Question

Two consecutive even positive integers, sum of the squares of which is $1060$ are.

• A. $16$ and $18$
• B. $12$ and $14$
• C. $22$ and $24$
• D. $20$ and $22$

Answer 1

The correct option is C $22$ and $24$

Let the two consecutive positive even integers be $x$ and $x+2$.

It is given that the sum of the squares of the integers is $1060$, therefore, we have:

$x^2+(x+2{)}^2=1060\Rightarrow x^2+[x^2+2^2+(2\times x\times 2)]=1060(\because (a+b{)}^2=a^2+b^2+2ab)\Rightarrow x^2+x^2+4+4x=1060\Rightarrow 2x^2+4x+4=1060\Rightarrow 2{(x}^2+2x+2)=1060\Rightarrow x^2+2x+2=\frac{1060}{2}\Rightarrow x^2+2x+2=530\Rightarrow x^2+2x+2-530=0\Rightarrow x^2+2x-528=0$

Now, we factorize the resulting equation as shown below:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-2\pm \sqrt{2^2-(4\times 1\times -528)}}{2\times 1}=\frac{-2\pm \sqrt{4+2112}}{2}=\frac{-2\pm \sqrt{2116}}{2}=\frac{-2\pm 46}{2}\Rightarrow x=\frac{-2+46}{2},x=\frac{-2-46}{2}\Rightarrow x=\frac{44}{2},x=-\frac{48}{2}\Rightarrow x=22,x=-24$

We always consider the positive value, therefore, $x=22$ and the other integer is:

$x+2=22+2=24$

Hence, the integers are $22$ and $24$.