If x-7-4-3 then x-1-x - a 8-3 b 14 c 49 d 48

X=(7+4√(3)) 1/x = 1/(7+4√(3)) = 1(7 4√(3)) /(7+4√(3)) (7 4√(3)) =7 4√(3)/7^2 (4√(3))^2 =7 4√(3)/49 48 = 7 4√(3) x + 1/x = 7+4√(3) + 7 4√(3) = 14 H ...

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Standard IX

Mathematics

Rationalisation

If x=7+4√3 th...

Question

If $x = (7 + 4 \sqrt{3})$ then $(x + \frac{1}{x})$ =?

(a) $8 \sqrt{3}$

(b) 14

(c) 49

(d) 48

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Solution

$x = (7 + 4 \sqrt{3}) \frac{1}{x} = \frac{1}{(7 + 4 \sqrt{3})} = \frac{1 (7 - 4 \sqrt{3})}{(7 + 4 \sqrt{3}) (7 - 4 \sqrt{3})} = \frac{7 - 4 \sqrt{3}}{7^{2} - (4 \sqrt{3})^{2}} = \frac{7 - 4 \sqrt{3}}{49 - 48} = 7 - 4 \sqrt{3} x + \frac{1}{x} = 7 + 4 \sqrt{3} + 7 - 4 \sqrt{3} = 14$

Hence, the correct answer is option (b).

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If x=(7+4√3) then (x+1x)=? (a) 8√3 (b) 14 (c) 49 (d) 48

x=(7+4√3)1x=1(7+4√3)=1(7−4√3)(7+4√3)(7−4√3)=7−4√372−(4√3)2=7−4√349−48=7−4√3 x+1x=7+4√3+7−4√3=14 Hence, the correct answer is option (b).

If $x = (7 + 4 \sqrt{3})$ then $(x + \frac{1}{x})$ =?

(a) $8 \sqrt{3}$

(b) 14

(c) 49

(d) 48