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If one root of the equation x^2 - x - k is square that of other find the other root . Find the value of K.

please give a detailed solution .

Darkside  Sep 2, 2018
 #1
avatar+93289 
+2

Mmm

 

If one root of the equation x^2 - x - k is square that of other find the other root . Find the value of K.

 

Roots are

 

    \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ = {1 \pm \sqrt{1+4k} \over 2}\\ We \;\;want\\ \left({1 - \sqrt{1+4k} \over 2}\right)^2 = {1 + \sqrt{1+4k} \over 2}\\ \)

 

The right hand side is clearly bigger than 1 so the left side (before it is squared) must be less than 0

 

\(1-\sqrt{1+4k}<0\\ \sqrt{1+4k}>1\\ 1+4k>1\\ 4k>0\\ k>0\)

 

 

 

 

\( \left({1 - \sqrt{1+4k} \over 2}\right)^2 = {1 + \sqrt{1+4k} \over 2}\\ {1+1+4k - 2\sqrt{1+4k} \over 4} = {2 +2 \sqrt{1+4k} \over 4}\\ 2+4k-2\sqrt{1+4k}=2 +2 \sqrt{1+4k}\\ 2k-\sqrt{1+4k}= \sqrt{1+4k}\\ 2k= 2\sqrt{1+4k}\\ k= \sqrt{1+4k}\\ k^2=1+4k\\ k^2-4k-1=0\\ k=\frac{4\pm\sqrt{16+4}}{2}=2\pm\sqrt5\)


\(\\but\;\; k>0\\ so\\ k=2+\sqrt5\)

Melody  Sep 2, 2018
 #2
avatar+93289 
+2

Here is the graph

 

Melody  Sep 2, 2018
 #3
avatar+93289 
+2

Maybe I should also consider it the other way around

 

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ = {1 \pm \sqrt{1+4k} \over 2}\\ \;\;maybe\;\;we \;\;want\\ {1 - \sqrt{1+4k} \over 2} =\left( {1 + \sqrt{1+4k} \over 2}\right)^2\\\)

 

 

 

For this to be true the left hand side has to be positive and we know it is less then \(\frac{1}{2}\)

So that means the RHS before it is squared must be between  \(\pm\frac{1}{\sqrt2}\)

 

\(2(1-\sqrt{1+4k})=1+1+4k+2\sqrt{1+4k}\\ 1-\sqrt{1+4k}=1+2k+\sqrt{1+4k}\\ -\sqrt{1+4k}=2k+\sqrt{1+4k}\\ -2\sqrt{1+4k}=2k\\ k=-\sqrt{1+4k}\\ \text{If k is a real number then there is not solution here}\)

 

What if k is complex ??

\(k^2=1+4k\\ k^2-4k-1=0\\ k=\frac{4\pm\sqrt{16+4}}{2}\\ k=2\pm\sqrt{5}\\\)

Well that k is not complex so there is no solution.

 

So I will stick to my original answer            \(k=2+\sqrt{5}\)

Melody  Sep 2, 2018
edited by Melody  Sep 2, 2018
 #4
avatar+247 
+2

thank you melody it is correct

Darkside  Sep 2, 2018
 #5
avatar+93289 
+1

How do you know it is correct?

I assume you are comparing it to an answer that you have, like in an answer sheet.

That does not necessarily mean it is correct, it just means that they are the same.

They might both be wrong    wink

 

The answer is fine though   cool

Melody  Sep 2, 2018

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