Thread: Combining inequalities

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.

Originally Posted by WSox24

I realize that x>=2

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


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

Originally Posted by FernandoRevilla

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


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

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

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

Originally Posted by WSox24

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

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

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

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

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

Originally Posted by Plato

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

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

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

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

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

Originally Posted by WSox24

Sorry for the triviality of this question, but what's confusing me here is $\displaystyle \sqrt{(x-3)^2}=3-x$. Is it just true that $\displaystyle \sqrt{(x-3)^2}= \pm (3-x) $ ...but I still don't see what difference that makes because $\displaystyle \(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 $\displaystyle f(x)=\sqrt{g(x)}$ and $\displaystyle g(x)\geq 0$ we consider the corresponding positive root (or $\displaystyle 0$ ) and for $\displaystyle f(x)=-\sqrt{g(x)}$ we consider the corresponding negative root (or $\displaystyle 0$ ) . In our case $\displaystyle x\leq 2$ hence $\displaystyle 3-x> 0$ , for that reason $\displaystyle f(x)=\sqrt{(x-3)^2}=3-x$ (the corresponding positive root) .

Originally Posted by FernandoRevilla

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

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

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

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 $\displaystyle \sqrt{9}=-3$.

Neither do I.

Originally Posted by FernandoRevilla

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

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