Math, asked by lalsv8913, 4 months ago

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

Answers

Answered by ItzMissKomal
0

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–√}

Answered by anindyaadhikari13
4

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
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