Question

The sum of the squares of two consecutive negative integers is 61. Find the smaller of the two integers

Answer 1

$x^2+(x+1)^2=61\\ x^2+x^2+2x+1-61=0\\ 2x^2+2x-60=0\ \ /:2\\ x^2+x-30=0\\ \Delta=1^2-4\cdot(-40)=1+120=121\ \ \Rightarrow\ \ \sqrt{\Delta} =11\\ \\ x_1= \frac{-1-11}{2} = \frac{-12}{2} =-6,\ \ \ \ x_2= \frac{-1+11}{2} = \frac{10}{2}=5\\ \\Ans.:x=-6$