September 20th, 2013, 10:53 PM  #1 
Member Joined: Sep 2013 Posts: 75 Thanks: 0  number system
Given f(x) = square root of (3  x)(2 + x), determine the values of x for which 1) f(x) = 0, 2) f(x) will be a real number. I seem not able to interpret what I am given and what I am bieng asked very well. Can I expand the brackets under square root sign and find discriminant? (1) b^2  4ac = 0 (2) b^2  4ac >= 0 Is this correct? Criticism welcome. Thank you in advance. 
September 21st, 2013, 04:13 AM  #2  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Re: number system Quote:
Square both sides We get x = {2 , 3} < This is a list of two solutions 2) The expression under the radical must be non  negative so we set up an inequality We already have the zeros, so this sets up 3 intervals Test convenient values in these intervals to figure out where the function is non  negative. Test x = 3 FALSE Test x = 0 TRUE Test x = 4 FALSE So , when x is in the interval [2 , 3] , f(x) is a real number. [2 , 3] < This is a continuous interval made up of an infinite amount of real numbers.  

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