the square if the sum of two positive consecutive even numbers exceeds the sum of their squares by 336

Question



Answer 1

Let x = 1st of two consecutive positive even integers.

Then x + 2 represents the next.

Their sum squared is (x+x+2)² which is 4x² + 8x + 4.

Their squares are respectively x² and (x + 2)², so the sum of the squares is x² + x² + 4x + 4, or 2x²+4x +4.

The former is 336 more than the latter (another way of saying exceeds by 336.)

So, 4x²+8x +4 = 2x² + 4x + 4 + 336

Now you have a quadratic equation. Set it equal to zero, solve by factoring (rejecting any negative result). Check to see if you answers satisfy the requirements of the problem.

Answer 2

Given by the question we can let n represent an arbitrary positive, even number and n+2 represents the consecutively next positive, even number. We can also interpret the exceeding of sum as meaning that it is 336 greater than the sum of their squares. With this representation, we can write the equation as the following:

(n+n+2)² -336=n² +(n+2)²

Then, all that is left to do is to solve for n.

(n+n+2)² -336=n² +(n+2)² Given

4n² +8n+4 -336 = 2n² +4n+4 Expand out terms

2n² +4n-336=0 Move all terms to LHS

2(n² +2n -168) = 0 Factor out common constant (has no impact on answer and makes numbers nicer)

Then we can plug the auxiliary equation (n² +2n -168) into the quadratic formula to find the factors.

This yields the following:

[-2±√2² - 4(1)(-168)]/(2(1)).

This gives us two answers for n because the discriminant ≠ 0 and is > 0 meaning our answers are real numbers. In order to be the correct n, it must be a positive even number. Only one of the factors will satisfy this, that is our smaller number and the larger number is 2 greater than that and we are done.