Question

The square of the sum of two consecutive positive even integers is 4048 more than the sum of the squares of these two numbers. Find the two numbers.
PLZZZZZ I NEED LE HELP FROM LE FELLOW STUDENTS!!!

Answer 1

Answer:

The numbers are 44 and 46

Step-by-step explanation:

Let

x, x+2 ----> the two consecutive positive even integers

we know that

$(x+x+2)^{2} =4,048+x^{2} +(x+2)^{2} \\ \\(2x+2)^{2} =4,048+x^{2} +x^{2}+4x+4\\ \\4x^{2}+8x+4=2x^{2}+4x+4,052\\ \\2x^{2} +4x-4,048=0$

Solve the quadratic equation using a graphing calculator

The solution is x=44

see the attached figure

x+2=44+2=46

therefore

The numbers are 44 and 46