Answer
True
Work Step by Step
The equation $x^{2}+4x=0$ is actually $x^{2}+4x+0=0$ as its constant is zero. Since $x^{2}+4x+0=0$ 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 $x^{2}+4x+0=0$ to the standard form of a quadratic equation $ax^{2}+bx+c=0$:
$a=1$, $b=4$, and $c=0$
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:
$x=\frac{-(4) \pm \sqrt {(4)^{2}-4(1)(0)}}{2(1)}$
Step 4: $x=\frac{-4 \pm \sqrt {16+0}}{2}$
Step 5: $x=\frac{-4 \pm \sqrt {16}}{2}$
Step 6: $x=\frac{-4 \pm 4}{2}$
Step 7: $x=\frac{-4+4}{2}$ or $x=\frac{-4-4}{2}$
Step 8: $x=\frac{0}{2}$ or $x=\frac{-8}{2}$
Step 9: $x=0$ or $x=-4$
Step 10: Therefore, the solution set is {$-4,0$}.
Therefore, the question statement is true.
