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Algebra

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5 Mar 2009 01:34:21 IST

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QUADRATIC

None

If a and b are integers and has discriminant as a perfect square, then prove that its roots are integers.

Please give the proof.

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Sahil Gupta

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5 Mar 2009 01:58:55 IST

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Case 1: both a and b are odd; D (a²-4b)^1/2 is odd, roots are (odd +/- odd)/2 i.e. integer

Case 2: a odd, b even; same as above

Case 3; a even, b odd; D even, roots are (even +/- even)/2 again integer

Case 4: both are even; same as above

SMARTY

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5 Mar 2009 02:29:16 IST

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If Discriminant =0(perfect square)sqrt(a^2-4b)=0a^2=4bRoots are -a/2,-a/2 which are integerssince discriminant is zeroMethod:ax^2+bx+c=0 is quadratic in xRoot of quadratic=(-b+sqrt(b^2-4ac))/2aand (-b-sqrt(b^2-4ac))/2a

SMARTY

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5 Mar 2009 02:30:49 IST

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If Discriminant =0(perfect square)

sqrt(a^2-4b)=0

a^2=4b

Roots are -a/2,-a/2 which are integers since discriminant is zero

Method:

ax^2+bx+c=0 is quadratic in x

Then Root of quadratic= ((-b+sqrt(b^2-4ac))/2a) and ((-b-sqrt(b^2-4ac))/2a)

XYZ

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5 Mar 2009 03:37:31 IST

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SMARTY, first of all it is given that the discriminant is a perfect square. It is not given that it is equal to zero. You have taken a case when the discriminant is zero. To prove something, we have to prove it for every case, not just by taking a specific case when D=0.

Also you have got the equal roots -a/2 and you say that it is an integer.

Consider a=1 (which is an integer)

Then root is -1/2, which is not an integer.

Sahil has given the correct answer. Thanks a lot for helping me. A hat from my side.

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