Answer
True
Work Step by Step
Step 1: We write $-2x^{2}-3x-1=0$ as $2x^{2}+3x+1=0$.
Comparing $2x^{2}+3x+1=0$ to the standard form of a quadratic equation $ax^{2}+bx+c=0$, we obtain:
$a=2$, $b=3$ and $c=1$
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:
$x=\frac{-(3) \pm \sqrt {(3)^{2}-4(2)(1)}}{2(2)}$
Step 4: $x=\frac{-3 \pm \sqrt {9-8}}{4}$
Step 5: $x=\frac{-3 \pm \sqrt {1}}{4}$
Step 6: $x=\frac{-3 \pm 1}{4}$
Step 7: $x=\frac{-3-1}{4}$ or $x=\frac{-3+1}{4}$
Step 8: $x=\frac{-4}{4}$ or $x=\frac{-2}{4}$
Step 9: $x=-1$ or $x=\frac{-1}{2}$
Step 10: Therefore, the solution set is {$-\frac{1}{2},-1$}.
Therefore, it is true that the solution set consists of two rational numbers.
