+296 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 .
+118687 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
\(1-\sqrt{1+4k}<0\\ \sqrt{1+4k}>1\\ 1+4k>1\\ 4k>0\\ k>0\)
\( \left({1 - \sqrt{1+4k} \over 2}\right)^2 = {1 + \sqrt{1+4k} \over 2}\\ {1+1+4k - 2\sqrt{1+4k} \over 4} = {2 +2 \sqrt{1+4k} \over 4}\\ 2+4k-2\sqrt{1+4k}=2 +2 \sqrt{1+4k}\\ 2k-\sqrt{1+4k}= \sqrt{1+4k}\\ 2k= 2\sqrt{1+4k}\\ k= \sqrt{1+4k}\\ k^2=1+4k\\ k^2-4k-1=0\\ k=\frac{4\pm\sqrt{16+4}}{2}=2\pm\sqrt5\)
\(\\but\;\; k>0\\ so\\ k=2+\sqrt5\)
+118687 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 \(k=2+\sqrt{5}\)
