Question

find the zeroes of the polynomial 2x^2-9-3x and find the relation between zeroes and coefficients

Answer 1

Solution :

Given p(x) = 2x² - 3x - 9

Let p(x) = 0

2x² - 3x - 9 = 0

Splitting the middle term, we get

=> 2x² - 6x + 3x - 9 = 0

=> 2x( x - 3 ) + 3( x - 9 ) = 0

=> ( x - 3 )( 2x + 3 ) = 0

=> x - 3 = 0 or 2x + 3 = 0

=> x = 3 or x = -3/2

Therefore ,

m = 3 ,n = -3/2 are two zeroes of p(x)

2 ) Compare p(x) with ax²+bx+c , we get

a = 2 , b = -3 , c = -9

i ) Sum of the zeroes = 3 - 3/2

= ( 6 - 3 )/2

= 3/2

= -b/a

ii ) Product of the zeroes = 3×(-3/2)

= -9/2

= c/a

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