Discriminate carefully
The following equation has two solutions x 1 and x 2 : x 2 + b x + c = 0 , where b and c are real numbers. Someone claims,
"In order for both x 1 and x 2 to be rational numbers, it is necessary and sufficient that the discriminant D = b 2 − 4 c is the square of a rational number."
It this true or false?
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1 solution
Perhaps I'm not understanding this, but I can't understand why you say D = ( x 1 − x 2 ) 2 . It doesn't seem to be true at all in general. For example if you take 2 x 2 + x − 1 , you get the roots − 1 and 1 / 2 . Therefore by your formula D = 2 . 2 5 but b 2 − 4 c = 5 . (also should it not be b 2 − 4 a c ? But that also != 2.25)
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The problem was limited to quadratics of the form x 2 + b x + c = 0 , i.e. with a = 1 .
More generally, for a x 2 + b x + c = 0 we have D = a 2 ( x 1 − x 2 ) 2 . With the example you give, D = a 2 ( x 1 − x 2 ) 2 = 2 2 ⋅ ( − 1 − ( − 2 1 ) ) 2 = 4 ⋅ 2 4 1 = 9 , and D = b 2 − 4 a c = 1 2 − 4 ⋅ 2 ⋅ ( − 1 ) = 1 − ( − 8 ) = 9 .
Fact: D = ( x 1 − x 2 ) 2 .
If x 1 , x 2 are rational, then so is x 1 − x 2 , so that D = ( x 1 − x 2 ) 2 is the square of a rational number. Thus the condition is necessary ; if D is not the square of a rational number, then x 1 and x 2 are not both rational.
But the converse is not true. Consider the equation x 2 − 2 2 x + 1 = 0 . The discriminant is D = 4 = 2 2 is the square of a rational number, but the solutions x = 2 ± 1 are not rational. Thus the condition is not sufficient .
A necessary and sufficient condition would be, that D is the square of a rational number, and b is a rational number . Proof: If D = d 2 , then x 1 , 2 = 2 1 ( − b ± d ) , which is obviously rational if b and d are.