# Thread: Roots of an equation

1. ## Roots of an equation

If $\displaystyle \alpha, \beta$ are the roots of
$\displaystyle (x - 2)(x - \sqrt3) = \sqrt7$,
find the roots of
$\displaystyle (x - \alpha)(x - \beta) + \sqrt7 = 0$

2. If $\displaystyle \alpha, \ \beta$ are the roots of equation $\displaystyle (x-2)(x-\sqrt{3})-\sqrt{7}=0$ then

$\displaystyle (x-2)(x-\sqrt{3})-\sqrt{7}=(x-\alpha)(x-\beta)$

Then, $\displaystyle (x-\alpha)(x-\beta)+\sqrt{7}=(x-2)(x-\sqrt{3})-\sqrt{7}+\sqrt{7}=$

$\displaystyle =(x-2)(x-\sqrt{3})=0\Rightarrow x_1=2, \ x_2=\sqrt{3}$

3. Hello, alexmahone!

Here's my approach . . .

If $\displaystyle \alpha, \beta$ are the roots of: .$\displaystyle (x - 2)(x - \sqrt3) \:=\: \sqrt7$,

. . find the roots of: .$\displaystyle (x - \alpha)(x - \beta) + \sqrt7 \:=\: 0$

If $\displaystyle \alpha\text{ and }\beta$ roots of: .$\displaystyle x^2 - (2+\sqrt{3})x + (2\sqrt{3}-\sqrt{7}) \:=\:0$

. . then: .$\displaystyle \begin{array}{cccc}\alpha + \beta &=& 2+\sqrt{3} & {\color{blue}[1]} \\ \\[-4mm]\alpha\beta &=& 2\sqrt{3}-\sqrt{7} & {\color{blue}[2]}\end{array}$

We want the roots of: .$\displaystyle x^2 - (\alpha + \beta)x + (\alpha\beta + \sqrt{7}) \:=\:0$

Substitute [1] and [2]: .$\displaystyle x^2 -(2+\sqrt{3})x + [(2\sqrt{3}-\sqrt{7}) + \sqrt{7}] \:=\:0$

. . and we have: .$\displaystyle x^2 - (2 + \sqrt{3})x + 2\sqrt{3} \:=\:0$

The roots of this equation are $\displaystyle p\text{ and }q$ where: .$\displaystyle \begin{Bmatrix}p+q &=& 2+\sqrt{3} \\ pq &=& 2\sqrt{3} \end{Bmatrix}$

Therefore: .$\displaystyle p \,=\,2,\;q \,=\,\sqrt{3}$