If x = 5−2,x+1x\sqrt{5} - 2, x + \dfrac{1}{x}5−2,x+x1 is equal to :
252\sqrt{5}25
4
454\sqrt{5}45
-4
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Given,
x = 5−2\sqrt{5} - 25−2
1x=15−2\dfrac{1}{x} = \dfrac{1}{\sqrt{5} - 2}x1=5−21
Rationalizing,
⇒1x=15−2×5+25+2=5+2(5)2−(2)2=5+25−4=5+21=5+2.⇒x+1x=5−2+5+2=25.\Rightarrow \dfrac{1}{x} = \dfrac{1}{\sqrt{5} - 2} \times \dfrac{\sqrt{5} + 2}{\sqrt{5} + 2} \\[1em] = \dfrac{\sqrt{5} + 2}{(\sqrt{5})^2 - (2)^2} \\[1em] = \dfrac{\sqrt{5} + 2}{5 - 4} \\[1em] = \dfrac{\sqrt{5} + 2}{1} \\[1em] = \sqrt{5} + 2. \\[1em] \Rightarrow x + \dfrac{1}{x} = \sqrt{5} - 2 + \sqrt{5} + 2 \\[1em] = 2\sqrt{5}.⇒x1=5−21×5+25+2=(5)2−(2)25+2=5−45+2=15+2=5+2.⇒x+x1=5−2+5+2=25.
Hence, Option 1 is the correct option.
Answered By
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Write a pair of irrational numbers whose product is irrational.
Write a pair of irrational numbers whose product is rational.
If x = 1 + 2, then (x+1x)2\sqrt{2}, \text{ then } \Big(x + \dfrac{1}{x}\Big)^22, then (x+x1)2 is :
222\sqrt{2}22
8
424\sqrt{2}42
227+31243\dfrac{2\sqrt{27} + 3\sqrt{12}}{4\sqrt{3}}43227+312 is equal to :
232\sqrt{3}23
323\sqrt{2}32
3
3+2\sqrt{3} + \sqrt{2}3+2
