A) \[\{6\}\]
B) \[\left\{ 6,-\frac{5}{2} \right\}\]
C) \[\{5,6\}\]
D) \[\left\{ 5,\frac{7}{2} \right\}\]
Correct Answer: C
Solution :
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 \[\{5,6\}\].You need to login to perform this action.
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