Question

If x + 1/x. =4 then find x2 +1|/x2

Answer 1

$x^2+\dfrac{1}{x^2}=14$

Step-by-step explanation:

We have,

$x+\dfrac{1}{x}=4$         .....(1)

To find, $x^2+\dfrac{1}{x^2}=?$

Squaring (1) in both sides, we get

$(x+\dfrac{1}{x})^{2}=4^{2}$

⇒ $x^2+\dfrac{1}{x^2}+2.x.\dfrac{1}{x} =16$

⇒ $x^2+\dfrac{1}{x^2}+2 =16$

⇒$x^2+\dfrac{1}{x^2}=16-2=14$

∴ $x^2+\dfrac{1}{x^2}=14$

Hence, $x^2+\dfrac{1}{x^2}=14$

Answer 2

The value of $\bf {x}^{2} + \frac{1}{ {x}^{2} } = 14$ if $x + \frac{1}{x} = 4$

Given:

• $x + \frac{1}{x} = 4$

To find:

• Value of ${x}^{2} + \frac{1}{ {x}^{2} }$

Solution:

Identity used:$\bf ( {a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab$

Step 1:

Take the given equation.

$x + \frac{1}{x} = 4$

Step 2:

Squaring both sides by using Identity.

$\left( {x + \frac{1}{x}} \right)^{2} = ( {4)}^{2}$

or

Apply identity in LHS

${x}^{2} + \frac{1}{ {x}^{2} } + 2 \times \red x \times \frac{1}{ \red x} = 16$

cancel x in red

${x}^{2} + \frac{1}{ {x}^{2} } + 2 = 16$

or

${x}^{2} + \frac{1}{ {x}^{2} } = 16 - 2$

Thus,

$\bf {x}^{2} + \frac{1}{ {x}^{2} } = 14$

Learn more:

1) if x-1/x =5 , find the value of x^2+1/x^2

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

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