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Algebra Level 3

If b > 1 b > 1 , find the minimum value of 9 b 2 − 18 b + 13 b − 1 \large \frac{9b^{2} - 18b + 13}{b - 1}


The answer is 12.

Benedict Dimacutac by Benedict Dimacutac , 21, Philippines

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1 solution

Chew-Seong Cheong
Oct 27, 2017

X = 9 b 2 − 18 b + 13 b − 1 = 9 ( b − 1 ) 2 + 4 b − 1 = 9 ( b − 1 ) + 4 b − 1 Since b > 1 , AM-GM inequality applies ≥ 2 9 × 4 = 12 \begin{aligned} X &= \frac{9b^2-18b+13}{b-1} \\ &= \frac{9(b-1)^2+4}{b-1} \\ &= 9(b-1)+\frac{4}{b-1} & \small \color{#3D99F6} \text{Since }b >1 \text{, AM-GM inequality applies} \\ &\ge 2\sqrt{9\times4}=\boxed{12} \end{aligned}

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