two numbers differ by 5, if their product is 336 then what is the sum of two numbers
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Let the first number be a
Second number = a - 5
According to question,
a ( a - 5) = 336
=> a^2 - 5a - 336 = 0
=> a^2 - 21a + 16a - 336 = 0
=> a ( a - 21) + 16(a - 21) = 0
=> (a - 21) (a + 16) = 0
a = 21 and - 16
Neglecting negative value, we get
a = 21
First number = 21
Second number = 21 - 5 = 16
Sum of numbers = 21 +16 = 37
Second number = a - 5
According to question,
a ( a - 5) = 336
=> a^2 - 5a - 336 = 0
=> a^2 - 21a + 16a - 336 = 0
=> a ( a - 21) + 16(a - 21) = 0
=> (a - 21) (a + 16) = 0
a = 21 and - 16
Neglecting negative value, we get
a = 21
First number = 21
Second number = 21 - 5 = 16
Sum of numbers = 21 +16 = 37
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