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If a(a + 2) = 24 and b(b + 2) = 24, where a ≠ b, then a + b = [#permalink]
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14 Sep 2015, 05:29
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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, 06:00
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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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16 Sep 2015, 22:31
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a^2+2a24=0 b^2+2b24=0 (a+6)(a4) (b+6)(b4) a=6, 4 b= 6, 4 A=6 b=4
(6)+4=2



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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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17 Sep 2015, 15:23
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pmklings wrote: a^2+2a24=0 b^2+2b24=0 (a+6)(a4) (b+6)(b4) a=6, 4 b= 6, 4 A=6 b=4
(6)+4=2 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, 14:18
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a(a+2) = 24eq1 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+2x24=0 Sum of roots of a quadratic equation is (coefficient of X)/(coefficient of x^2) = 2



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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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20 Sep 2015, 11:53
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If a(a+2)=24 and b(b+2)=24 a(a+2)=b(b+2) a^2+2a=b^2+2b a^2b^2=2b2a (ab)(a+b)=2(ba) a+b=2 (No need for smart numbers)
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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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18 Oct 2015, 11:59
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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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18 Oct 2015, 12:06
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Bunuel 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
Kudos for a correct solution. a(a + 2) = 24 and b(b + 2) = 24 => a, b must be integers and if a is 6 or 4, b will be 4 and 6 respectively => a+b = 2 Ans: 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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18 Oct 2015, 12:09
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Answer: B
Start by distributing a and b terms:
a^2 + 2a = 24 b^2 + 2b = 24
when
a = 4 b = 6
16 + 8 = 24 36  12 = 24
4  6 = 2



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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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18 Oct 2015, 14:06
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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+2ab^22b = 0 a^2  b^2 can be written as (a+b)(ab) +2a2b = 0 factor 2a2b> +2(ab) rewrite: (a+b)(ab)=2(ab) divide by (ab) and get a+b = 2



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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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18 Oct 2015, 20:52
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Bunuel 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
Kudos for a correct solution. as RHS is equal so we can equate LHS of both the eqns. \(a^2 + 2a = b^2 + 2b\) \(a^2  b^2 = 2(ba)\) \((a+b)(ab) = 2(ba)\) \(a+b =2\)
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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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18 Oct 2015, 21:49
Bunuel 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
Kudos for a correct solution. a(a + 2) = 24 => we have a = 4 or 6 also b(b + 2) = 24 => b = 4 or 6 given a ≠ b 1) when a =4, b= 6 and a + b = 2 1) when a = 6, b= 4 and a + b = 2 Answer choice 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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18 Jul 2016, 04:59
a(a+2)=24....(i) b(b+2)=24....(ii) from (i) & (ii) a(a+2)=b(b+2) or a^2b^2=2(ab) or, (a+b)(ab)=2(ab) or, (a+b)=2 ANS: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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12 Jul 2017, 10:11
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we can do in this way also 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 _________________ Please award kudos if you like my explanation. Thanks



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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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17 Sep 2017, 21:29
a(a + 2) = 24 a^2+2a24 = 0 a = 6 or 4 same way b= 6 or 4 since a not equal to b a+b must be 6+4 = 2 answer 2
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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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22 Sep 2017, 06:44
Quote: a(a + 2) = 24
a^2+2a24 = 0 a = 6 or 4
Let’s solve for a and b. a(a + 2) = 24 a^2 + 2a  24 = 0 (a + 6)(a  4) = 0 a = 6 or a = 4 and b(b + 2) = 24 b^2 + 2b  24 = 0 (b + 6)(b  4) = 0 b = 6 or b = 4 Since a ≠ b, if a = 6, then b = 4, and if a = 4, then b = 6. Either way, we see that a + b is equal to 2. Answer: 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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02 Dec 2017, 22:32
Hi everyone, just want to confirm my thinking here.
If a(a+2)=24, can't we just say that a=0 or 2? Thus the same for b, which a+b gives us 2. No need to solve the long way, or am I missing something?



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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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Re: If a(a + 2) = 24 and b(b + 2) = 24, where a ≠ b, then a + b =
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