Polynomials Rational Roots
Find all positive integers $ k $ with the following property:
For all positive integers $ a, $ $ b, $ and $ c $ that make the roots of
\[ax^2 + bx + c = 0\]rational, the roots of $ 4ax^2 + 12bx + kc = 0 $ will also be rational.
Enter all the possible values of $ k, $ separated by commas.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
The roots of $ ax^2 + bx + c $ are rational if and only if the discriminant
\[b^2 - 4ac\]is a perfect square.
Similarly, the roots of $ 4ax^2 + 12bx + kc = 0 $ are rational if and only if its discriminant
\[(12b)^2 - 4(4a)(kc) = 144b^2 - 16kac\]is a perfect square.
To narrow down the possible values of $ k, $ we look at specific examples. Take $ a = 1, $ $ b = 6, $ and $ c = 9 $. Then $ b^2 - 4ac = 0 $ is a perfect square, and we want
\[144b^2 - 16kac = 5184 - 144k = 144 (36 - k)\]to be a perfect square, which means $ 36 - k $ is a perfect square. We can check that this occurs only for $ k = 11, $ 20, 27, 32, 35, and 36.
Now, take $ a = 3, $ $ b = 10, $ and $ c = 3 $. Then $ b^2 - 4ac = 64 $ is a perfect square, and we want
\[144b^2 - 16kac = 14400 - 144k = 144 (100 - k)\]to be a perfect square, which means $ 100 - k $ is a perfect square. We can check that this occurs only for $ k = 19, $ 36, 51, 64, 75, 84, 91, 96, 99, 100. The only number in both lists is $ k = 36 $.
And if $ b^2 - 4ac $ is a perfect square, then
\[144b^2 - 16kac = 144b^2 - 576ac = 144 (b^2 - 4ac)\]is a perfect square. Hence, the only such value of $ k $ is $ \boxed{36} $.
Freeflows