We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

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**+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

#3**+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