Question

Form quadratic polynomial whose Zeiss are 2&-3/2

Answer 1

Heya !!!

Let Alpha = 2 and beta = -3/2


Therefore,



Sum of zeroes = Alpha + Beta = 2 + (-3/2)


=> 2 - 3/2

=> 4 - 3/2 = 1/2




And,



Product of zeroes = Alpha × Beta = 2 × -3/2



=> -6/2 = -3


Therefore,



Required Quadratic polynomial = X²-(Sum of zeroes)X + Product of zeroes


=> X²-(1/2)X + (-3)


=> X²-X/2 - 3



=> 2X²-X-6 = 0



HOPE IT WILL HELP YOU.... :-)

Answer 2

Heya here ,

Solution
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$let \: \: \alpha = 2 \: \: and \: \: \beta = \frac{ - 3}{2}$
$sum \: of \: \: zeroes \: ( \alpha + \beta ) = 2 + \frac{ - 3}{2} = \frac{1}{2} \\ \\ product \: \: of \:zeroes \: ( \alpha \beta ) = 2 \times \frac{ - 3}{2} = \frac{ - 6}{2} = - 3$
Required polynomial
$= {x}^{2} - ( \alpha + \beta )x \: + \: \alpha \beta \\ = >2 {x}^{2} -x - 6$=0

Hence , 2x^2-x-6=0 is the polynomial .
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HOPE IT'S HELPS YOU.
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