# Quod Erat Demonstrandum

## 2017/06/21

### 兩題二次方程

Filed under: mathematics,NSS — johnmayhk @ 6:28 下午
Tags: ,

1.

Refer to the figure below.

If $\alpha$ and $\beta$ are x-coordinates of P and Q respectively such that $\alpha^2+\beta^2=13$, find the value(s) of m.

$-x^2+3x-2=mx-8$
$x^2+(m-3)x-6=0$

$\alpha^2+\beta^2=13$
$(\alpha+\beta)^2-2\alpha\beta=13$

$(3-m)^2-2(-6)=13$
$3-m=\pm 1$
$m=4$ or $m=2$

＝　＝　＝　停一停，想一想　＝　＝　＝

$m=4$，解出 P 的 y-coordinate 為 0；故 $m=4$ 不合，即原題答案只有 $m=2$

2.

Suppose $x^2+kx+1 > k$ for any real value of $x$, find the range of values of $k$,

$\Delta > k$

（學生常犯錯誤之一）時，十居其十是錯的。

Suppose two tangents to the graph $y=x^2+kx+1$ from O are perpendicular to each other, find the value(s) of $k$.

$\Delta=-1$

$\Rightarrow k^2-4=-1$

$\Rightarrow k=\pm\sqrt{3}$

$\Delta=-1$

$ax^2+bx+c=mx$

$(b-m)^2-4ac=0$

$\Rightarrow m^2-2bm+b^2-4ac=0$ ………………….. (*)

$m_1m_2$ 是 product of roots of (*)

$b^2-4ac=-1$

$\Delta=-1$