b^2(b+2m)^2=(b+m)^4 ,,,,,,, find b(m)

Question

b²(b+2m)²=(b+m)⁴

,,,,,,,

find b(m)

Answer 1

Hope this is helpful!

Answer 2

(b^2)(b+2m)^2 = (b+m)^4 solve for m, m= 0 then (b^2)(b^2) = b^4 and b(m) = b(0) = 0

so, b(m) = 0

but 4th degree equations have possibly 4 solutions

n degree equations have potentially n solutions

Answer 3

I found this problem on "bymath dot com" progressions ans sequences problems.. 6.018. It is a great math site.

a good way to solve b²(b+2m)²=(b+m)⁴ is:[(b+m)/b]²=[(b+2m)/(b+m)]²

[1+m/b]=±[1+(m/(b+m)]

[1+m/b]=+[1+(m/(b+m)],,,b²=0 trivial answer

[1+m/b]=-[1+(m/(b+m)]

2+(m/(b+m))+m/b=0

2b²+2bm+bm+bm+m²=0

b₁=[-1+(√2)/2]m

b₂=[-1-(√2)/2]m

for either b, for any m, have b and y=mx+b line equation. So would have points (0,b),

(1,b+m), (2,b+2m) which form an arithmetic sub sequence.

notice (0,b²), (1,(b+m)²), (2,(b+2m)²) form a geometric sub sequence

for the exponential equations y₁=b₁²(3+2√2)^x and

y₂=b₂²(3-2(2√)^x

for b₁=(((√2)/2))-1)m

(0,b)=(0,(((√2)/2))-1)m),,,

(0,b²)=(0,[3/2]-√2])m²)

(1,b+m)=(1,(m√2)/2),,,,(

1,(b+m)²)=(1,m²/2)

(b+m)²/b²=(1/2)(3/2]-√2)^-1=(3+2√2)

(2,(b+2m))=(2,(((√2)/2))+1)m),,,,

(2,(b+2m)²)=(2,[3/2]+√2])m²)

(b+2m)²/(b+m)²=[(3/2)+√2]/[1/2]=(3+2√2)

y₁=b₁²(3+2√2)^x