# Thread: Roots of an equation

1. ## Roots of an equation

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

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

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

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

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

3. Hello, alexmahone!

Here's my approach . . .

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

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

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

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

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

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

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

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

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