Question

If x⁴ + 1\x⁴ = 2 find x + 1\x of Maths.​

Answer 1

Answer:

x4+1x4=47

⟹x4+1x4+2=49

⟹(x2+1x2)2=49

⟹x2+1x2=±7

⟹x2+1x2+2=2±7

⟹(x+1x)2=2±7

⟹x+1x=±2±7−−−−√={3,−3,i5–√,−i5–√}

Answer 2

Required Answer:-

Given:

• $\sf {x}^{4} + \dfrac{1}{ {x}^{4} } = 2$

To find:

• $\sf x + \dfrac{1}{x} = ?$

Solution:

This question can be solved by using the identity (a + b)² = a² + b² + 2ab. Let's solve.

We have,

$\sf \implies {x}^{4} + \dfrac{1}{ {x}^{4} } = 2$

Adding 2 to both sides, we get,

$\sf \implies {x}^{4} + \dfrac{1}{ {x}^{4} } + 2= 2 + 2$

$\sf \implies {({x}^{2})}^{2} + \bigg(\dfrac{1}{ {x}^{2} } \bigg) ^{2} + 2 \times {x}^{2} \times \dfrac{1}{ {x}^{2} } = 4$

This is in the form of (a + b)². So,

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

$\sf \implies \sqrt{\bigg( {x}^{2} + \dfrac{1}{ {x}^{2} } \bigg) ^{2}} = \sqrt{4}$

$\sf \implies {x}^{2} + \dfrac{1}{ {x}^{2} } = \pm 2$

We omit the negative value because x² must be positive if it is a real number. So x² + 1/x² must be positive.

$\sf \implies {x}^{2} + \dfrac{1}{ {x}^{2} } \neq - 2$

Therefore,

$\sf \implies {x}^{2} + \dfrac{1}{ {x}^{2} } = 2$

Now, adding 2 to both sides, we get,

$\sf \implies {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 = 2 + 2$

$\sf \implies {(x)}^{2} + \bigg( \dfrac{1}{x } \bigg)^{2} + 2 \times x \times \dfrac{1}{x} = 4$

Again, it is in (a + b)² form. So,

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

$\sf \implies \sqrt{\bigg( x + \dfrac{1}{x } \bigg)^{2} } = \sqrt{4}$

$\sf \implies \bigg( x + \dfrac{1}{x } \bigg) = \pm 2$

★ Hence, the value of x + 1/x is 2.

Answer:

• The value of x + 1/x is ±2

Identity Used:

• (a + b)² = a² + 2ab + b²

More Identities:

• (a - b)² = a² - 2ab + b²
• a² - b² = (a + b)(a - b)
• (a + b)² = (a - b)² + 4ab
• (a - b)² = (a + b)² - 4ab
• (a + b)³ = a³ + 3a²b + 3ab² + b³
• (a - b)³ = a³ - 3a²b + 3ab² + b³