Question

Find two consecutive even integers such that the square of the smaller is 10 more than the larger

Answer 1

`(2n)^2=2n+2+10\\ 4n^2=2n+12\\ 4n^2-2n-12=0\\ 2n^2-n-6=0\\ 2n^2-4n+3n-6=0\\ 2n(n-2)+3(n-2)=0\\ (2n+3)(n-2)=0\\ n=-(3)/(2) \vee n=2\\ -(3)/(2)\\ot \in \mathbb{Z}\Rightarrow n=2\\\\ 2n=4\\ 2n+2=6\\\\ \text{These numbers are 4 and 6.}`

Answer 2

x, x+2 - two consecutive even integers
x is the smaller one


`x^2=(x+2)+10 \\ x^2=x+12 \\ x^2-x-12=0 \\ x^2-4x+3x-12=0 \\ x(x-4)+3(x-4)=0 \\ (x+3)(x-4)=0 \\ x+3=0 \ \lor \ x-4=0 \\ x=-3 \ \lor \ x=4`

-3 isn't an even integer, so x=4.


`x=4 \\ x+2=4+2=6`

The two integers are 4 and 6.