# Quod Erat Demonstrandum

## 2011/10/12

### 無聊改卷後

Filed under: NSS — johnmayhk @ 2:40 下午

Solve $2(x^6+\frac{1}{x^6})-5(x^2+\frac{1}{x^2})=21$ for real $x$.

(Answer: $\pm \frac{1\pm \sqrt{5}}{2}$)

$2^3$

$=2\times 2\times 2$

$(x^2+\frac{1}{x^2})^3$

$\equiv (x^2+\frac{1}{x^2})(x^2+\frac{1}{x^2})(x^2+\frac{1}{x^2})$

$(x^2+\frac{1}{x^2})^3$

$=(x^2+\frac{1}{x^2})(x^4-x^2(\frac{1}{x^2})+\frac{1}{x^4}$) ………. (1)

$=x^6+\frac{1}{x^6}$ ………. (2)

$a^3+b^3=(a+b)(a^2-ab+b^2)$

$x^2+\frac{1}{x^2}=3$

$\Rightarrow x^4-3x^2+1=0$

$\Rightarrow x^2=\frac{3\pm\sqrt{5}}{2}=\frac{6\pm 2\sqrt{5}}{4}=\frac{(1\pm\sqrt{5})^2}{4}$

$\Rightarrow x=\pm\frac{1\pm\sqrt{5}}{2}$

$x^2+\frac{1}{x^2}=3$

$\Rightarrow (x+\frac{1}{x})^2=5$

$\Rightarrow x+\frac{1}{x}=\pm\sqrt{5}$

$\Rightarrow x^2\pm\sqrt{5}x+1=0$

$\Rightarrow x=\pm\frac{1\pm\sqrt{5}}{2}$

Solve

$-12\sin\theta\cos\theta+9\cos^2\theta+2\sin\theta-4\cos\theta+3-2\sqrt{5}=15$ such that $(\sin\theta-2\cos\theta)^2$ is rational.

$3x^2+2x-2\sqrt{5}-15=0$

$\Rightarrow (x-\sqrt{5})(3x+2+3\sqrt{5})=0$

$\Rightarrow x=\sqrt{5}$ or $x=\frac{-2-3\sqrt{5}}{3}$ (rejected since $x^2$ is rational)

$3x^2+2x-2\sqrt{5}-15=0$

Solve $(\log x)^{\log x}=x$.

（當然要求 $x$ 是實數。）

Solve $\sqrt{x^2-4x+4}=-(x-2)$ (2 marks)

Find a rational number $x$ such that

$\sqrt[3]{1+\sqrt{x}}+\sqrt[3]{1-\sqrt{x}}=\sqrt[3]{5}$.