Java program to find integer root of a quadratic equation

Question

So, here is my requirement. If a quadratic equation has two roots(an int & a float), I want to take only integer value for further manipulation. I can't figure it out how it's made. Can anyone tell me please. (Java would be better).

Answer 1

[I do not use Java regularly. Here is a solution using C. As only elementary concepts are used, a Java practitioner should be able to translate it readily.]

Searching the web for “sum of cubes” reveals this page which tells us the sum of k³ for k from 1 to n is n²•(n+1)²/4.

That is a quartic equation, for which closed-form solutions are known, but we easily see that, for positive n, n²•(n+1)²/4 is between n⁴/4 and (n+1)⁴/4. Then, if m is the sum of the first n cubes, n = floor((4•m)^1/4). So, if we have a pow implementation that is faithfully rounded using round-to-nearest (the computed result is one of the two representable values nearest the mathematical result), we can find n with floor(pow(4*m, .25)). If pow is not faithfully rounded, then round(pow(4*m, .25)) will serve over the domain for which pow returns some reasonable result without too much error. (round works because (4•m)^1/4 never exceeds n by more than ½. Proof omitted, although Wolfram Alpha shows us the limit as n goes to ∞ is ½, and the excess is monotonic.)

Thus, if m is the sum of the first n cubes, then n is the result of round(pow(4*m, .25)). So we can compute this value for n, then compute the sum of the first n cubes as n*n*(n+1)*(n+1)/4 and test whether that equals m. If it does, we found a solution and return it. If it does not, m is not a sum of cubes, and we return −1:

#include <math.h>
#include <stdio.h>

static double findNb(double m)
{
    double n = round(pow(4*m, .25));
    double sum = n * n * (n+1) * (n+1) / 4;
    return m == sum ? n : -1;
}

static void Test(double m)
{
    printf("findNb(%.99g) -> %.99g.\n", m, findNb(m));
}

int main(void)
{
    Test(0);
    Test(1);
    Test(2);
    Test(8);
    Test(9);
    Test(10);
    Test(250500249999.);
    Test(250500250000.);
    Test(250500250001.);
}

Output:

findNb(0) -> 0.
findNb(1) -> 1.
findNb(2) -> -1.
findNb(8) -> -1.
findNb(9) -> 2.
findNb(10) -> -1.
findNb(250500249999) -> -1.
findNb(250500250000) -> 1000.
findNb(250500250001) -> -1.

Of course, the limits of floating-point precision will cause this code to fail once m is larger than can be represented in double.

Answer 2

Use the basic quadatic formula to find the roots.

Set the roots to different values (both doubles)

Use modulous (%) by 1 and cast the value to a double. If the double calculated is !=0, then it is not an int.