If =2−√3, then find the value of (−1/) , (+ 1/) and (^2+ 1/^2).
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Let √x - 1/√x = a
Squaring both the sides,
x + 1/x - 2 = a^2
Putting the value,
3–2√2 + 1/(3–2√2) - 2 = a^2
a^2 = 1 - 2√2 + 1/(3–2√2)
= [(3–2√2) (1–2√2) + 1] / 3–2√2
= {3 - 8√2 + 9} / 3–2√2
= [12 - 8√2] / 3–2√2
Rationalising both the sides
= {(12 - 8√2)(3+2√2)} ÷ (9–8)
= 36 + 24√2 - 24√2 + 16(2)
= 36 - 32
=> 4
a^2 = 4
a = √4
a = 2, -2
So,
√x - 1/√x = a = 2, -2.
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Explanation:
If =2−√3, then find the value of (−1/) , (+ 1/) and (^2+ 1/^2).
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