## Elementary Algebra

The equation $x^{2}=27$ is actually $x^{2}+0x-27=0$. Since $x^{2}+0x-27=0$ is in the standard form of the quadratic equation $ax^{2}+bx+c=0$, it can be solved by the quadratic formula. We now solve this equation through the quadratic formula to illustrate our conclusion: Step 1: Comparing $x^{2}+0x-27=0$ to the standard form of a quadratic equation $ax^{2}+bx+c=0$, we obtain: $a=1$, $b=0$, and $c=-27$ Step 2: The quadratic formula is: $x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$ Step 3: Substituting the values of a,b and c in the formula: $x=\frac{-(0) \pm \sqrt {(0)^{2}-4(1)(-27)}}{2(1)}$ Step 4: $x=\frac{0 \pm \sqrt {0+108}}{2}$ Step 5: $x=\frac{0 \pm \sqrt {108}}{2}$ Step 6: $x=\frac{\pm \sqrt {36\times3}}{2}$ Step 7: $x=\frac{\pm 6\sqrt 3}{2}$ Step 8: $x=\frac{-6\sqrt 3}{2}$ or $x=\frac{+6\sqrt 3}{2}$ Step 9: $x=-3\sqrt 3$ or $x=3\sqrt 3$ Step 10: Therefore, the solution set is {$-3\sqrt 3, 3\sqrt 3$}. Therefore, the question statement is true.