Question

23. If x+ 1/x = 4, then find the value of x² + 1/x?​

Answer 1

Step-by-step explanation:

ANSWER

x

4

+

x

4

1

=119

Adding 2 on both the sides,

x

4

+

x

4

1

+2=119+2

(x

2

+

x

2

1

)

2

=121⇒x

2

+

x

2

1

=11

Subtracting 2 on both sides.

x

2

+

x

2

1

−2=11−2

(x−

x

1

)

2

=9⇒x−

x

1

=±3

Hence, x−

x

1

=3.

Answer 2

Solution:-

Given

$\rm \: \to x + \dfrac{1}{x} = 4$

To find the value of

$\rm \to \: {x}^{2} + \dfrac{1}{ {x}^{2} }$

Now using this identity

$\rm \to \: (a + b)^{2} = {a}^{2} + {b }^{2} + 2ab$

Now take

$\rm \: \to x + \dfrac{1}{x} = 4$

Squaring on both side

$\rm \: \to \bigg(x + \dfrac{1}{x} \bigg)^{2} = {4}^{2}$

$\rm \to \: {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 \times x \times \dfrac{1}{x} = 16$

$\rm \to \: {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 \times \cancel{x} \times \dfrac{1}{ \cancel{x}} = 16$

$\rm \to \: {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 = 16$

$\rm \to \: {x}^{2} + \dfrac{1}{ {x}^{2} } = 16 - 2$

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

So Value of

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