Question

If x^2 + 1/x^2 = 14, then find the value of x + 1 /x
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Answer 1

Answer:

Tap on the image for the answer.

Answer 2

Answer:

4

Step-by-step explanation:

$\sf{x^{2} + \dfrac{1}{x^{2} } = 14}\\\\\\\Rightarrow \: \sf{\bigg(x + \dfrac{1}{x}\bigg)^{2} - 2 \big(x)\bigg(\dfrac{1}{x}\bigg) = 14}$

$\Rightarrow \sf{\bigg(x + \dfrac{1}{x}\bigg)^{2} - 2 = 14}$

$\Rightarrow \: \sf{\bigg(x + \dfrac{1}{x}\bigg)^{2} = 14 + 2}$

$\Rightarrow \: \sf{\bigg(x + \dfrac{1}{x}\bigg)^{2} = 16}$

$\sf{Taking \: square \: root \: on \: both \: sides,}$

$\sf{\sqrt{\bigg(x + \dfrac{1}{x}\bigg)^{2} } = \sqrt{16} }$

$\therefore \boxed{\bf{\bigg(x + \dfrac{1}{x}\bigg) = 4}}$

Know more:

$\boxed{\begin{minipage}{6 cm}\bf{\dag}\:\:\underline{\textsf{Fraction Rules :}}\\\\\bigstar\:\:\sf\dfrac{A}{C} + \dfrac{B}{C} = \dfrac{A+B}{C} \\\\\bigstar\:\:\sf{\dfrac{A}{C} - \dfrac{B}{C} = \dfrac{A-B}{C}}\\\\\bigstar\:\:\sf\dfrac{A}{B} \times \dfrac{C}{D} = \dfrac{AC}{BD}\\\\\bigstar\:\:\sf\dfrac{A}{B} + \dfrac{C}{D} = \dfrac{AD}{BD} + \dfrac{BC}{BD} = \dfrac{AD+BC}{BD} \\\\\bigstar\:\:\sf\dfrac{A}{B} - \dfrac{C}{D} = \dfrac{AD}{BD} - \dfrac{BC}{BD} = \dfrac{AD-BC}{BD}\\\\\bigstar \:\:\sf \dfrac{A}{B} \div \dfrac{C}{D} = \dfrac{A}{B} \times \dfrac{D}{C} = \dfrac{AD}{BC}\end{minipage}}$