Question

Prove that 1903<(√13+3)^4<1904
√ is surd
^ is exponent

Answer 1

Answer on Question #43904 – Math – Other

Prove that $1903 < \left(3 + \sqrt{13}\right)^4 < 1904$.

Solution.

Firstly, compute $\left(3 + \sqrt{13}\right)^4$:

$$\begin{array}{l} \left(3 + \sqrt{13}\right)^4 = \left(9 + 2 \cdot 3 \cdot \sqrt{13} + 13\right)^2 = 4 \cdot \left(11 + 3\sqrt{13}\right)^2 = \\ = 4 \cdot \left(121 + 2 \cdot 11 \cdot 3\sqrt{13} + 9 \cdot 13\right) = 4 \cdot \left(238 + 66\sqrt{13}\right) = 952 + 264\sqrt{13}; \end{array}$$

Now simplify our double inequality:

$$\begin{array}{l} 1903 < \left(3 + \sqrt{13}\right)^4 < 1904 \Leftrightarrow 951 < 264\sqrt{13} < 952 \Leftrightarrow \frac{951}{264} < \sqrt{13} < \frac{952}{264} \Leftrightarrow \\ \Leftrightarrow \frac{317}{88} < \sqrt{13} < \frac{119}{33} \Leftrightarrow 3\frac{53}{88} < \sqrt{13} < 3\frac{20}{33} \Leftrightarrow \left(3 + \frac{53}{88}\right)^2 < 13 < \left(3 + \frac{20}{33}\right)^2 \Leftrightarrow \\ \Leftrightarrow 9 + 6 \cdot \frac{53}{88} + \frac{53^2}{88^2} < 13 < 9 + 6 \cdot \frac{20}{33} + \frac{20^2}{33^2} \Leftrightarrow \frac{27}{44} + \frac{53^2}{88^2} < 1 < \frac{7}{11} + \frac{20^2}{33^2}; \end{array}$$

Consider these equations separately:

$$\begin{array}{l} \frac{27}{44} + \frac{53^2}{88^2} < 1 \Leftrightarrow \frac{53^2}{88^2} < \frac{17}{44} \Leftrightarrow 53^2 < 88 \cdot 34 \Leftrightarrow 2809 < 2992 - \text{it's true}; \\ 1 < \frac{7}{11} + \frac{20^2}{33^2} \Leftrightarrow \frac{4}{11} < \frac{400}{33^2} \Leftrightarrow 33 \cdot 12 < 400 \Leftrightarrow 396 < 400 - \text{it's true}; \end{array}$$

Hence, both inequalities are true, so the whole double inequality is true.

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