Question

If x=3+2√2 , find the value of x2 + (1/x2 ).
pls help me​

Answer 1

Answer:

Given,

x=3+2√2

Inverse equatio,

1/x=1/3+2√2

=>1/x=(3–2√2){(3+2√2)(3–2√2)}

=>1/x=(3–2√2)/{(3)²-(2√2)²}

=>1/x=(3–2√2)/(9–8)

=>1/x=(3–2√2)/1

=>1/x=3–2√2

=>x+1/x=3+2√2+3–2√2

=>x+1/x=6

=>x²+1/x²=(x+1/x)² -2.x.1/x

=>x²+1/x² =(6)²-2

=>x²+1/x²=36–2

=>x²+1/x²=34

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Answer 2

Answer:

Given,

x=3+2√2

• Inverse equatio,

1/x=1/3+2√2

⇒ 1/x=(3–2√2){(3+2√2)(3–2√2)}

⇒ 1/x=(3–2√2)/{(3)²-(2√2)²}

⇒ 1/x=(3–2√2)/(9–8)

⇒ 1/x=(3–2√2)/1

⇒ 1/x=3–2√2

⇒ x+1/x=3+2√2+3–2√2

⇒ x+1/x=6

⇒ x²+1/x²= (x+1/x)² -2.x.1/x

⇒ x²+1/x² = (6)²-2

⇒ x²+1/x²= 36–2

x²+1/x² = 34