Question

Ans: x €(-0, -2] [1,0)
13. The A.M: between two numbers exceeds their G.M. by 2 and
the G.M. exceeds the H. M by 1.8. Find the numbers.​

Answer 1

Answer:

Suppose am=gm+u and gm=hm+v with u,v≠0 and u≠v.

Since hm≤gm≤am, u≥0 and v≥0.

Since gm2=am⋅hm, gm2=(gm+u)(gm−v)=gm2+gm(u−v)−uv so gm(u−v)=uv. Therefore u>v and gm=uvu−v.

Then am=gm+u=uvu−v+u=uv+u(u−v)u−v+u=u2u−v and hm=gm−v=uvu−v−v=uv−v(u−v)u−v=v2u−v.

If there are only two values, a and b, then a+b2=u2u−v and ab−−√=uvu−v so b=u2v2a(u−v)2 and u2u−v=a+u2v2a(u−v)22=a2(u−v)2+u2v22a(u−v)2 or 2au2(u−v)=a2(u−v)2+u2v2.

Solving a2(u−v)2−2u2(u−v)a+u2v2=0,

aandb=2u2(u−v)±4u4(u−v)2−4(u−v)2u2v2−−−−−−−−−−−−−−−−−−−−−−√2(u−v)2=u2(u−v)±u(u−v)u2−v2−−−−−−√(u−v)2=u2±uu2−v2−−−−−−√u−v=uu±u2−v2−−−−−−√u−v=2u2u−v−a=2u2u−v−u2±uu2−v2−−−−−−√u−v=2u2−u2∓uu2−v2−−−−−−√u−v=u2∓uu2−v2−−−−−−√u−v

For this case, u=2 and v=1.6.

gm=2⋅1.62−1.6=3.2.4=8.

To get a and b, u2−v2−−−−−−√=4−2.56−−−−−−−√=1.44−−−−√=1.2 so a=22±1.2.4=23.2,.8.4=2(8,2)=(16,4) and b=(4,16).