Question

please answer this question...
with solution...​

Answer 1

Answer:

$\underline{\bold{\huge{2\sqrt{35}}}}$

Step-by-step explanation:

$\text{If }x+\frac{1}{x} = 12.\: \text{To find }x-\frac{1}{x}\\\text{Now } (x-\frac{1}{x})^2 = (x+\frac{1}{x})^2 - 4\\= (12)^2 - 4 \\=144-4= 140\\x-\frac{1}{x} = \sqrt{140} = \sqrt{7\times2\times2\times5} = 2\sqrt{35} \\x+\frac{1}{x}=12\text{ ...(i)} \\x - \frac{1}{x} = 2\sqrt{35} \text{ ...(ii)}\\\text{Add (i) and (ii)}\\2x = 12+2\sqrt{35} \\\implies x = 6+\sqrt{35} \\\text{For} \frac{1}{x} = \frac{1}{6+\sqrt{35}} = \frac{1}{6+\sqrt{35}} \times \frac{6-\sqrt{35}}{6- \sqrt{35}}$

$= \frac{6-\sqrt{35}}{(6)^2 - (\sqrt{35})^2} = \frac{6-\sqrt{35}}{36-35} = 6 - \sqrt{35}\\\text{So } x-\frac{1}{x} = 6+\sqrt{35}-(6-\sqrt{35}) = \bold{2\sqrt{35}}$