Question

The sum of the squares of two consecutive even integers is 452. find the two integers.

Answer 1

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Let the two consecutive even integers are:-
x and x+2

according to the question ,we have
x^2 + (x+2)^2 =452
2x^2 + 4 + 4x =452
x^2 +2x -224=0

Now splitting the middle terms
x^2 + 16x -14x -224=0
we get
x=14
So the integers are 14 and 16.

I Hope it helps you!

Answer 2

Hi there! We can create a number equation to find your solution. Already, we know that two consecutive integers can be represented by $x+x+1$. All we have to do is square that expression and make it equal to 452. This gives us:
$(x)^{2} +(x+1)^{2} =452$
We can simplify that down to $2x^{2}+4x+4 =452$. Now, we can subtract 452 from both sides to make the whole left side equation equal to zero. Doing that gives us $2x^{2}+4x-448=0$. We see that 2 can be factored out of the whole equation, giving us $2(x^{2}+2x-224)=0$. Next, we can factor the equation to get $(x-16)(x+14)=0$. We can use the Zero Product Property to find the x values of -16 and 14. Therefore, the two integers are -16 and 14. Hope this helped!