SSC CGL 2024 PYQ — Question 72

The correct answer is option 3 i.e. $3\sqrt5$.

$(x+{1\over x})=7$

⇒ $(x+{1\over x})^2=7^2=49$   [Squaring both sides]

⇒ $(x-{1\over x})^2+\left( 4\times x\times \frac{1}{x}\right)=49$

⇒ $(x-{1\over x})^2+4=49$

⇒ $(x-{1\over x})^2=49-4=45$

⇒ $(x-{1\over x})=\sqrt{45}=3\sqrt5$

Therefore, the required value is 3√5.