Waste less time on Facebook — follow Brilliant.
×

Polynomial Sprint extra question -2

Prove that if the co-efficients of the quadratic equation \(ax^{2}+bx+c = 0\) are odd integers,the roots of the equations cannot be rational numbers.

Note by Eddie The Head
3 years, 5 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

Let a = 2d + 1, b = 2e + 1, c = 2f + 1 where d,e,f are integers.

Using the quadratic formula we get x = (-b +- sqrt{b^2 - 4ac})/(2a). We will show that b^2 - 4ac cannot be a square number.

Substituting our above equations (a = 2d + 1, b = 2e + 1, c = 2f + 1) into b^2 - 4ac we get 4e^2 + 4e - 16df - 8d - 8f - 3. We can rewrite this as (8)(-2df - d - f) + (4)(e)(e + 1) - 3. Since one of (e) and (e+1) must be even, it follows that 4(e)(e+1) must be divisible by 8. Therefore b^2 - 4ac is congruent to 5 modulo 8. This means that b^2 - 4ac is not a square number, as square numbers must be congruent to 0, 1, or 4 mod 8. This shows that x is irrational.

Colin Tang - 3 years, 5 months ago

Log in to reply

Suppose that p/q is a root of the quadratic equation and that p,q are coprime, then c is divisible by p and a is divisible by q. Hence both p and q are odd. Substituting p/q for x, we get: \(\frac{ap^2}{q^2} + \frac{p}{q} + c = 0 \). However, this is clearly impossible as both p, q are odd. Hence there are no rational roots for the equation.

Jianzhi Wang - 3 years, 5 months ago

Log in to reply

There are 3 odd terms in the final equation. If all of them are positive, then we are done. If one is negative, then the LHS will still be an odd number (odd + odd - odd = odd != 0).

Jianzhi Wang - 3 years, 5 months ago

Log in to reply

Why is it impossible?

Eddie The Head - 3 years, 5 months ago

Log in to reply

Could you clarify please? Why is this "clearly impossible"? You can rewrite it as ((a/q)(p)(p) + p)/q = -c where a/q is an odd integer (a is divisible by q, a and q are both odd) and p and q and c are all odd integers--how do you know that the fraction isn't simplifiable to -c

Never mind, I think I've figured it out. Since a/q is odd and p is odd, (a/q)(p)(p) is odd and p is odd, which makes (a/q)(p)(p) + p even. You want (a/q)(p)(p) + p to equal -cq, which is a product of two odd integers and therefore odd. But you can't have an even number equal to an odd number.

Colin Tang - 3 years, 5 months ago

Log in to reply

He means that if we multiply by \(q^2\), LHS will be odd and RHS =0, clearly a contradiction.

Bogdan Simeonov - 3 years, 5 months ago

Log in to reply

@Bogdan Simeonov If the coefficients are odd integers, they can be negative, right? We could somehow choose the coefficients so that LHS is zero. However, if we write the equation as \(cq^{2} = -(ap^{2} + bpq)\) and see different cases by taking \(p, q\) to be even or odd, we see that the equation may be satisfied only when both \(p, q\) are even. But that contradicts the condition that \(p, q\) are coprime.

This is what Colin says.

Siladitya Basu - 3 years, 5 months ago

Log in to reply

@Bogdan Simeonov Ok that makes sense now Thanks

Colin Tang - 3 years, 5 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...