A) \[-7-4\sqrt{3}\]
B) \[-2-\sqrt{3}\]
C) \[-1\]
D) \[-2+\sqrt{3}\]
Correct Answer: D
Solution :
[d] \[a{{x}^{2}}+bx+c=0\] has 1 real root, so the discriminant is zero or\[{{b}^{2}}=4ac\]. Given that a, b, c are in arithmetic progression or \[b=\frac{a+c}{2}.\] We need to find the unique root or \[\frac{-b}{2a}\]discriminant is zero). From \[{{b}^{2}}=4ac,\]we have \[\frac{-b}{2a}=\frac{-2c}{b}.\] Ignoring the negatives, we have \[\frac{2c}{b}=\frac{2c}{\frac{a+c}{2}}=\frac{4c}{a+c}=\frac{1}{\frac{a}{4c}+\frac{1}{4}}\] From \[b=\frac{a+c}{2}\]and \[{{b}^{2}}=4ac,\]we have \[{{a}^{2}}+2ac+{{c}^{2}}=16ac\] or \[{{a}^{2}}-14ac+{{c}^{2}}=0\] or \[{{\left( \frac{a}{c} \right)}^{2}}-14\left( \frac{a}{c} \right)+1=0\] So, \[\frac{a}{c}=\frac{14\pm \sqrt{192}}{2}=7\pm 4\sqrt{3}\] But \[7-4\sqrt{3}<1,\] violating the assumption that\[a\ge c\]. \[\therefore \,\,\,\,\,\,\,\,\frac{a}{c}=7+4\sqrt{3}\] \[\therefore \,\,\,\,\,\,\,\,\frac{1}{\frac{a}{4c}+\frac{1}{4}}=\frac{1}{2+\sqrt{3}}=2-\sqrt{3}\] But we need the negative of this, so the required root is \[-2+\sqrt{3}\] .
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