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September 20th, 2013, 10:53 PM   #1
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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.
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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



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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