Question

the base of triangle is( x²+y²) and its height is 12 xy find the area of a triangle

Answer 1

Let's find some restraints. So be:

a=x2+x+1a=x2+x+1

b=2x+1b=2x+1

c=x2−1c=x2−1

For a<b+ca<b+c :

x2+x+1<2x+1+x2−1∴0<x−1∴x>1x2+x+1<2x+1+x2−1∴0<x−1∴x>1

For b<a+cb<a+c:

2x+1<x2+x+1+x2−1∴2x2−x−1>02x+1<x2+x+1+x2−1∴2x2−x−1>0

The roots for the quadratic equation are:

x=−(−1)±(−1)2−4×2×(−1)−−−−−−−−−−−−−−−−−√2×2=1±34={1,−12}x=−(−1)±(−1)2−4×2×(−1)2×2=1±34={1,−12}

So,  x<−12∪x>1x<−12∪x>1

For c<a+bc<a+b

x2−1<x2+x+1+2x+1x2−1<x2+x+1+2x+1

−3<3x−3<3x

x>−1x>−1

So, it is a triangle for all x>1