Respuesta :
The given equation is
[tex]x^2+x+9=0[/tex]To complete the square, first, we have to move the independent term to the other side.
[tex]x^2+x=-9[/tex]Then, we divide the linear coefficient by 2, then we apply a square power to it.
[tex](\frac{1}{2})^2=\frac{1}{4}[/tex]We add this fraction to each side.
[tex]x^2+x+\frac{1}{4}=-9+\frac{1}{4}[/tex]Then, we factor the trinomial and sum the numbers on the right side.
[tex]\begin{gathered} (x+\frac{1}{2})^2=\frac{-36+1}{4} \\ (x+\frac{1}{2})^2=\frac{-35}{4} \end{gathered}[/tex]Then, to solve for x, we use a square root on both sides.
[tex]\begin{gathered} \sqrt[]{(x+\frac{1}{2})^2}=\pm\sqrt[]{-\frac{35}{4}} \\ (x+\frac{1}{2})=\pm\sqrt[]{-\frac{35}{4}} \end{gathered}[/tex]As you can observe, the equation has no real solutions because the square root of a negative number can be solved in the real numbers.
