Question

Please solve this ASAP..Q.no 3..And get 100 points

Answer 1

zeroes of 2x^2 - 5x - 3

=> 2x^2 - 6x + x - 3 = 0

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

=> ( x - 3 ) ( 2x + 1 ) = 0

=> x = 3 Or x = -1/2

According to the question,

zeroes of x^2 + px + q are 2(3) = 6 and 2(-1/2) = -1

taking 6 as zero,

x^2 + px + q = 0

6^2 + 6p + q = 0

36 + 6p + q = 0

q = - 36 - 6p

taking -1 as zero,

x^2 + px + q = 0

(-1)^2 -p + q = 0

1 - p + q = 0

q = p - 1

========================

q = q

=> -36 - 6p = p - 1

=> -36 + 1 = p + 6p

=> -35 = 7p

=> -5 = p

q = p - 1

= -5 - 1

= - 6

Hence, p = - 5 and q = -6

Answer 2

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