# If alpha and beta are the zeros of x^2+pc+q, find (Alpha/beta +2)(beta/alpha+2) ?

## If $\alpha \mathmr{and} \beta$ are the zeros of ${x}^{2} + p x + q ,$ find (alpha/beta +2)(beta/alpha+2) ?

Feb 22, 2018

The answer is $= \frac{2 {p}^{2} + q}{q}$

#### Explanation:

${x}^{2} + p x + q = 0$

The roots are $\alpha$ and $\beta$

Therefore,

$\alpha + \beta = - p$... .....$\left(1\right)$

and

$\alpha \beta = q$..........$\left(2\right)$

We need

$\left(\frac{\alpha}{\beta} + 2\right) \left(\frac{\beta}{\alpha} + 2\right) = 1 + 2 \frac{\alpha}{\beta} + 2 \frac{\beta}{\alpha} + 4$

$\left(\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{{\alpha}^{2} + {\beta}^{2}}{\alpha \beta}\right)$

$= \frac{{\left(\alpha + \beta\right)}^{2} - 2 \alpha \beta}{q}$

$= \frac{{p}^{2} - 2 q}{q}$

Finally,

$\left(\frac{\alpha}{\beta} + 2\right) \left(\frac{\beta}{\alpha} + 2\right) = 5 + 2 \cdot \frac{{p}^{2} - 2 q}{q}$

$= \frac{5 q + 2 {p}^{2} - 4 q}{q}$

$= \frac{2 {p}^{2} + q}{q}$

Feb 22, 2018

$2 \cdot {p}^{2} / q + 1 , \mathmr{and} , \frac{1}{q} \left(2 {p}^{2} + q\right)$.

#### Explanation:

Given that $\alpha \mathmr{and} \beta$ are the zeroes of ${x}^{2} + p x + q$, we have

$\alpha + \beta = - p , \mathmr{and} , \alpha \cdot \beta = q \ldots \ldots \ldots \ldots \ldots . . \left(\ast\right)$.

$\text{Now, the reqd. value} = \left(\frac{\alpha}{\beta} + 2\right) \left(\frac{\beta}{\alpha} + 2\right)$,

$= 4 + 2 \left(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\right) + \left(\frac{\alpha}{\beta}\right) \left(\frac{\beta}{\alpha}\right)$,

$= 4 + \frac{2 \left({\alpha}^{2} + {\beta}^{2}\right)}{\alpha \cdot \beta} + 1$,

$= \frac{4 \alpha \cdot \beta + 2 \left({\alpha}^{2} + {\beta}^{2}\right)}{\alpha \cdot \beta} + 1$,

$= \frac{2 {\left(\alpha + \beta\right)}^{2}}{\alpha \cdot \beta} + 1$,

$= \frac{2 {\left(- p\right)}^{2}}{q} + 1. \ldots \ldots \ldots \ldots \ldots \left[\because , \left(\ast\right)\right]$,

$= 2 \cdot {p}^{2} / q + 1 , \mathmr{and} , \frac{1}{q} \left(2 {p}^{2} + q\right)$.

Feb 22, 2018

Thus,
$\left(\frac{\alpha}{\beta} + 2\right) \left(\frac{\beta}{\alpha} + 2\right) = \frac{q + 2 {p}^{2}}{q}$

#### Explanation:

Given:
$\alpha$is a root of ${x}^{2} + p x + q = 0$
$\beta$is a root of ${x}^{2} + p x + q = 0$
$\text{If ax^2+bx+c=0, } t h e n '$
$\alpha + \beta = - \frac{b}{a}$
Here, $a = 1 b = p , c = q ,$
Substituting
$\alpha + \beta = - \frac{p}{1} = - p$
$\alpha \beta = \frac{c}{a} = \frac{q}{1} = q$
$\left(\frac{\alpha}{\beta} + 2\right) \left(\frac{\beta}{\alpha} + 2\right) = \frac{\alpha + 2 \beta}{\beta} \times \frac{\beta + 2 \alpha}{\alpha}$
$= \frac{\left(\alpha + 2 \beta\right) \left(\beta + 2 \alpha\right)}{\alpha \beta}$

$= \frac{\alpha \beta + 2 {\beta}^{2} + 2 {\alpha}^{2} + 4 \alpha \beta}{\alpha \beta}$

$= \frac{\alpha \beta + 2 {\left(\alpha + \beta\right)}^{2}}{\alpha \beta}$

$= \frac{q + 2 {\left(- p\right)}^{2}}{q}$

$\left(\frac{\alpha}{\beta} + 2\right) \left(\frac{\beta}{\alpha} + 2\right) = \frac{q + 2 {p}^{2}}{q}$

Thus,
$\left(\frac{\alpha}{\beta} + 2\right) \left(\frac{\beta}{\alpha} + 2\right) = \frac{q + 2 {p}^{2}}{q}$