Question

If $x=\frac{\left[\begin{array}{rl}& 729+6(2)(243)+15(4)(81)+20(8)(27)\\ & +15(16)(9)+6(32)3+64\end{array}\right]}{1+4(4)+6(16)+4(64)+256}$ then $\sqrt{x}-\frac{1}{\sqrt{x}}$ is equal to:

• A. 0.2
• B. 4.8
• C. 1.02
• D. 5.2
• E. 25

Answer 1

Answer: (B)

$x=\frac{\left[\begin{array}{rl}& 729+6(2)(243)+15(4)(81)+20\\ & \times (8)(27)+15(16)(9)+6(32)(3)+64\end{array}\right]}{1+4(4)+6(16)+4(64)+256}$ $=\frac{\left[\begin{array}{rl}& {}^6{C}_0{(3)}^6{+}^6{C}_1{3}^5.2{+}^6{C}_2{3}^4{.2}^2{+}^6{C}_3\\ & \times 3^3{2}^2{+}^6{C}_4{.3}^2{.2}^4{+}^6{C}_5{3.2}^5{+}^6{C}_6{2}^6\end{array}\right]}{\left[\begin{array}{rl}& {}^4{C}_0{1}^4{+}^4{C}_{1}4{+}^4{C}_2{4}^2{+}^4{C}_3{4}^3\\ & {+}^4{C}_4{4}^4\end{array}\right]}$ $\Rightarrow$ $x=\frac{{(3+2)}^6}{{(1+4)}^4}=5^2$ $\therefore$ $\sqrt{x}=5$ $\therefore$ $\sqrt{x}-\frac{1}{\sqrt{x}}=5-\frac{1}{5}$ $=\frac{24}{5}=4.8$