Math, asked by sm0itmoharias, 1 year ago

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.

Answers

Answered by qais
4
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
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