The problem asks us to express the fraction $\frac{\sqrt{2}}{\sqrt{2}-1}$ in the form $a + b\sqrt{2}$, where $a$ and $b$ are rational numbers. Also, find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$ given the quadratic equation $2x^2 - 4x + 5 = 0$ with roots $\alpha$ and $\beta$.

1. Problem Description

The problem asks us to express the fraction $\frac{\sqrt{2}}{\sqrt{2}-1}$ in the form $a + b\sqrt{2}$, where $a$ and $b$ are rational numbers. Also, find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$ given the quadratic equation $2x^2 - 4x + 5 = 0$ with roots $\alpha$ and $\beta$.

2. Solution Steps

Part d:

To express $\frac{\sqrt{2}}{\sqrt{2}-1}$ in the form $a + b\sqrt{2}$, we need to rationalize the denominator. We can do this by multiplying the numerator and denominator by the conjugate of the denominator, which is $\sqrt{2}+1$.

$\frac{\sqrt{2}}{\sqrt{2}-1} = \frac{\sqrt{2}}{\sqrt{2}-1} \cdot \frac{\sqrt{2}+1}{\sqrt{2}+1}$

$= \frac{\sqrt{2}(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)}$

$= \frac{2 + \sqrt{2}}{2 - 1}$

$= \frac{2 + \sqrt{2}}{1}$

$= 2 + \sqrt{2}$

Therefore, $a = 2$ and $b = 1$.

Part a:

Given the quadratic equation $2x^2 - 4x + 5 = 0$, with roots $\alpha$ and $\beta$, we need to find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$.

We know that the sum of the roots is given by $\alpha + \beta = -\frac{b}{a}$, and the product of the roots is given by $\alpha\beta = \frac{c}{a}$, where $a, b, c$ are the coefficients of the quadratic equation $ax^2 + bx + c = 0$.

In this case, $a = 2$, $b = -4$, and $c = 5$.

Thus, $\alpha + \beta = -\frac{-4}{2} = 2$ and $\alpha\beta = \frac{5}{2}$.

We want to find $\frac{1}{\alpha} + \frac{1}{\beta}$. We can rewrite this expression as:

$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}$

Substituting the values of $\alpha + \beta$ and $\alpha\beta$, we get:

$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{2}{\frac{5}{2}} = 2 \cdot \frac{2}{5} = \frac{4}{5}$

3. Final Answer

d. $2 + \sqrt{2}$

a. $\frac{4}{5}$