# Quod Erat Demonstrandum

## 2011/10/16

### 無聊改卷後2

Filed under: NSS — johnmayhk @ 2:40 下午

https://johnmayhk.wordpress.com/2011/10/12/boring-chatting-after-marking-paper/

Solve $3x^2+2x-2\sqrt{5}-15=0$ and express the answers in the form of $a+b\sqrt{c}$ where $a,b,c$ are rational numbers and $\sqrt{c}$ is irrational.

Solution

$3x^2+2x-2\sqrt{5}-15=0$ ………. (*)

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

$3x^2+2x=3(\sqrt{5})^2+2(\sqrt{5})$

Hence, $\sqrt{5}$ is a root of (*).

Let $\alpha$ be another root, then

$\alpha+\sqrt{5}=-\frac{2}{3}$

$\alpha=-\frac{2}{3}-\sqrt{5}$

Thus, the roots of (*) are

$\sqrt{5}$ and $-\frac{2}{3}-\sqrt{5}$

## 2 則迴響 »

1. $3(x^2-5)+2(x-\sqrt{5})=0$
$3(x+\sqrt[5])(x-\sqrt{5})+2(x-\sqrt{5})=0$
$(x-\sqrt{5})(3x+3\sqrt{5}+2)=0$

迴響 由 abc — 2011/10/16 @ 11:25 下午 | 回覆