Question

If one root of $ {{x}^{2}}-x-k=0 $ is square of the other, then k is equal to

• A. $ 2\pm \sqrt{3} $
• B. $ 3\pm \sqrt{2} $
• C. $ 2\pm \sqrt{5} $
• D. $ 5\pm \sqrt{2} $

Answer 1

Answer: (C)

Let the roots of the equation $ {{x}^{2}}-x-k=0 $ are $ \alpha $ and $ {{\alpha }^{2}} $ . Then, $ \alpha +{{\alpha }^{2}}=1 $ ..(i) and $ \alpha .{{\alpha }^{2}}={{\alpha }^{3}}=-k $ $ \Rightarrow $ $ \alpha ={{(-k)}^{1/3}} $ On putting this value of a in Eq. (i), we get $ {{(-k)}^{1/3}}+{{(-k)}^{2/3}}=1 $ ?(ii) On cubing both sides, we get $ (-k)+{{(-k)}^{2}}+3k[{{(-k)}^{1/3}}+{{(-k)}^{2/3}}]=1 $ $ \Rightarrow $ $ -k+{{k}^{2}}-3k({{k}^{2/3}}-{{k}^{1/3}})=1 $ $ \Rightarrow $ $ {{k}^{2}}-k-3k(1)=1 $ [Using Eq.(ii)] $ \Rightarrow $ $ {{k}^{2}}-4k-1=0 $ $ \Rightarrow $ $ lk=\frac{4\pm \sqrt{20}}{2}=2\pm \sqrt{5} $