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705-805 (Hard)|   Algebra|      
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Bunuel
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If a and b are positive integers such that a^2 + 2b = b^2 + 2a + 5, fi 15 Jun 2021, 05:45
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75% (hard)

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55% (02:26) correct 45% (02:08) wrong based on 291 sessions

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If a and b are positive integers such that \(a^2 + 2b = b^2 + 2a + 5\), find the value of b.

(A) 0
(B) 1
(C) 2
(D) 3
(E) Cannot be determined
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D
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If a and b are positive integers such that a^2 + 2b = b^2 + 2a + 5, fi 15 Jun 2021, 09:22
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My approach :


Please find attached image, it was too much to type.

Posted from my mobile device
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If a and b are positive integers such that a^2 + 2b = b^2 + 2a + 5, fi 15 Jun 2021, 10:25
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Bunuel
If a and b are positive integers such that \(a^2 + 2b = b^2 + 2a + 5\), find the value of b.

(A) 0
(B) 1
(C) 2
(D) 3
(E) Cannot be determined
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\(a^2\) and \(b^2\) in the equation, look if you can use \(a^2-b^2\). Most of the time, the answer will be yes.

\(a^2 + 2b = b^2 + 2a + 5\)

\(a^2 -b^2 = 2a -2b+ 5\)

\((a-b)(a+b)=2(a-b)+5\)

\((a-b)(a+b)-2(a-b)=5\)

\((a-b)(a+b-2)=5=1*5\)

As a and b are positive integers, \(a+b\geq 2\), so \(a+b-2\geq{0}\). This means both (a-b) and (a+b-2) must be some positive integers too. and only possibility that their product is 5 is 1 and 5.

1) a-b=1 and a+b-2=5
Add the two => 2a=5+2+1=8...a=4, and b=3.

2) a-b=5 and a+b-2=1
Add the two => 2a=5+2+1=8...a=4, and b=a-5=-1, but b is positive...Not possible.

Thus b=3.

D
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If a and b are positive integers such that a^2 + 2b = b^2 + 2a + 5, fi 15 Jun 2021, 22:49
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\(a^2 + 2b = b^2 + 2a + 5\)

=> \(a^2 - b^2 + 2b - 2a = 5\)

=> (a - b) (a + b) - 2 (a - b) = 5

=> (a - b) [a + b - 2] = 5 * 1

Case I: a - b = 5 or a + b - 2 = 1

=> a - b = 5 and a + b = 3

On adding both , we get 2a = 8 or a = 4 and hence b = -1 (Not possible)



Case II: a - b = 1 or a + b - 2 = 5

=> a - b = 1 and a + b = 7

On adding both , we get 2a = 8 or a = 4 and hence b = 3

Answer D
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If a and b are positive integers such that a^2 + 2b = b^2 + 2a + 5, fi 05 Feb 2025, 13:16
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\(a^2+2a-5=b^2-2b\)

We can substitute the value of b and check for LHS=RHS for the above equation

checking all the 4 no. takes less time than following the algebraic process.

substituting 3 in the above equation:

\(a^2+2a-5=(3)^2-2*3\)

\(a^2+2a-5=9-6=3\)

\(2^2+2*2-5=3\)

\(3=3\)

same holds true for value of 'a' we need to check any value of 'a' that can give the equivalent value as LHS
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