(x⋅y)=x+y2(x \cdot y) = \frac{x+y}{2}(x⋅y)=2x+y
(3⋅b)=3+b2(3 \cdot b) = \frac{3+b}{2}(3⋅b)=23+b
x∘y=x2yx \circ y = \frac{x^{2}}{y}x∘y=yx2
(3+b2)∘48=(3+b2)248=13\left( \frac{3+b}{2} \right) \circ 48 = \frac{ \left( \frac{3+b}{2} \right)^{2} }{48} = \frac{1}{3}(23+b)∘48=48(23+b)2=31
(3+b)248×4=13\frac{(3+b)^{2}}{48 \times 4} = \frac{1}{3}48×4(3+b)2=31
(3+b)2=48×43=64(3+b)^{2} = \frac{48 \times 4}{3} = 64(3+b)2=348×4=64
b2+6b+9=64 ⟹ b2+6b+9−64=0b^{2} + 6b + 9 = 64 \implies b^{2} + 6b + 9 - 64 = 0b2+6b+9=64⟹b2+6b+9−64=0
b2+6b−55=0 ⟹ b2−5b+11b−55=0b^{2} + 6b - 55 = 0 \implies b^{2} - 5b + 11b - 55 = 0b2+6b−55=0⟹b2−5b+11b−55=0
b(b−5)+11(b−5)=0 ⟹ (b−5)=0 or ( b+11)=0b(b - 5) + 11(b - 5) = 0 \implies (b - 5) = \text{0 or (}\, b + 11 ) = 0b(b−5)+11(b−5)=0⟹(b−5)=0 or (b+11)=0
Since b > 0, b - 5 = 0
b = 5.