Three students solved the inequality: $$ 2x^2+12x+20 \geq 0 $$
Luke remembered that $(x+6)^2=x^2+12x+36$ and so he rewrote the inequality into the form: $$ (x+6)^2−36+20 \geq 0 $$ Then, he solved it as follows: $$ \begin{align} (x+6)^2 \geq 16 \cr x+6 \geq 4 \cr x\geq−2 \end{align} $$ Adam decided to determine the discriminant of the quadratic polynomial: $$ D =12^2−4\cdot 2 \cdot 20=−16 $$ The discriminant came out negative, from which he concluded that the inequality has no solution.
Eva also found out that the discriminant was negative, so she decided to graph the quadratic function $f(x)=2x^2+12x+20$. She knew that the graph was a parabola and determined its vertex using the method of completing the square: $$ \begin{align} 2(x^2+6x+9)−9\cdot 2+20 \geq 0 \cr 2(x+3)^2+2 \geq 0 \cr V=(−3;−2) \end{align} $$ Now she reasoned as follows: The vertex of the parabola lies below the $x$-axis, the discriminant of the quadratic trinomial is negative, so this means that the parabola opens downwards, and the function cannot take on positive values. The given inequality has no solution.
Which one of them proceeded correctly in solving?
None of them.
Eva
Luke
Adam
By completing the square, we get: $$ 2(x+3)^2+2\geq0 $$ The coordinates of the vertex of the parabola are: $x=−3$, $y=2$, but there is no need to determine them. The next two pieces of information are enough: The discriminant of the quadratic trinomial is negative, which means that the parabola has no $x$-intercepts. The coefficient of the quadratic term is positive ($a=2$), so the parabola opens upwards. All that together means that the quadratic function takes only positive values, and the solution of the given inequality is all $x \in \mathbb{R}$.