Question

The sum of the squares of two consecutive positive even numbers is 20.
What is the smaller number? @Noura11

Answer 1

Let a and b the two consecutive positive even numbers.
Then :
\[a=2n\text{ and } b=a+2=2n+2\]
So we have :
\[a^2+b^2=20\]
Then :
\[4n^2+(2n+2)^2=20\]
Then :
\[4n^2+4n^2+8n+4=20\]
Then :
\[8n^2+8n-16\]
We divide by 8 :
\[n^2+n-2=0\]
Then :
\[\Delta= 1-4\times(-2)=9\]
So :
\[n=\frac{1+\sqrt9}{2}=\frac{4}{2}=2\]
So :
\[a=2n=2\times2=4\]

Answer 2

but whts the answer

Answer 3

Ok there is an error

Answer 4

wht is the answer

Answer 5

I have to write :
\[n=\frac{-1+\sqrt9}{2}=1\]
So the smaller number will be :
\[a=2\times 1=2\]

Answer 6

so the answer is 2 right am i correct or no

Answer 7

Yes, it is 2

Answer 8

i posted another one