Consecutive members

The block has a volume of 1728 cm³. Determine the lengths of the edges a, b, and c of the blocks for which a < b < c and a + b + c = 38 cm and whose numerical values in cm represent three consecutive members of the geometric sequence.

Correct answer:

a =  8 cm
b =  12 cm
c =  18 cm

Step-by-step explanation:

$$\begin{gathered} V=1728{\text{cm}}^3 \\ 1728=2^6\times 3^3 \\ V=abc \\ b=qa \\ c=q^{2}a \\ a+b+c=38 \\ V=a\cdot aq\cdot a{q}^2=a^3\cdot q^3=(aq{)}^3 \\ a\cdot q=\sqrt[3]{V}=\sqrt[3]{1728}=12 \\ a\cdot aq\cdot (38-a-aq)=V \\ a\cdot 12\cdot (38-a-12)=V \\ a\cdot 12\cdot (38-a-12)=1728 \\ -12a^2+312a-1728=0 \\ 12a^2-312a+1728=0 \\ 12=2^2\cdot 3 \\ 312=2^3\cdot 3\cdot 13 \\ 1728=2^6\cdot 3^3 \\ \text{GCD}(12,312,1728)=2^2\cdot 3=12 \\ {a}^2-26a+144=0 \\ p=1;q=-26;r=144 \\ D=q^2-4pr=26^2-4\cdot 1\cdot 144=100 \\ D>0 \\ {a}_{1,2}=\frac{-q\pm \sqrt{D}}{2p}=\frac{26\pm \sqrt{100}}{2} \\ {a}_{1,2}=\frac{26\pm 10}{2} \\ {a}_{1,2}=13\pm 5 \\ {a}_1=18 \\ {a}_2=8 \\ a=a_2=8=8\text{cm} \\ q=12/a=12/8=\frac{3}{2}=1.5 \end{gathered}$$

Our quadratic equation calculator calculates it.

$b=q\cdot a=1.5\cdot 8=12\text{cm}$

$c=q\cdot b=1.5\cdot 12=18\text{cm}$

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