CareerCup

public static double[] findNumbers(int sum, int product){
		/**
		 * Given the sum of two numbers & their product, return the two numbers.
		 * a * b = 20 
		 * a + b = 10
		 * b = 10 - a
		 * 
		 * 10a - a^2 = 20
		 * a^2 -10a + 20, now solve with quadratic formula
		 */
		double[] result = new double[2];
		double tmp = Math.sqrt(Math.pow(sum , 2) - 4*(product));
	  	result[0] = (sum + tmp)/2;
	  	result[1] = (sum - tmp)/2;
		return result;
	}

Hello,

let the 2 nos be a and b so the given is

a*b = (say)8 -- equation1
a+b=6. -- equation 2

You can keep 8 and 6 and x and y too.

so a = 6-b => substitute this in equation 1

=> (6-b)*b = 8
=>6b-b^2 = 8
=> b^2 - 6b + 8 = 0

Now this becomes a quadratic equation and will always be in such problems. solve get the value of b and by substituting that into either of equations above of ur choice will get the value of a.

So in this case
(b-4)(b-2)=0 so the o/p is 4 and 2

Edit:::::
In short, in such problems try to reduce the number of variables to one and try to represent others in term of only one variable. Then that variable will help you to find out rest of the values.

I think all you need to do is get the second number and find it's factors and also add them to check if there sum is the first number.

For eg:8,12

1*12=12
1+12!=8

Continue...

2*6=12
2+6=8
Break...

The solution can be modified to include negative integers as well...

package CareerCup;

/**
 * Created by Deven on 17/08/15.
 */
public class SumProduct {

    public static void main(String[] args) {
        int sum=6,product=8;

        int a,b;


        a=(sum +(int)Math.sqrt((sum*sum)-(4*product)))/2;
        b=(sum -(int)Math.sqrt((sum*sum)-(4*product)))/2;
        System.out.println("Answers are:" + a + " "+ b);
    }
}

package CareerCup;

/**
 * Created by Deven on 17/08/15.
 */
public class SumProduct {

    public static void main(String[] args) {
        int sum=6,product=8;

        int a,b;


        a=(sum +(int)Math.sqrt((sum*sum)-(4*product)))/2;
        b=(sum -(int)Math.sqrt((sum*sum)-(4*product)))/2;
        System.out.println("Answers are:" + a + " "+ b);
    }
}

Simplify the equations:
a+b=x --> a=x-b
a*b=y --> a=y/b

Substituting (x-b) for 'a' in the equation and solving the quadratic equation yields:
(x-b)*b=y --> b^2 -x*b + y =0 --> b=(x+-(x^2-4*y)^(1/2))/2

a, b == (x-(x^2-4*y)^(1/2))/2, (x+(x^2-4*y)^(1/2))/2

Validating with the numbers from the example
a=(6-(36 - 32)^(1/2))/2 --> a=(6-2)2 -->a=2
b=(6+(36 - 32)^(1/2))/2 --> b=(6+2)2 -->b=4

Solve for 'a' in the equations:
a+b=x --> a=x-b
a*b=y --> a=y/b

Substituting (x-b) for 'a' in the equation and solving the quadrtic equation yeilds:
(x-b)*b=y --> b^2 -xb + y =0 --> b=(x+-(x^2-4*y)^(1/2))/2
a, b == (x-(x^2-4*y)^(1/2))/2, (x+(x^2-4y)^(1/2))/2

Validating with the numbers from the example
a=(6-(36 - 32)^(1/2))/2 --> a=(6-2)2 --> a=2
b=(6+(36 - 32)^(1/2))/2 --> b=(6+-2)2 --> b=4

Hi
Here's the code

package CareerCup;

/**
 * Created by Deven on 17/08/15.
 */
public class SumProduct {

    public static void main(String[] args) {
        int sum=1037,product=29232;

        int a,b;


        a=(sum +(int)Math.sqrt((sum*sum)-(4*product)))/2;
        b=(sum -(int)Math.sqrt((sum*sum)-(4*product)))/2;
        System.out.println("Answers are:" + a + " "+ b);
    }
}

Compute the quadratic result for values:

public static void printComponents(int sum, int product){
    int sqrRt = (int)Math.sqrt((sum * sum) - (4 * product));
    int posA = (-sum + sqrRt) / 2;
    int posB = product / posA;
    if(posA + posB != sum ){
        posA = (-sum - sqrRt) / 2;
        posB = product / posA;
    }
    java.lang.System.out.println(""+posA+", "+posB);
}