By testing the critical points derived from the discriminant conditions ensuring the inner square root and the overall quadratic are valid, the valid intervals for a are found to be the set from negative infinity to one fourth inclusive union with the set from three fourths to infinity inclusive.