Question

if @(alpha) and beta are the zeroes of the quadratic polynomial f(x)=2x square-5x+7 .find a polynomial whose zeroes are {2@(alpha)+3 beta} and {3@(alpha)+2 beta}

Answer 1

Heya !!!!

here is the answer ,

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$let \: the \: given \: quadratic \: polynomial \: be \: f(x) = 2 {x}^{2} - 5x + 7$
given : @ and b ( beta ) are zeros of given quadratic polynomial .

Therefore,

$sum \: of \: roots = - ( \frac{ - 5}{2} )$
@+b = 5/2 - - - - - - (1)

$and \: product \: of \: roots \: = \: \frac{7}{2}$
@b = 7/2 - - - - - - - (2)


now, when zeros of  polynomial are (3α+2β) and (2α+3β), then 

sum of roots= (3α+2β) + (2α+3β) =5α+5β=5(α+β) 

= 5 (@+b) [ from (1) ]

$= 5 \times \frac{5}{2}$
$= \frac{25}{2}$

and product of roots = (3@+2b) (2@+3b)

= 6@^2+9@b +4@b +6b^2. [from (3)]

= 6(@^2+b^2) +13@b

= 6 [(@+b)^2-2@b] + 13@b

$= 6( ( { \frac{5}{2}) }^{2} - 2 \times \frac{7}{2} ) + 13 \times \frac{5}{2} \\ \\ = 6( \frac{25}{4 } - 7 ) + \frac{65}{2}$
$= 6( \frac{25 - 28}{4} ) + \frac{65}{2} \\ \\ = 6( \frac{ - 3}{4} ) + \frac{65}{2} \\ \\ = \frac{ - 9}{2} + \frac{65}{2} = \frac{ - 9 + 65}{2} = \frac{56}{2} = 28$

thus, the polynomial obtained is 

$g(x) = {x}^{2} - (sum \: ofzeros)x + product \: of \: zeros \\ \\ = {x}^{2} - \frac{25}{2} x + 28$

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Hope it helps u !!!!

Cheers :))

☺☺

# Nikky