If x -5-2- then find the value of x -1- x-A- 4B- 2 - 2 -5D- -5

The correct option is A 4 x = √(5) + 2 ∴1/x =1/√(5)+ 2=1 × (√(5) 2)/(√(5)+2)(√(5) 2) =√(5) 2/(√(5))^2 (2)^2=√(5) 2/5 4 =√(5) 2/1 =√(5) 2 ∴ x 1/x=(√(5) + 2) ...

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Byju's Answer

Standard IX

Mathematics

Addition and Subtraction of Numbers with Square Root

If x =√5+2, t...

Question

If $x = \sqrt{5} + 2$ , then find the value of $x - \frac{1}{x}$ .

$2 \sqrt{5}$

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$\sqrt{5}$

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Solution

The correct option is B
4

$x = \sqrt{5} + 2$

$∴ \frac{1}{x} = \frac{1}{\sqrt{5} + 2} = \frac{1 \times (\sqrt{5} - 2)}{(\sqrt{5} + 2) (\sqrt{5} - 2)}$

$= \frac{\sqrt{5} - 2}{(\sqrt{5})^{2} - (2)^{2}} = \frac{\sqrt{5} - 2}{5 - 4} = \frac{\sqrt{5} - 2}{1}$

$= \sqrt{5} - 2$

$∴ x - \frac{1}{x} = (\sqrt{5} + 2) - (\sqrt{5} - 2)$

$= \sqrt{5} + 2 - \sqrt{5} + 2 = 4$

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If x=√5+2, then find the value of x−1x.

The correct option is B 4 x=√5+2 ∴1x=1√5+2=1×(√5−2)(√5+2)(√5−2) =√5−2(√5)2−(2)2=√5−25−4=√5−21 =√5−2 ∴x−1x=(√5+2)−(√5−2) =√5+2−√5+2=4

If $x = \sqrt{5} + 2$ , then find the value of $x - \frac{1}{x}$ .