Question

18. If a and B are the zeros of the quadratic polynomial f(x) = x2 – 3x - 2, find a quadratic
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polynomial whose zeros are 2a +b and 2B + a​

Answer 1

Evaluating Quadratic Polynomial

Answer: Required quadratic polynomial  is $x^{2} -9x + 16$ .

Explanation:

given that a and b are zeros of following quadratic polynomial

$x^{2} -3x-2$

Need to determine a quadratic polynomial whoes zeros are 2a + b and 2b + a.

here we will be using relation between zeros of quadratic polynomial and coefficient of quadratic polynomial

lets first consider given polynomial

$x^{2} -3x-2$

Sum of zeros = - ( coefficient of $x$ ) / (coefficient of $x^{2}$)

=> a + b = - (-3) / (1) = 3

=> a + b = 3    ----(1)

Product of zeros =  ( constant term ) / (coefficient of $x^{2}$)

=> a × b that is ab = -2/1   ----(2)

lets say sum of zeros of required polynomial  be M and product of zeros of required polynomial = N

As zeros of required polynomial is 2a + b and 2b + a

=> M = (2a +b ) + ( 2b + a ) = 2a + a + b + 2b = 3a + 3b = 3 ( a + b)

=> M = 3 × ( 3 )            [ from (1) a + b = 3 ]

=> M = 9     ------(3)

=> N = (2a +b ) × ( 2b + a ) = 2a( 2b + a ) + b ( 2b + a )

=>N = 4ab + 2$a^{2}$ + 2$b^{2}$ + ab

=>N = 5ab + 2 ( $a^{2} +b^{2}$)

=>N = 5ab + $(2((a+b)^{2} - 2ab))$  

using algebraic identity $(a+b)^{2} = a^{2} +b^{2} + 2ab => a^{2} +b^{2} = (a+b)^{2} - 2ab$

on substituting ab = -2 and a + b = 3 , we get

N = 5 x ( -2) + ( 2 ( (3 x 3 ) - 2 x (-2) ) ) = -10 + 2 x ( 9 + 4) = -10 + 26 = 16

=> N = 16  -----(4)

So required quadratic polynomial is

$k(x^{2} - Mx + N)$  

where k is any constant. lets for sake of simplicity assume k to be 1

so required quadratic polynomial will be

$x^{2} -9x + 16$             [ using equation 3 and 4 where M = 9 and N = 16 ]

so required quadratic polynomial  is $x^{2} -9x + 16$ .

#answerwithquality

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