Paul solved the inequality $$ 4x^2+4\sqrt{2}x+3\leq 4x+2\sqrt{2} $$ this way:
(1) He moved all the terms to the left side of the inequality: $$ 4x^2+4\sqrt{2}x−4x+3−2\sqrt{2}\leq 0. $$
(2) He combined like terms: $$ 4x^2+4(\sqrt{2}−1)x+3−2\sqrt{2}\leq 0. $$
(3) He calculated the discriminant of the quadratic trinomial: $$\begin{aligned} D &=4(\sqrt{2}−1)^2−4(3−2\sqrt{2})\cr D &=4(2−2\sqrt{2}+1)-12+8\sqrt{2}\cr D &=8-8\sqrt{2}+4-12+8\sqrt{2}\cr D &=0 \end{aligned}$$
(4) Paul found that the discriminant is zero, so the quadratic polynomial on the left side of the inequality has only one root (a double root). He knew that in this case the root of the quadratic polynomial $ax^2+bx+c$ is given by the formula $x=−b/2a$, i.e. $$ x=−\frac{4(\sqrt{2}−1)}{8}=\frac{1−\sqrt{2}}{2}. $$
(5) Moreover, Paul realized that a quadratic trinomial whose discriminant is equal to zero is a perfect square and simplified the inequality to the form $$ 4\left(x−\frac{1−\sqrt{2}}{2}\right)^2 \leq 0. $$
(6) From the above inequality, he concluded that the inequality is satisfied only when $$ x=\frac{1−\sqrt{2}}{2}. $$
The teacher asked Paul’s fellow students to comment on his solution. Here are some comments. Which one is correct?
Jane: The result is correct, but Paul incorrectly substituted the coefficients into the formula for the discriminant $D$. He got the right value of the discriminant $D$ just by chance. The error is in step (3).
Rick: The whole solution is correct. All of Paul's steps are correct.
Bob: The error is in step (4). It should have been $x=b/2a$, i.e. $x=\frac{\sqrt{2}−1}2$.
Mary: The error is in step (5). The inequality cannot be simplified into the form $$ 4\left(x−\frac{1−\sqrt{2}}{2}\right)^2\leq 0. $$
Tom: The error in in step (6). The inequality has no solution.
The discriminant of a quadratic trinomial $ax^2+bx+c=0$ is given by the formula $$ D =b^2-4ac. $$ In our case $a=4$, $b=4(\sqrt{2}−1)$ and $c=3−2\sqrt{2}$, i.e. $$\begin{aligned} D =(4(\sqrt{2}−1))^2&−4\cdot 4(3−2\sqrt{2})=16(2−2\sqrt{2}+1)−16(3-2\sqrt{2})= \cr&=16(3-2\sqrt{2})−16(3-2\sqrt{2})=0. \end{aligned}$$