Question

find a quadratic polynomial whose zeroes are -9 and -1/9​

Answer 1

let,

$\alpha = - 9$

$\beta = \frac{ - 1}{9}$

Then,

Sum of the zeroes,

$\alpha + \beta = ( - 9 ) + ( \frac{ - 1}{9} ) = \frac{ - 81 - 1}{9} = \frac{ - 82}{9}$

Product of the zeroes,

$\alpha \beta = (- 9)( \frac{ - 1}{9} ) = 1$

Now,

Quadratic polynomial =

$k( {x}^{2} - sx + p)$

$= k( {x}^{2} - ( \frac{ - 82}{9}) x + 1)$

$= k( \frac{9 {x}^{2} + 82x + 9}{9})$

$= \frac{1}{9} (9 {x}^{2} + 82x + 9)$

therefore, the required polynomial is

$9 {x}^{2} + 82x + 9$

Answer 2

Step-by-step explanation:

α = -9   β = -1/9

quadratic polynomial

= k {x ²- (α + β)x + αβ}

= k { x² - (-9 - 1/9)x -9 x -1/9}

= k { x² - (-19/9)x + 1}

= k { x² + 19/9 x + 1}

when k = 1

the quadratic polynomial is x² + 19/9 x + 1

Hope it helps!!

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