2 degree polynomial

Using algebraic manipulation,

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

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

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

Substituting $\alpha + \beta$ = 5 and $\alpha \beta = 25$ into the equation, we get $\frac {25}{25} - 2 = -1$ .

Relevant wiki: Vieta's Formula Problem Solving - Basic

Firstly, This question is incorrect..It must be x^2 - 5x + 25 = 0..

You forgot to put x in 5x..

Here, (a/b) + (b/a) is equal to (a + b)^2 - 2ab/ab..

Now, By Vietta's Sum,

a + b = 5 and ab = 25.

Putting this in (a+b)^2 - 2ab/ab we get :-

= 25 - 2(25)/25

= -25/25 = -1

Hence, Answer = -1.

${ x }^{ 2 }-5x+25=0$

$x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$

$x=\frac { 5\pm \sqrt { 25-100 } }{ 2 }$

$x=\frac { 5\pm \sqrt { -75 } }{ 2 } \Rightarrow x=\frac { 5\pm i5\sqrt { 3 } }{ 2 }$

$x=\frac { 5(1\pm i\sqrt { 3 } ) }{ 2 }$

then $\alpha =\frac { 5(1+i\sqrt { 3 } ) }{ 2 }$ and $\beta =\frac { 5(1-i\sqrt { 3 } ) }{ 2 }$

$\frac { \alpha }{ \beta } =\frac { \frac { 5(1+i\sqrt { 3 } ) }{ 2 } }{ \frac { 5(1-i\sqrt { 3 } ) }{ 2 } }$

$\frac { \alpha }{ \beta } =\frac { (1+i\sqrt { 3 } ) }{ (1-i\sqrt { 3 } ) }$

$\frac { \alpha }{ \beta } =\frac { (1+i\sqrt { 3 } ) }{ (1-i\sqrt { 3 } ) } \times \frac { (1+i\sqrt { 3 } ) }{ (1+i\sqrt { 3 } ) }$

$\frac { \alpha }{ \beta } =\frac { (1+i\sqrt { 3 } )^{ 2 } }{ 4 }$

And do the same thing with $\frac { \beta }{ \alpha }$ and you will get $\frac { \beta }{ \alpha } =\frac { (1-i\sqrt { 3 } )^{ 2 } }{ 4 }$

$\frac { \beta }{ \alpha } +\frac { \alpha }{ \beta } =\frac { (1-i\sqrt { 3 } )^{ 2 }+(1+i\sqrt { 3 } )^{ 2 } }{ 4 }$

$\frac { (1-2i\sqrt { 3 } -3)+(1+2i\sqrt { 3 } -3) }{ 4 } \Rightarrow \frac { -4 }{ 4 }$

$\boxed{-1}$

with vieta, we got $a+b=5$ , $ab=25$ and $(\frac{a}{b}$ + $\frac{b}{a}$ = $\frac{a^2+b^2}{ab}$ ) $a^2+b^2=(a+b)^2-2ab =5^2-2*25 =25-50 =-25. so, \frac{a^2+b^2}{ab}$ = $\frac{-25}{25}$ $=-1$