Exponents on polynomials

May 2010
22
0
I'm having trouble with this problem.

The only way I can get the answer the book gives is to illegally remove the exponent of 1/2 from the 1st term.

Problem: (2x+1)^(1/2) + (x+3)(2x+1)^(-1/2)

my attempt:
=(2x+1)^(1/2) + ((x+3) / ((2x + 1)^(1/2))
(I don't see any like terms here unless I remove the exponent)

Book's answer: (3x+4) / (sqrt(2x+1))

Another question: I see that I should move
(2x+1)^(-1/2) to the denominator because of the negative exponent. When I do this, does it go in the denominator of all the other terms, or just in the den of (x+3)?

Please explain what I'm doing wrong. Thanks alot!!

 

pickslides

MHF Helper
Sep 2008
5,237
1,625
Melbourne
\(\displaystyle (2x+1)^{\frac{1}{2}} + (x+3)(2x+1)^{\frac{-1}{2}}=(2x+1)^{\frac{1}{2}} + \frac{(x+3)}{(2x+1)^{\frac{1}{2}}}=\frac{(2x+1) +(x+3)}{(2x+1)^{\frac{1}{2}}}
\)

now simplify the numerator
 
Oct 2009
4,261
1,836
I'm having trouble with this problem.

The only way I can get the answer the book gives is to illegally remove the exponent of 1/2 from the 1st term.

Problem: (2x+1)^(1/2) + (x+3)(2x+1)^(-1/2)

my attempt:
=(2x+1)^(1/2) + ((x+3) / ((2x + 1)^(1/2))
(I don't see any like terms here unless I remove the exponent)

Book's answer: (3x+4) / (sqrt(2x+1))

Another question: I see that I should move (2x+1)^(-1/2) to the denominator because of the negative exponent. When I do this, does it go in the denominator of all the other terms, or just in the den of (x+3)?

Please explain what I'm doing wrong. Thanks alot!!

\(\displaystyle \sqrt{2x+1}+\frac{x+3}{\sqrt{2x+1}}=\frac{2x+1+x+3}{\sqrt{2x+1}}=\frac{3x+4}{\sqrt{2x+1}}\) , via \(\displaystyle \sqrt{a}+\frac{1}{\sqrt{a}}=\frac{a+1}{\sqrt{a}}\) ... common denominator and stuff.

Tonio
 
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May 2010
22
0
That's new to me

That formula you gave, Tonio, was totally new to me. I never knew you could do that with radicals. I tested it out and it works. Thanks alot. Very helpful.
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
I'm having trouble with this problem.

The only way I can get the answer the book gives is to illegally remove the exponent of 1/2 from the 1st term.

Problem: (2x+1)^(1/2) + (x+3)(2x+1)^(-1/2)

my attempt:
=(2x+1)^(1/2) + ((x+3) / ((2x + 1)^(1/2))
(I don't see any like terms here unless I remove the exponent)

Book's answer: (3x+4) / (sqrt(2x+1))

Another question: I see that I should move
(2x+1)^(-1/2) to the denominator because of the negative exponent. When I do this, does it go in the denominator of all the other terms, or just in the den of (x+3)?

Please explain what I'm doing wrong. Thanks alot!!

I expect you've been asked to simplify this... To simplify expressions involving fractions, you need a common denominator.

\(\displaystyle \sqrt{2x + 1} + \frac{x + 3}{\sqrt{2x + 1}} = \frac{\sqrt{2x + 1}\sqrt{2x + 1}}{\sqrt{2x + 1}} + \frac{x + 3}{\sqrt{2x + 1}}\)

\(\displaystyle = \frac{2x + 1}{\sqrt{2x + 1}} + \frac{x + 3}{\sqrt{2x + 1}}\)

\(\displaystyle = \frac{2x + 1 + x + 3}{\sqrt{2x + 1}}\)

\(\displaystyle = \frac{3x + 4}{\sqrt{2x + 1}}\).


I choose to write fractions with rational denominators though, so to clean it up even more...

\(\displaystyle = \frac{(3x + 4)\sqrt{2x + 1}}{2x + 1}\).
 
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