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Olympiad workbook
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Olympiad workbook
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IOQM
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Quadratic Equation

Olympiad workbook · IOQM · Quadratic Equation

Let a, b, c be reals satisfying 3 a b+2=6 b, 3 b c+2=5 c, 3 c a+2=4...

Let a,b,ca, b, c be reals satisfying 3ab+2=6b,3bc+2=5c,3ca+2=4a3 a b+2=6 b, 3 b c+2=5 c, 3 c a+2=4 a Let QQ denote the set of all rational numbers. Given that the product abca b c can take two values rs∈Q\frac{r}{s} \in Q and tu∈Q\frac{t}{u} \in Q, in lowest form, find r+s+t+ur+s+t+u.

Correct answer

18

Step-by-step solution

3ab+2=6b…(i),3bc+2=5c…(ii),3ac+2=4a.3abc+2c=6bc…(1)3abc+2a=5ac…(2)3abc+2b=4ab…(3)\begin{array}{ll} 3 a b+2=6 b \ldots(\mathrm{i}), & 3 b c+2=5 c \ldots(\mathrm{ii}), 3 a c+2=4 a . \\ 3 a b c+2 c=6 b c & \ldots(1) \\ 3 a b c+2 a=5 a c & \ldots(2) \\ 3 a b c+2 b=4 a b & \ldots(3) \end{array} Substitute bcb c from (ii) in (1), 3abc+2c=10c−43abc=8c−4...(4)\begin{aligned} & 3 a b c+2 c=10 c-4 \\ & 3 a b c=8 c-4 ...(4) \end{aligned} Substitute ac from (iii) in (2), 3abc+2a=5(4a−23)9abc+6a=20a−109abc=14a−10....(5)\begin{aligned} & 3 a b c+2 a=5\left(\frac{4 a-2}{3}\right) \\ & 9 a b c+6 a=20 a-10 \\ & 9 a b c=14 a-10 ....(5) \end{aligned} Substitute aba b from (i) in (3), 3abc+2b=4[6b−23]9bac+6b=24b−89abc=18b−8...(6)\begin{aligned} & 3 a b c+2 b=4\left[\frac{6 b-2}{3}\right] \\ & 9 b a c+6 b=24 b-8 \\ & 9 a b c=18 b-8 ...(6) \end{aligned} ∴\therefore \quad From (4), (5), (6), 24c−12=14a−10=18b−8=λc=12+λ24a=10+λ14b=8+λ18\begin{aligned} & 24 c-12=14 a-10=18 b-8=\lambda \\ & c=\frac{12+\lambda}{24} \quad a=\frac{10+\lambda}{14} \quad b=\frac{8+\lambda}{18} \end{aligned} Substitute in (i), β̸(10+λ14)(8+λ186)+2=∅(8+λ183)(λ+8)(λ+10)+84×2=28(8+λ)λ2+18λ+80+168=224+28λλ2−10λ+24=0(λ−6)(λ−4)=0λ=6,4 For λ=6a=1614b=1418c=1824\begin{aligned} & \not \beta\left(\frac{10+\lambda}{14}\right)\left(\frac{8+\lambda}{186}\right)+2=\varnothing\left(\frac{8+\lambda}{183}\right) \\ & (\lambda+8)(\lambda+10)+84 \times 2=28(8+\lambda) \\ & \lambda^2+18 \lambda+80+168=224+28 \lambda \\ & \lambda^2-10 \lambda+24=0 \\ & (\lambda-6)(\lambda-4)=0 \\ & \lambda=6,4 \\ & \text { For } \lambda=6 \\ & a=\frac{16}{14} \quad b=\frac{14}{18} \quad c=\frac{18}{24} \end{aligned} ∴abc=1614×1418×1824=1624=23\therefore \quad a b c=\frac{16}{14} \times \frac{14}{18} \times \frac{18}{24}=\frac{16}{24}=\frac{2}{3} For λ=4\lambda=4, a=1b=1218c=1624abc=49r+s+t+u=18\begin{aligned} & a=1 \quad b=\frac{12}{18} \quad c=\frac{16}{24} \\ & a b c=\frac{4}{9} \\ & r+s+t+u=18 \end{aligned}
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