# SOLUTION: I am stuck on this problem. My answer is different from what the textbook answer says. Find the domain of each function. y=&#8730;2x^2 + 5x - 12 (The whole right side is und

Algebra ->  Algebra  -> Radicals -> SOLUTION: I am stuck on this problem. My answer is different from what the textbook answer says. Find the domain of each function. y=&#8730;2x^2 + 5x - 12 (The whole right side is und      Log On

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 Question 597389: I am stuck on this problem. My answer is different from what the textbook answer says. Find the domain of each function. y=√2x^2 + 5x - 12 (The whole right side is under the square root sign. This is how I did it: y=√2x^2 + 5x - 12 y-(x+4)(2x-3) x+4 is greater or equal to 0 x is greater or equal to -4 2x-3 is greater or equal to 0 2x is greater or equal to 3 x is greater or equal to 3/2 However, the textbook answer says the answer is: x is less than or equal to -4 OR x is greater or equal to 3/2 Please explain how the textbook got the answer. Thanks!!Answer by edjones(7569)   (Show Source): You can put this solution on YOUR website!x is less than or equal to -4 OR x is greater or equal to 3/2 Test to see what 2x^2+5x-12 equals when -4 < x < 3/2 and you will find that it is a negative number. The sqrt of a negative number is not allowed in the domain. . Ed