Question

given x²+1/x²=7 and x not equal to 0 find x²-1/x² write the steps also.​

Answer 1

Step-by-step explanation:

Given:

${x}^{2} + \frac{1}{ {x}^{2} } = 7$

To find:

${x}^{2} - \frac{1}{ {x}^{2} }$

Solution:

${x}^{2} + \frac{1}{ {x}^{2} } + 2 = 7 + 2 \\ \\ or \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } + 2.x. \frac{1}{x} = 9 \\ \\ ( {x + \frac{1}{x}) }^{2} = 9$

take square root both sides

$x + \frac{1}{x}=3\\\\\frac{x^2+1}{x}=3\\\\x^2-3x+1=0$

Apply Quadratic formula to find value of x

$x_{1,2}=\frac{-b±\sqrt{b^2-4ac}}{2a}\\\\x_{1,2}=\frac{3±\sqrt{9-4}}{2}\\\\x_{1,2}=\frac{3±\sqrt{5}}{2}\\\\x_1=\frac{3+\sqrt{5}}{2}\\\\x_2=\frac{3-\sqrt{5}}{2}$

Put value of x

1) $x_1=\frac{3+\sqrt{5}}{2}$

${x}^{2} - \frac{1}{ {x}^{2} } = \left({\frac{3+\sqrt{5}}{2}}\right)^{2} - \frac{1}{ \left({\frac{3+\sqrt{5}}{2}}\right)^{2} }\\\\= \frac{14+6\sqrt{5}}{4}-\frac{4}{14+6\sqrt{5}}\\\\=\frac{7+3\sqrt{5}}{2}-\frac{2}{7+3\sqrt{5}}\\\\=\frac{(7+3\sqrt{5})^2-4}{2(7+3\sqrt{5})}$

$=\frac{49+45+42\sqrt{5}-4}{14+6\sqrt{5}}\\\\=\frac{90+42\sqrt{5}}{14+6\sqrt{5}}\\\\=\frac{45+21\sqrt{5}}{7+3\sqrt{5}}$

$\bold{\red{ {x}^{2} - \frac{1}{ {x}^{2} } =\frac{45+21\sqrt{5}}{7+3\sqrt{5}}}}$

By this way another value of ${x}^{2} - \frac{1}{ {x}^{2} }=\frac{45-21\sqrt{5}}{7-3\sqrt{5}}$

by taking another value of x.

Hope it helps you.