Just lost my battle
Find the sum of all integral values of a for which the quadratic equation ( x − a ) ( x − 1 0 ) + 1 = 0 has integral roots.
The answer is 20.
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2 solutions
Brilliant approach! (+1)
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Haha... You must have done the same way.. :-)
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No, this time I didn't :P
I took the long way.
When a is even, x = 2 1 0 + a is the integral solution if
( 2 1 0 + a ) 2 = 1 0 a + 1 .
From this, we get a = 8 , 1 2 .
( x − a ) ( x − 1 0 ) + 1 ( x − a ) ( x − 1 0 ) = 0 = − 1
For integer solutions, we have:
( x − a ) ( x − 1 0 ) = ⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎧ ( 1 ) ( − 1 ) ( − 1 ) ( 1 ) ⇒ { x − 1 0 = − 1 x − a = 1 ⇒ x = 9 ⇒ a = 8 ⇒ { x − 1 0 = 1 x − a = − 1 ⇒ x = 1 1 ⇒ a = 1 2
The required answer is 8 + 1 2 = 2 0 .
Let a integral root be α .
∴ ( α − a ) ( α − 1 0 ) + 1 = 0 ⟹ − 1 ( α − a ) 1 ( α − 1 0 ) = − 1 O R 1 ( α − a ) − 1 ( α − 1 0 ) = − 1 ( ∗ ∗ ) ⟹ a = 8 , 1 2 8 + 1 2 = 2 0
( ∗ ∗ Here I have used the fact that product of two integers can be − 1 only when they are + 1 and − 1 or vice versa.)