Do You Know About Roots?
a x 2 + b x + c = 0 is called a quadratic equation if a = 0 . Which of the following statements is/are true concerning quadratic equations?
[ 1 ] A quadratic equation always has 2 distinct roots.
[ 2 ] If a , b and c are all odd numbers, it is possible for a x 2 + b x + c = 0 to have a rational solution.
[ 3 ] If a x 2 + b x + c = 0 has real solutions and a , b , c > 0 , then both solutions must be negative.
Note : This problem is a part of the set "I Don't Have a Good Name For This Yet". See the rest of the problems here . And when I say I don't have a good name for this yet, I mean it. If you like problems like these and have a cool name for this set, feel free to comment here .
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2 solutions
2) If b 2 − 4 a c = p 2 , where p is an integer, then if a, b, c are odd, p must be odd. Let b = 2B + 1 and p = 2P + 1, and rearranging and dividing by 4, we end up with B 2 + B − P 2 − P = a c . The left side is always even, but the right is odd, which is not possible.
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Nice!
I also went for a contradiction proof. If there existed a rational solution q p where p and q are co-prime positive integers, then a ( q p ) 2 + b q p + c = 0 .
Or, a p 2 + b p q + c q 2 = 0 .
Since p and q are co-prime, one of them has to be odd, so the left hand side will always be odd and never be equal to zero, a contradiction.
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@Michael Mendrin and @Mursalin Habib , since the solution both of you provided are clear and nice, would you mind if I edit my [2] to be the same as yours?
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@Christopher Boo – Boo, I think that's the idea of the Brilliant format, collaboration on getting at least one good, readable solution for others to check up on, so go right ahead. And, anyway, copyrighting this stuff costs too much.
if d =0 then the roots will be same
Okay, is it impossible for a x 2 + b x + c = 0 to have a rational solution if a , b , c are odd numbers?
My statement wasn't clear enough. I have edited it. Now can you post a solution?
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Oh sorry if I misunderstood your problem, I will try to edit the solution.
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Actually, it's my fault. The wording was ambiguous. Try it now.
I failed on the second statement, I couldn't prove it but thanks.I only proved it for 1 number and got rational solution. @Christopher Boo @Michael Mendrin and @Mursalin Habib
Here's another solution for b). A solution requires that d 2 = b 2 − 4 a c for some odd integers a , b , c , d . But it is not difficult to prove that a difference of odd squares must be divided by 8, which is impossible since a c is odd.
Another way to look at [3] is of course the Descartes' Rule of Signs, which tells us there are no positive solutions and 2 or 0 negative solutions, but, since only real solutions exist, there are 2 negative solutions.
I don't think this is a Level 4 problem. More like Level 1 or 2.
This solution will only be completed with the help from Michael Mendrin and Mursalin Habib .
[1] When b 2 − 4 a c = 0 , a quadratic equation will only have one root, where the vertex of the parabola is tangent to the x-axis. [1] is wrong.
[2] Assume that there exists an integer p , such that b 2 − 4 a c = p 2 , then if a , b , c are odd, p must be odd. Let b = 2 B + 1 and p = 2 P + 1 , and rearranging the expression we get B 2 + B − P 2 − P = a c . L H S is always even but R H S is odd. This leads to a contradiction. [2] is wrong.
[3] If a , b , c > 0 and x is a positive number, the expression a x 2 + b x + c must be a positive number. Even if x = 0 , the expression will also be number since c > 0 . Hence x must be a negative number if the equation has real solutions. [3] is correct.