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The product of two numbers is 120 and the sum of their squares is 289. The sum of the two number is -
Answer & Solution
Correct Answer:
Option
A
Let the two number are a and b
According to question,
$$\eqalign{ & {a^2} + {b^2} = 289 \cr & ab{\text{ }} = {\text{ }}120 \cr & {\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab \cr & {\left( {a + b} \right)^2} = {\left( {289} \right)^2} + \left( {2 \times 120} \right) \cr & {\left( {a + b} \right)^2} = 529 \cr & \left( {a + b} \right) = 23 \cr} $$
According to question,
$$\eqalign{ & {a^2} + {b^2} = 289 \cr & ab{\text{ }} = {\text{ }}120 \cr & {\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab \cr & {\left( {a + b} \right)^2} = {\left( {289} \right)^2} + \left( {2 \times 120} \right) \cr & {\left( {a + b} \right)^2} = 529 \cr & \left( {a + b} \right) = 23 \cr} $$
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