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Re: If a(a + 2) = 24 and b(b + 2) = 24, where a ≠ b, then a + b = [#permalink]
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14 Sep 2015, 07:00
9
2
ske wrote:
If a(a + 2) = 24 and b(b + 2) = 24, where a ≠ b, then a + b =
(A) −48 (B) −2 (C) 2 (D) 46 (E) 48
press 1 +kudos if you like my post and for motivation
a(a + 2) = 24 and b(b + 2) = 24
i.e. Product of two consecutive Even Integers = 24 i.e. Either 4*(4+2) = 24 or -6*(-6+2) = 24
i.e. if a = 4 then b = -6 or if a = -6 then b = 4
But in each case a+b = -6+4 = -2
Answer: option B
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Re: If a(a + 2) = 24 and b(b + 2) = 24, where a ≠ b, then a + b = [#permalink]
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19 Sep 2015, 15:18
8
1
a(a+2) = 24--eq1 b(b+2) = 24 --eq2
this can be written as X(x+2) = 24 where a,b are roots of this equation. We just need to find sum of roots for the equation x^2+2x-24=0 Sum of roots of a quadratic equation is -(coefficient of X)/(coefficient of x^2) = -2
Re: If a(a + 2) = 24 and b(b + 2) = 24, where a ≠ b, then a + b = [#permalink]
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18 Oct 2015, 15:06
1
I did the long way: a^2 +2a = 24 b^2 +2b = 24 these 2 equations are equal a^2+2a=b^+2b substract a^2+2a-b^2-2b = 0 a^2 - b^2 can be written as (a+b)(a-b) +2a-2b = 0 factor 2a-2b-> +2(a-b) rewrite: (a+b)(a-b)=-2(a-b) divide by (a-b) and get a+b = -2
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79% (01:17) correct 
