Question

The solution set of the equation $ \sqrt{3{{x}^{2}}-7x-30}-\sqrt{2{{x}^{2}}-7x-5}=x-5 $ is

• A. $ \{6\} $
• B. $ \left\{ 6,-\frac{5}{2} \right\} $
• C. $ \{5,6\} $
• D. $ \left\{ 5,\frac{7}{2} \right\} $

Answer 1

Answer: (C)

Given equation can be rewritten as $ \sqrt{3{{x}^{2}}-7x-30}=(x-5)+\sqrt{2{{x}^{2}}-7x-5} $ On squaring both sides, we get $ 3{{x}^{2}}-7x-30={{x}^{2}}+25-10x+(2{{x}^{2}}-7x-5) $ $ +2(x-5)\sqrt{2{{x}^{2}}-7x-5} $
$ \Rightarrow $ $ 10x-50=2(x-5)\sqrt{2{{x}^{2}}-7x-5} $
$ \Rightarrow $ $ (x-5)\,(5-\sqrt{2{{x}^{2}}-7x-5)}=0 $
$ \Rightarrow $ either $ x-5=0 $ or $ 5-\sqrt{2{{x}^{2}}-7x-5}=0 $
$ \Rightarrow $ $ x=5 $ or $ 5=\sqrt{2{{x}^{2}}-7x-5} $
$ \Rightarrow $ $ x=5 $ or $ 2{{x}^{2}}-7x-30=0 $
$ \Rightarrow $ $ x=5 $ or $ x=-5/2,\,\,6 $ But $ x=-\frac{5}{2} $ is not satisfied the given equation. Hence, required roots are $ \{5,6\} $ .