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Given that -3/2 is one of the root of the equation 2y^2+6y+p=0, find the value of p

1 Answer

3 votes

Answer:

p =
(9)/(2)

Explanation:

given y = -
(3)/(2) is a root of the equation , then this value makes the equation true.

substitute y = -
(3)/(2) into the equation and solve for p

2( -
(3)/(2) )² + 6(-
(3)/(2) ) + p = 0

2(
(9)/(4)) - 9 + p = 0


(9)/(2) -
(18)/(2) + p = 0

-
(9)/(2) + p = 0 ( add
(9)/(2) to both sides )

p =
(9)/(2)

User Jamie Buchanan
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