How do you solve using completing square method $ {{x}^{2}}+3x+1=0 $ ? \[\]

Hint: We recall that in completing the square method to solve the quadratic equation $ a{{x}^{2}}+bx+c=0 $ we begin by subtracting $ c $ both sides. And the divide by both sides by if $ a\ne 1 $ . We then add $ {{\left( \dfrac{-b}{2a} \right)}^{2}} $ both sides and we make a complete square on the left hand side. We square root both sides to find the solutions of the quadratic equation. \[\]

Complete step by step answer:
We know that the general form of the quadratic equation is given by $ a{{x}^{2}}+bx+c=0 $ where $ a\ne 0 $ . We are given the following quadratic equation in the question
\[{{x}^{2}}+3x+1=0\]
We compare the given equation with the general form of the quadratic equation $ a{{x}^{2}}+bx+c=0 $ and find $ a=1,b=3,c=1 $ . We begin solving by completing square method by first subtracting $ c=1 $ both sides of the given equation to have;
\[{{x}^{2}}+3x=-1\]
Since we already have $ a=1 $ we do not need to divide both sides by $ a $ . We add both sides of the above step by $ {{\left( \dfrac{-b}{2a} \right)}^{2}}={{\left( \dfrac{-3}{2\cdot 1} \right)}^{2}}={{\left( \dfrac{3}{2} \right)}^{2}} $ to have;
\[\begin{align}
  & {{x}^{2}}+3x+{{\left( \dfrac{3}{2} \right)}^{2}}=-1+{{\left( \dfrac{3}{2} \right)}^{2}} \\
 & \Rightarrow {{x}^{2}}+2\cdot x\cdot \dfrac{3}{2}+{{\left( \dfrac{3}{2} \right)}^{2}}=-1+\dfrac{9}{4} \\
\end{align}\]
We make complete square in the left hand side of the above step using algebraic identity $ {{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} $ for $ a=x,b=\dfrac{3}{2} $ and simultaneously simplify right hand side have;

\[\begin{align}
  & \Rightarrow {{\left( x+\dfrac{3}{2} \right)}^{2}}=\dfrac{-4+9}{4} \\
 & \Rightarrow {{\left( x+\dfrac{3}{2} \right)}^{2}}=\dfrac{5}{4} \\
\end{align}\]
We take square root both sides of the above equation to have;
\[\begin{align}
  & \Rightarrow x+\dfrac{3}{2}=\sqrt{\dfrac{5}{4}} \\
 & \Rightarrow x+\dfrac{3}{2}=\pm \dfrac{\sqrt{5}}{2} \\
 & \Rightarrow x=\pm \dfrac{\sqrt{5}}{2}-\dfrac{3}{2} \\
\end{align}\]
So we have the solutions of the given quadratic equation as
\[\Rightarrow x=-\dfrac{3}{2}+\dfrac{\sqrt{5}}{2}.,-\dfrac{3}{2}-\dfrac{\sqrt{5}}{2}\].

Note:
We note that the types of solutions otherwise known as roots of a quadratic equation $ a{{x}^{2}}+bx+c=0 $ depends upon the discriminant $ D={{b}^{2}}-4ac $ . If D is a perfect square we get rational roots and if D is not a perfect square we get irrational roots like in this problem since here $ D={{3}^{2}}-4\cdot 1\cdot 1=9-4=5 $ is not a perfect square. The irrational roots always occur as conjugates which means in the form $ a+\sqrt{b},a-\sqrt{b} $ where $ a $ is a rational number and $ b $ is an irrational number. Here we have $ a=\dfrac{-3}{2},b=\dfrac{\sqrt{5}}{2} $ .