Question

if x^2+1/x^2=6,find the value of x^4+1/x^4​

Answer 1

★ Given :-

• x² + 1/x² = 6

★ To find :-

♦ x⁴ + 1/x⁴ = ?

★ Solution :-

$\bf{ {x}^{2} + \dfrac{1}{ {x}^{2} } = 6}$

Squaring on both sides,

$\Rightarrow \sf{ \bigg( {x}^{2} + \dfrac{1}{ {x}^{2} } \bigg) {}^{2} = {6}^{2} }$

Split using (a + b)² = a² + b² + 2ab

$\Rightarrow \: \sf{ ({x}^{2}) {}^{2} + \bigg(\dfrac{1}{ {x}^{2} } \bigg) {}^{2} + \bigg(2. x . \dfrac{1}{x} \bigg) = 36 }$

$\Rightarrow \: \sf{x {}^{4} + \dfrac{1}{ {x}^{4} } + 2 = 36 }$

$\Rightarrow \: \sf{ {x}^{4} + \dfrac{1}{ {x}^{4} } = 36 - 2}$

$\therefore \: \boxed{ \red{ \sf{{x}^{4} + \dfrac{1}{ {x}^{4} } = 34}}}$

Know more:

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