The quadratic equation $x^2+4mx+m = 2x - 6$ has exactly one real root. Find the positive value of $m$.
x^2+4mx+m = 2x - 6 has exactly one real root means the discriminant of the quadratic = 0
Put in quadratic form
x^2 + ( 4m-2) x + m+ 6 = 0
discriminant = b^2 - 4 ac = 0
= (4m-2)^2 - 4 (1)(m+6)
= 16m^2 - 16m + 4 - 4m - 24
= 16 m^2 - 20 m - 20 Use quadratic formula with a = 16 b = -20 c = -20
to find the positive valueof m = 5/8 + sqrt(105) / 8 ( or m = 1.90587 )
x^2+4mx+m = 2x - 6 has exactly one real root means the discriminant of the quadratic = 0
Put in quadratic form
x^2 + ( 4m-2) x + m+ 6 = 0
discriminant = b^2 - 4 ac = 0
= (4m-2)^2 - 4 (1)(m+6)
= 16m^2 - 16m + 4 - 4m - 24
= 16 m^2 - 20 m - 20 Use quadratic formula with a = 16 b = -20 c = -20
to find the positive valueof m = 5/8 + sqrt(105) / 8 ( or m = 1.90587 )