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

Sagot :

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

[tex]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[/tex]

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

[tex]x=4 \\ x+2=4+2=6[/tex]

The two integers are 4 and 6.
konrad509
[tex](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=-\dfrac{3}{2} \vee n=2\\ -\dfrac{3}{2}\not \in \mathbb{Z}\Rightarrow n=2\\\\ 2n=4\\ 2n+2=6\\\\ \text{These numbers are 4 and 6.} [/tex]
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