Combining inequalities

**WSox24** · December 18th 2011, 12:31 AM

1st post here.

I came across this problem in a GMAT reviewer and I naturally got B, but the correct answer is listed as A.

There was a brief explanation about why the answer is A: (2-x)^(1/2), but the brief explanation confused me. I realize that x>=2, but I don't see how I can use that to get the correct answer.

**FernandoRevilla** · December 18th 2011, 01:16 AM

Originally Posted by WSox24

I realize that x>=2

It should be $x\leq 2$ . Now, $\sqrt{x^2-6x+9}=\sqrt{(x-3)^2}=3-x$ , because $x\leq 2$ and it is supposed to choose the positive root . So, for all $x\leq 2$

$\sqrt{x^2-6x+9}\;+\;\sqrt{2-x}\:+x-3=3-x+\;\sqrt{2-x}\;+x-3=\sqrt{2-x}$

**WSox24** · December 18th 2011, 05:06 AM

Originally Posted by FernandoRevilla

It should be $x\leq 2$ . Now, $\sqrt{x^2-6x+9}=\sqrt{(x-3)^2}=3-x$ , because $x\leq 2$ and it is supposed to choose the positive root . So, for all $x\leq 2$

$\sqrt{x^2-6x+9}\;+\;\sqrt{2-x}\:+x-3=3-x+\;\sqrt{2-x}\;+x-3=\sqrt{2-x}$

Sorry for the triviality of this question, but what's confusing me here is $\sqrt{(x-3)^2}=3-x$ . Is it just true that $\sqrt{(x-3)^2}= \pm (3-x)$ ...but I still don't see what difference that makes because $\(x-3)^2$ will be positive regardless of x.

I guess I need to review some basic properties of abs value, roots, and inequalities

**Plato** · December 18th 2011, 05:16 AM

Originally Posted by WSox24

Sorry for the triviality of this question, but what's confusing me here is $\sqrt{(x-3)^2}=3-x$ . Is it just true that $\sqrt{(x-3)^2}= \pm (3-x)$ ...but I still don't see what difference that makes because $\(x-3)^2$ will be positive regardless of x.

It is never true that $\sqrt{a^2}=\pm a$ .

This is true $\sqrt{a^2}=|a|\ge 0.$

Thus $\sqrt{(x-3)^2}=|3-x|$ .

If $x<3$ then $|x-3|=3-x$

**FernandoRevilla** · December 18th 2011, 06:12 AM

Originally Posted by Plato

It is never true that $\sqrt{a^2}=\pm a$ .

This is true $\sqrt{a^2}=|a|\ge 0.$

Thus $\sqrt{(x-3)^2}=|3-x|$ .

If $x<3$ then $|x-3|=3-x$

(i) $\sqrt{9}=\pm 3$
(ii) Consider the function $f(x)=\sqrt{x}$ . Then $f(9)=3$
(iii) Consider the function $f(x)=-\sqrt{x}$ . Then $f(9)=-3$

**FernandoRevilla** · December 18th 2011, 06:25 AM

Originally Posted by WSox24

Sorry for the triviality of this question, but what's confusing me here is $\sqrt{(x-3)^2}=3-x$ . Is it just true that $\sqrt{(x-3)^2}= \pm (3-x)$ ...but I still don't see what difference that makes because $\(x-3)^2$ will be positive regardless of x.

I guess I need to review some basic properties of abs value, roots, and inequalities

There is an universal mathematical convention. For the function $f(x)=\sqrt{g(x)}$ and $g(x)\geq 0$ we consider the corresponding positive root (or $0$ ) and for $f(x)=-\sqrt{g(x)}$ we consider the corresponding negative root (or $0$ ) . In our case $x\leq 2$ hence $3-x> 0$ , for that reason $f(x)=\sqrt{(x-3)^2}=3-x$ (the corresponding positive root) .

**ILikeSerena** · December 18th 2011, 06:28 AM

Originally Posted by FernandoRevilla

(i) $\sqrt{9}=\pm 3$

From wikipedia:
"Every non-negative real number x has a unique non-negative square root, called the principal square root, which is denoted by $\sqrt$ , where $\sqrt$ is called radical sign."

I'm not aware of any convention where it is possible that $\sqrt{9}=-3$ .

**FernandoRevilla** · December 18th 2011, 06:40 AM

Originally Posted by ILikeSerena

"Every non-negative real number x has a unique non-negative square root,

Right, and a unique non-positive square root.

I'm not aware of any convention where it is possible that $\sqrt{9}=-3$ .

Neither do I.

**Plato** · December 18th 2011, 07:41 AM

Originally Posted by FernandoRevilla

(i) $\sqrt{9}=\pm 3$

I suppose that depends upon one's mathematics education community.
In North America we stress that $\sqrt{a}\ge 0$ , if defined.
That includes the GMAT test.

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