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

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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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New post 14 Sep 2015, 06:29
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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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New post 14 Sep 2015, 07:00
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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


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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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New post 20 Sep 2015, 12:53
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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^2-b^2=2b-2a
(a-b)(a+b)=2(b-a)
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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New post 16 Sep 2015, 23:31
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a^2+2a-24=0 b^2+2b-24=0
(a+6)(a-4) (b+6)(b-4)
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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New post 17 Sep 2015, 16:23
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pmklings wrote:
a^2+2a-24=0 b^2+2b-24=0
(a+6)(a-4) (b+6)(b-4)
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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New post 19 Sep 2015, 15:18
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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
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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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New post 18 Oct 2015, 12:59
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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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New post 18 Oct 2015, 13:06
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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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New post 18 Oct 2015, 13:09
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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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New post 18 Oct 2015, 15:06
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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+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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Re: If a(a + 2) = 24 and b(b + 2) = 24, where a ≠ b, then a + b =  [#permalink]

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New post 18 Oct 2015, 21:52
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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.

\(a^2 + 2a = b^2 + 2b\)
\(a^2 - b^2 = 2(b-a)\)
\((a+b)(a-b) = 2(b-a)\)
\(a+b =-2\)
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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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New post 18 Oct 2015, 22:49
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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

=> 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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New post 18 Jul 2016, 05:59
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a(a+2)=24....(i)
b(b+2)=24....(ii)
from (i) & (ii)
a(a+2)=b(b+2)
or a^2-b^2=-2(a-b)
or, (a+b)(a-b)=-2(a-b)
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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New post 12 Jul 2017, 11:11
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we can do in this way also :-D

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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New post 17 Sep 2017, 22:29
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a(a + 2) = 24
a^2+2a-24 = 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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New post 22 Sep 2017, 07:44
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Quote:
a(a + 2) = 24

a^2+2a-24 = 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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New post 02 Dec 2017, 23:32
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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New post 02 Dec 2017, 23:37
emilycwaters wrote:
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?


Does a = 0 or a = -2 satisfy a(a + 2) = 24. Foe both of these values a(a + 2) is 0, NOT 24.

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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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New post 27 Jan 2019, 05:01
Here's my solution
If a(a + 2) = 24 and b(b + 2) = 24
Then \(a^2\) +2a =\(b^2\) +2b
Therefore, \(a^2\) - \(b^2\) = 2b-2a
(a+b) (a-b) = 2(b-a)
so, a+b = -2
Is this a right approach? Bunuel ScottTargetTestPrep
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If a(a + 2) = 24 and b(b + 2) = 24, where a ≠ b, then a + b =   [#permalink] 27 Jan 2019, 05:01
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