Question
If $${x^2} + \frac{1}{5}x + {a^2}$$ is a perfect square, then a is?
Answer: Option C
$$\eqalign{
& {x^2} + \frac{1}{5}x + {a^2} \cr
& {{\text{A}}^2} + {\text{2}} \times {\text{AB}} + {{\text{B}}^2} = {\left( {{\text{A}} + {\text{B}}} \right)^2} \cr
& {x^2} + 2 \times \frac{1}{{10}} \times x + {a^2} = {\left( {x + \frac{1}{{10}}} \right)^2} \cr
& {\text{A}} = x \cr
& {\text{B}} = \frac{1}{{10}} \cr
& {\text{B}} = a = \frac{1}{{10}} \cr} $$
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$$\eqalign{
& {x^2} + \frac{1}{5}x + {a^2} \cr
& {{\text{A}}^2} + {\text{2}} \times {\text{AB}} + {{\text{B}}^2} = {\left( {{\text{A}} + {\text{B}}} \right)^2} \cr
& {x^2} + 2 \times \frac{1}{{10}} \times x + {a^2} = {\left( {x + \frac{1}{{10}}} \right)^2} \cr
& {\text{A}} = x \cr
& {\text{B}} = \frac{1}{{10}} \cr
& {\text{B}} = a = \frac{1}{{10}} \cr} $$
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