Question

a+1/a =6. find a-1/a​

Answer 1

Answer:

HENCE,

(a - \frac{1}{a} ) = 4 \sqrt{2} \: \: or \: \: - 4 \sqrt{2} \\ \\ and \\ \\ (a {}^{2} - \frac{1}{a {}^{2} } ) = 24 \sqrt{2} \: \: or \: \: - 24 \sqrt{2}

Step-by-step explanation:

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Answer 2

Answer:

a - 1/a = ±4√2

Step-by-step explanation:

Given,

$\sf a + \dfrac{1}{a}=6$

To find,

$\sf a-\dfrac{1}{a} = \ ?$

Solution,

we know,

  (x + y)² = x² + y² + 2ab

Put x = a , y = 1/a

 $\sf (a+\dfrac{1}{a})^2=a^2+(\dfrac{1}{a})^2 +2(a)(\dfrac{1}{a}) \\\\ 6^2=a^2+\dfrac{1}{a^2}+2 \\\\ a^2+\dfrac{1}{a^2}=36-2 \\\\ a^2+\dfrac{1}{a^2}=34$

Also,

 (x - y)² = x² + y² - 2xy

Substitute,

 $\sf (a-\dfrac{1}{a})^2=a^2+(\dfrac{1}{a})^2-2(a)(\dfrac{1}{a}) \\\\ (a-\dfrac{1}{a})^2=a^2+\dfrac{1}{a^2}-2 \\\\ (a-\dfrac{1}{a})^2=34-2 \\\\ (a-\dfrac{1}{a})^2=32 \\\\ (a-\dfrac{1}{a})=\sqrt{32} \\\\ a-\dfrac{1}{a}=\sqrt{16 \times 2} \\\\ a-\dfrac{1}{a}=\pm 4\sqrt{2} \\\\ \boxed{\sf a-\dfrac{1}{a}=4\sqrt{2} , -4\sqrt{2}}$

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$\boxed{\begin{minipage}{7 cm}\boxed{\bigstar\:\:\textbf{\textsf{Algebraic\:Identity}}\:\bigstar}\\\\1)\sf\:(A+B)^{2} = \sf A^{2} + 2AB + B^{2}\\\\2)\sf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\3)\sf\: A^{2} - B^{2} = (A+B)(A-B)\\\\4)\sf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\5)\sf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\6)\sf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\7)\sf\:(A-B)^{3} = A^{3} - 3AB(A-B) + B^{3}\\\\8)\sf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\\end{minipage}}$