greg997 Messages 105 Reaction score 2 Thread starter May 20, 2010 #1 Can someone show me in steps how to manipulate this formula in attachement to obtain 1/(sqr(w^2+4)? Thanks Attachments Untitled.jpg 6.1 KB · Views: 492
Can someone show me in steps how to manipulate this formula in attachement to obtain 1/(sqr(w^2+4)? Thanks
jack action Science Advisor Insights Author Messages 3,641 Reaction score 10,062 May 20, 2010 #2 The two equations are not equal. Just replace w by any value and you'll see that both equations give different answers.
The two equations are not equal. Just replace w by any value and you'll see that both equations give different answers.
Mentallic Homework Helper Messages 3,802 Reaction score 95 May 20, 2010 #3 The best you can really get is: [tex]\frac{\sqrt{4-w^2}}{w^2+4}[/tex] Last edited: May 20, 2010
Mentallic Homework Helper Messages 3,802 Reaction score 95 May 20, 2010 #4 You must have made a typo. The expression should be, [tex]\sqrt{\left(\frac{2}{w^2+4}\right)^2+\left(\frac{w}{w^2+4}\right)^2}[/tex]
You must have made a typo. The expression should be, [tex]\sqrt{\left(\frac{2}{w^2+4}\right)^2+\left(\frac{w}{w^2+4}\right)^2}[/tex]
greg997 Messages 105 Reaction score 2 May 20, 2010 #5 Mentallic said: You must have made a typo. The expression should be, [tex]\sqrt{\left(\frac{2}{w^2+4}\right)^2+\left(\frac{w}{w^2+4}\right)^2}[/tex] Hmm, that's quite possible. Then how to get that solution?
Mentallic said: You must have made a typo. The expression should be, [tex]\sqrt{\left(\frac{2}{w^2+4}\right)^2+\left(\frac{w}{w^2+4}\right)^2}[/tex] Hmm, that's quite possible. Then how to get that solution?
Mentallic Homework Helper Messages 3,802 Reaction score 95 May 20, 2010 #6 Well I can't just give you the answer, you need to show an attempt at solving the problem first. But I can give you some hints, you can most certainly find the answer by using a combination of these rules: [tex]\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}[/tex] [tex]\sqrt{a^2b}=|a|\sqrt{b}[/tex] (the |a| means the positive of a, but remember that for any real number n, n2 is always positive). [tex]\frac{a}{c}+\frac{b}{c}=\frac{a+b}{c}[/tex] [tex]\frac{\sqrt{a}}{a}=\frac{1}{\sqrt{a}}[/tex] Good luck!
Well I can't just give you the answer, you need to show an attempt at solving the problem first. But I can give you some hints, you can most certainly find the answer by using a combination of these rules: [tex]\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}[/tex] [tex]\sqrt{a^2b}=|a|\sqrt{b}[/tex] (the |a| means the positive of a, but remember that for any real number n, n2 is always positive). [tex]\frac{a}{c}+\frac{b}{c}=\frac{a+b}{c}[/tex] [tex]\frac{\sqrt{a}}{a}=\frac{1}{\sqrt{a}}[/tex] Good luck!
greg997 Messages 105 Reaction score 2 May 21, 2010 #7 Hello, I am sorry but it just does not work. What am I doing work? Thanks Attachments 1.jpg 11.2 KB · Views: 437
Cyosis Homework Helper Messages 1,495 Reaction score 5 May 21, 2010 #8 You're adding fractions, but you're multiplying numerators.