If x−1x=8x - \dfrac{1}{x} = 8x−x1=8, the value of x2+1x2−8x^2 + \dfrac{1}{x^2} - 8x2+x21−8 is :
56
58
70
-4
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(x−1x)2=x2+1x2−(2×x×1x)⇒(x−1x)2=x2+1x2−2⇒x2+1x2=(x−1x)2+2\Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - \Big(2 \times x \times \dfrac{1}{x}\Big) \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = \Big(x - \dfrac{1}{x}\Big)^2 + 2 \\[1em](x−x1)2=x2+x21−(2×x×x1)⇒(x−x1)2=x2+x21−2⇒x2+x21=(x−x1)2+2
Subtracting 8 on both sides,
⇒x2+1x2−8=(x−1x)2+2−8⇒x2+1x2−8=82−6⇒x2+1x2−8=64−6⇒x2+1x2−8=58.\Rightarrow x^2 + \dfrac{1}{x^2} - 8 = \Big(x - \dfrac{1}{x}\Big)^2 + 2 - 8 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} - 8 = 8^2 - 6 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} - 8 = 64 - 6 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} - 8 = 58.⇒x2+x21−8=(x−x1)2+2−8⇒x2+x21−8=82−6⇒x2+x21−8=64−6⇒x2+x21−8=58.
Hence, Option 2 is the correct option.
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For x = 5 and y = 3, the value of x2 + y2 - 2xy + 7 is :
11
169
71
1
If x+1x=2x + \dfrac{1}{x} = 2x+x1=2, the value of x2+1x2+5x^2 + \dfrac{1}{x^2} + 5x2+x21+5 is :
-1
2
9
7
If x2+1x2=9x^2 + \dfrac{1}{x^2} = 9x2+x21=9, the value of x4+1x4+5x^4 + \dfrac{1}{x^4} + 5x4+x41+5 is :
78
86
84
81
If x2 - 3x + 1 = 0, the value of
x2+1x2+1x^2 + \dfrac{1}{x^2} + 1x2+x21+1 is :
8
10
5
109\dfrac{10}{9}910
