Answer
True
Work Step by Step
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.
