Answer:
The two consecutive positive integers are [tex]4[/tex] and [tex]6[/tex].
Step-by-step explanation:
Let [tex]x[/tex] be an even integer, then the next even integer is [tex](x+2)[/tex].
The sum of the squares of these even integers is [tex]52[/tex].
We can write the following equation and solve for [tex]x[/tex].
[tex]x^2+(x+2)^2=52[/tex].
We expand to get,
[tex]x^2+x^2+4x+4=52[/tex].
We simplify to get,
[tex]2x^2+4x+4-52=0[/tex].
This gives us,
[tex]2x^2+4x-48=0[/tex]
We divide through by 2 to get,
[tex]x^2+2x-24=0[/tex].
This is now a quadratic trinomial
We split the middle term to get,
[tex]x^2+6x-4x-24=0[/tex].
We factor to obtain,
[tex]x(x+6)-4(x+6)=0[/tex]
[tex]\Rightarrow (x+6)(x-4)=0[/tex]
[tex]\Rightarrow (x+6)=0 or (x-4)=0[/tex]
[tex]\Rightarrow x=-6 or x=4[/tex]
We discard [tex]x=-6[/tex] since it is not a positive integer.
[tex]\therefore x=4[/tex]
Hence the two consecutive positive even integers are,
[tex]4[/tex]
and
[tex]4+2=6[/tex]
Answer:
4 and 6
Step-by-step explanation:
one number = x
other number = x + 2
x² + (x + 2)² = 52
x² + x² + 4x + 4 = 52
2x² + 4x - 48 = 0
x² + 2x - 24 = 0
(x - 4)(x + 6) = 0
x = 4 or x = -6
Since x is positive
x = 4
x + 2 = 6
