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 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
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Re: number system

Quote:
 Originally Posted by bongantedd 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.
1) 

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

$0 \le (3 - x)(x + 2)$

We already have the zeros, so this sets up 3 intervals

$(- \infty \ , \ -2] \ , \ [-2 \ , \ 3] \ , \ [3 \ , \ \infty)$

Test convenient values in these intervals to figure out where the function is non - negative.

Test x = -3

$0 \le [3 - (-3)](-3 +2) \ \$ FALSE

Test x = 0

$0 \le (3-0)(0+ 2) \ \$ TRUE

Test x = 4

$0 \le (3 -4)(4 + 3) \ \$ 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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