My Math Forum Confusing wording for an inequality question.

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 July 7th, 2011, 03:56 PM #1 Newbie   Joined: Jul 2011 Posts: 5 Thanks: 0 Confusing wording for an inequality question. Determine the value of the variable for which the expression is defined as a real number. quadroot((1-x)/(2+x)) I hope you guys realize what I mean by quad root, like the square root, but with a small 4 in front of it. Now how do I find out what would the equation real and what wouldn't make it real?
 July 7th, 2011, 04:21 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond Re: Confusing wording for an inequality question. You have $$$\frac{1\,-\,x}{2\,+\,x}$$^{\frac{1}{4}}$. Since the denominator of the exponent is even, the expression in brackets must be greater than or equal to zero for the expression to be real. So you must solve the inequality $\frac{1\,-\,x}{2\,+\,x}\,\ge\,0$ 1 - x is negative when x > 1 and positive when x < 1. 2 + x is negative when x < -2 and positive when x > -2. Putting all this together, using the fact that a negative divided by a positive is negative (and so on), the solution is -2 < x ? 1. x can't be -2 or (1 - x)/(2 + x) is undefined, i.e. -2 is not in the domain of the given function.
 July 7th, 2011, 06:41 PM #3 Senior Member   Joined: Jul 2011 Posts: 245 Thanks: 0 Re: Confusing wording for an inequality question. Hai2u $\sqrt[4]{\frac{1-x}{2+x}} \\ 2 + x > 0 \\ x > -2 \\ 1 - x \geq 0 \\ -x \geq -1 \\ x \leq 1 \\ \\ \therefore -2 < x \leq 1$ bai2u(?)
 July 7th, 2011, 08:13 PM #4 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 $\frac{1-x}{2+x}\,\geq\,0\,\Rightarrow\,\frac{(1-x)(2+x)^2}{(2+x)}\,=\,(1-x)(2+x)\,\geq\,0\,\cdot\,(2+x)^2\,=\,0$ $(x-1)(x+2)\,\leq\,0$ $\therefore\,-2\,<\,x\,\leq\,1\text{ (because }2+x\,\neq\,0\text{)}$

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