find the values of M for which the equation (m^2-4)x^2+nx+p=0
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Answer:
("m" - 2)"x"^2 - (5 + "m")"x" + 16 = 0`
Here a = m - 2, b = -(5 + m) and c = 16
Given : equation has equal roots
Then D = 0
`=> "b"^2 - 4"ac" = 0`
`=> [-(5 + "m")]^2 - 4("m" - 2)(16) = 0`
`=> 25 + "m"^2 + 10"m" - 64"m" + 128 = 0`
`=> "m"^2 - 54"m" + 153 = 0`
`=> "m"^2 - 51"m" - 3"m" + 153 = 0`
=> "m"("m" - 51) - 3("m" - 51) = 0
`=> ("m" - 51)("m" - 3) = 0`
then m - 51 = 0 or m - 3 = 0
=> m = 51 or m = 3
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