Higher algebra problem -1

Let the roots of the equation be, $\alpha = \dfrac {-b \pm \sqrt{b^2-4c} } {2}$

Then, $\alpha - 2 = \dfrac {1} {\alpha}$

$\quad \Rightarrow \alpha^2 - 2\alpha - 1 = 0$

$\quad \Rightarrow \alpha = \dfrac {2 \pm \sqrt{4+4} } {2}$

Since $\alpha$ are the roots, therefore, $b = 2$ and $c = -1$ .

$\Rightarrow b^2 + c^2 = 2^2 + (-1)^2 = \boxed{5}$

Let the two roots of the equation be $\alpha$ and $\beta$ . We have that $\alpha-2=\frac{1}{\alpha}$ , and that $\beta-2=\frac{1}{\beta}$ .

Putting everything over a common denominator gives $\frac{\alpha^2-2\alpha-1}{\alpha}=0$ . Due to the Zero Product Property , $\alpha^2-2\alpha-1=0$ .

Since $\alpha$ is a root of our original equation, we know $\alpha^2+b\alpha+c=0$ , and this, in turn, is equal to $\alpha^2-2\alpha-1=0$ . The same can be achieved with $\beta$ .

As coefficients in two equal polynomials must also be equal , $b=-2$ , and $c=-1$ . $b^2+c^2={-2}^2+{-1}^2=5. _{\square}$

Let $r_1$ & $r_2$ be the roots of the equation, so $r_1 + r_2 = -b$ and $r_1r_2 = c$ by Vieta's formulas.

We know that.... $r_1 - 2 = \frac{1}{r_1} \ and \ \ r_2 - 2 = \frac{1}{r_2} \\ \Rightarrow r_1 - \frac{1}{r_1} = 2 \rightarrow (1) \\ r_2 - \frac{1}{r_2} = 2 \rightarrow (2)$

Setting $(1) = (2)$ lets us find the value for $c$ , which is $-1$ . Then to find $b$ we add our 2 equations....

$(1) + (2) = r_1 + r_2 - \left( \frac{1}{r_1} + \frac{1}{r_2} \right) = 4 \\ = r_1 + r_2 - \left( \frac{r_1 + r_2}{r_1r_2} \right) = 4\\ = (r_1 + r_2) \left(1 - \frac{1}{r_1r_2} \right) = 4$

Substituting the value for $c$ in for $r_1r_2$ and simplifying, we find that $b = -2$ . So $b^2 + c^2 = (-2)^2 + (-1)^2 = 5$ .