Asked by L.Bianchessi
Find the domain of f(x) = 1/ (square root 2x^2-7x-15)
I tried setting 2x^2-7x-15 greater or equal to zero but I don't know where to go from there to solve for my domain. Help please!:)
I tried setting 2x^2-7x-15 greater or equal to zero but I don't know where to go from there to solve for my domain. Help please!:)
Answers
Answered by
Damon
set it equal to zero
2 x^2 - 7 x - 15 = 0
(x-5)(2x+3) = 0
x = 5 or x = -1.5
between x = -1.5 and x = 5 the denominator is imaginary, the square root of a negative number
Thus the domain is x < -1.5 and x >5
2 x^2 - 7 x - 15 = 0
(x-5)(2x+3) = 0
x = 5 or x = -1.5
between x = -1.5 and x = 5 the denominator is imaginary, the square root of a negative number
Thus the domain is x < -1.5 and x >5
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