Math, asked by shristisharma1409, 4 months ago

if x=3+2√2 find the value of x^2+1/x​

Answers

Answered by jyoti27jyoti27
0

Answer:

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Step-by-step explanation:

x=3+22–√1x=13+22–√=3−22–√9−8=3−22–√(x2+1x2)=(x+1x)2−2(x2+1x2)=(3+22–√+3−23–√)2−2(x2+1x2)=36−2=34

If x = 4/ (2√3+3√2), then what is the value of x+1/x?

What is the value of x+1/x if x=7-4√3?

If x=3-2√2, then what is the value of (√x) - (1/√x)?

If x= 3+2√2, then what is the value of (x+1/x)?

If x=3+2√2, then what is the value of √x-1/√x?

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

Ans.

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x = (3 + 2√2)

1/x = 1/(3 + 2√2) = 1/(3 + 2√2) * (3 - 2√2)/(3 - 2√2) = (3 - 2√2)/(9 - 8) =>

1/x = 3 - 2√2

x^2 + 1/x^2:

x + 1/x = (3 + 2√2) + (3 - 2√2) = 6

(x + 1/x)^2 = 6^2

x^2 + 1/x^2 + 2(x)(1/x) = 36

x^2 + 1/x^2 + 2 = 36

x^2 + 1/x^2 = 34

Here x=3+2√2

Therefore, 1/x=1/3+2√2

Or,1/x=(3–2√2){(3+2√2)(3–2√2)}

Or,1/x=(3–2√2)/{(3)²-(2√2)²}

Or,1/x=(3–2√2)/(9–8)

Or,1/x=(3–2√2)/1

Or,1/x=3–2√2

x+1/x=3+2√2+3–2√2

Or,x+1/x=6

Now, x²+1/x²=(x+1/x)² -2.x.1/x

Or,x²+1/x² =(6)²-2

Or,x²+1/x²=36–2

Or,x²+1/x²=34

If x=3+2√3, then what is the value of √x+1/√x?

If ‘x=3-2√2’, can you find the value of ‘√x+1/√x’?

If x = 3+2√3, then what is the value of 1/x?

If x = √3 + √2, then what is the value of (x²+x+1/x + 1/x²)?

If x= (3-2√2), how do you show that (√x-1/√x) =2?

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

Ans

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If x = 3 +2√2, then 1/x = 1/(3+2√2) =(3–2√2), because 3^2 -(2√2)^2 =9–8=1. Hence x^2 +(1/x^2) = [x+(1/x)]^2 -2 = 6^2 -2 =36 -2 = 34.

x=3+ 2×(2^0.5) = ((2^0.5)^2) + (1^2) + 2×1×(2^0.5)

= ((2^0.5) + 1)^2 = A^2, say.

Note (x + (1/x) )^2 = (x^2)+(1/(x^2)) + 2 = V+2 where V is the Answer value.

V= (x+ (1/x) )^2 — 2 = ((A^2)+(1/(A^2))^2 — 2

= (A^4) + (1/(A^4)) — 2 =

((2^0.5)+1)^4) + (1/((2^0.5)+1)^4) — 2×A×(1/A)

= (x-(1/x))^2 = ((x^2)-1)/x)^2 = ((x+1)×(x-1)/x)^2

= ((2×(2+(2^0.5)))×(2×(1+(2^0.5))) )^2 ÷ (((2^0.5)+1)^4)

= 16 × ((2+(2^0.5))^2) ÷ (((2^0.5)+1)^2)

Now rationalize the denominator by multiplying with (((2^0.5)-1)^2) to get Denominator = (2-1)^2=1 and the

Numerator = 16× (((2^0.5)+2)×((2^0.5)-1) )^2

= 16×((2 + (2^0.5) -2) ^2)

= 16×2

If x = 4/ (2√3+3√2), then what is the value of x+1/x?

What is the value of x+1/x if x=7-4√3?

If x=3-2√2, then what is the value of (√x) - (1/√x)?

If x= 3+2√2, then what is the value of (x+1/x)?

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If x=3+2√3, then what is the value of √x+1/√x?

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