Question

27. If a and B are the zeroes of the polynomial f(x) = x² - 4x – 5 then find the value of a²+b²where a and B is alpha and beta ​

Answer 1

Given::

a,b are the zeroes of the polynomial

polynomial f(x) = x² - 4x – 5

Find::

a²+b²

➣If the polynomial is ax²+bx+c

and m,n are zeros then

m+n=-b/a

mn=c/a

So ➟a+b=-(-4/1)=4

ab=-5/1=-5

➣a²+b²=(a+b)²-2ab

➟a²+b²=16+10=26

therefore a²+b²=26

Answer 2

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$\begin{gathered} { \green{ \bold{given \: f(x) = 5 {x}^{2} - 7x + 1}}} \\ if \: \alpha \: and \: \beta \: are \: the \: zeroes \: of \: f(x) \\ = > \alpha + \beta = \frac{ - b}{a} \\ = \frac{ - ( - 7)}{5} = \frac{7}{5} \\ = > \alpha \beta = \frac{c}{a} \\ = \frac{1}{5} \\ \red{ \bold{to \: find... \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } }} \\ = \frac{ { \alpha }^{2} + { \beta }^{2} }{ \alpha \beta } \\ \green{first( \alpha + \beta ) {}^{2} = { \alpha }^{2} + 2 \alpha \beta + \beta {}^{2}} \\ = > ( \frac{7}{5} ) {}^{2} = \alpha {}^{2} + \beta {}^{2} + \frac{2}{5} \\ = > \frac{49}{25} = \frac{2}{5} + \alpha {}^{2} + \beta {}^{2} \\ = > \alpha {}^{2} + \beta {}^{2} = \frac{39}{25} \\ then \: \frac{ \alpha {}^{2} + \beta {}^{2} }{ \alpha \beta } = \frac{ \frac{39}{25} }{ \frac{1}{5} } = \frac{39}{5} \\ (\blue{answer...})\end{gathered}$