Question

solve this question: If x= root2 + 1, find x⁴_1/x⁴​

Answer 1

Simplify:

Step-by-step explanation:

$\ Given \ value: \\\\x=\sqrt{2} +1 \\\\\ find:\\\\x^4 -\frac{1}{x^4} = ?\\\\\ Solution: \\\\x=\sqrt{2} +1 \\\\x^2=(\sqrt{2} +1)^2 \\\\\ (a+b)^2 = a^2+b^2+2ab\\\\x^2 = (\sqrt{2})^2+1^2+2(\sqrt{2})1 \\\\x^2 = 2+1+2\sqrt{2} \\\\x^2 = 3+2\sqrt{2} \\\\(x^2)^2= (3+2\sqrt{2})^2\\\\(x^4)= (3)^2+(2\sqrt{2})^2+2\times 3 \times2\sqrt{2} \\\\(x^4)= 9+8+12\sqrt{2} \\\\\ (x^4)= 17+12\sqrt{2} \\\\\ find: \\\\x^4-\frac{1}{x^4} \\\\\ put \ the\ value \ of \ x^4 \ in \ above \ equation:$

$(17+12\sqrt{2}) - \frac{1}{(17+12\sqrt{2})}\\\\\frac{(17+12\sqrt{2})^2-1}{(17+12\sqrt{2})}\\\\\frac{(17)^2+(12\sqrt{2})^2+2\times 17 \times(12\sqrt{2}) -1}{(17+12\sqrt{2})}\\\\\frac{289+288+408\sqrt{2} -1}{(17+12\sqrt{2})}\\\\\frac{576+408\sqrt{2}}{(17+12\sqrt{2})}$

$\frac{576+408\sqrt{2}}{(17+12\sqrt{2})}\times \frac{(17-12\sqrt{2})}{(17-12\sqrt{2})}\\\\\frac{(576+408\sqrt{2})(17-12\sqrt{2})}{(17)^2-(12\sqrt{2})^2}\\\\\frac{(9792+6936\sqrt{2}-6912\sqrt{2}-9792)}{(289-288)}\\\\\frac{(24\sqrt{2})}{1}\\\\(24\sqrt{2})$

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