Find three consecutive positive numbers such that the square of the middle number exceeds the difference of the squares of the other two by 60.
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Let the numbers be (a-1), a and (a +1)
A/q
a² - [(a+1)² - (a-1)²] = 60
⇒a² - [(a+1 +a -1)(a+1-a +1)] =60
⇒a² -(2a×2) = 60
⇒a²- 4a -60 =0
⇒a² -10a + 6a - 60 =0
⇒a(a -10) + 6(a -10) =0
⇒(a-10) (a+6) =0
either a =10 or a = -6
ignoring a= -6 as the numbers are positive,
so when a =10, then numbers are 9,10 and 11
A/q
a² - [(a+1)² - (a-1)²] = 60
⇒a² - [(a+1 +a -1)(a+1-a +1)] =60
⇒a² -(2a×2) = 60
⇒a²- 4a -60 =0
⇒a² -10a + 6a - 60 =0
⇒a(a -10) + 6(a -10) =0
⇒(a-10) (a+6) =0
either a =10 or a = -6
ignoring a= -6 as the numbers are positive,
so when a =10, then numbers are 9,10 and 11
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