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The real numbers $c, b, a$ form an arithmetic sequence with $a \geq b \geq c \geq 0$. The quadratic $a x^{2}+b x+c$ has exactly one root. What is this root? 
Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Unary Quadratic Equations
It is given that $a x^{2}+b x+c=0$ has 1 real root, so the discriminant is zero, or $b^{2}=4 a c$. Because $a, b, c$ are in arithmetic progression, $b-a=c-b$, or $b=\frac{a+c}{2}$. We need to find the unique root, or $-\frac{b}{2 a}$ (discriminant is 0 ). From $b^{2}=4 a c$, we can get $-\frac{b}{2 a}=-\frac{2 c}{b}$ Ignoring the negatives(for now), we have $\frac{2 c}{b}=\frac{2 c}{\frac{a+c}{2}}=\frac{4 c}{a+c}=\frac{1}{\frac{1}{\frac{4 c}{a+c}}}=\frac{1}{\frac{a+c}{4 c}}=\frac{1}{\frac{a}{4 c}+\frac{1}{4}}$. Fortunately, finding $\frac{a}{c}$ is not very hard. Plug in $b=\frac{a+c}{2}$ to $b^{2}=4 a c$, we have $a^{2}+2 a c+c^{2}=16 a c$, or $a^{2}-14 a c+c^{2}=0$, and dividing by $c^{2}$ gives $\left(\frac{a}{c}\right)^{2}-14\left(\frac{a}{c}\right)+1=0$, so $\frac{a}{c}=\frac{14 \pm \sqrt{192}}{2}=7 \pm 4 \sqrt{3}$. But $7-4 \sqrt{3}\textless1$, violating the assumption that $a \geq c$. Therefore, $\frac{a}{c}=7+4 \sqrt{3}$. Plugging this in, we have $\frac{1}{\frac{a}{4 c}+\frac{1}{4}}=\frac{1}{2+\sqrt{3}}=2-\sqrt{3}$. But we need the negative of this, so the answer is (D). 