If x-1-x-5- find the values of x2-1-x2-and-x4-1-x4-

X+1/x=√(5) Squaring both sides, ∵ (x+1/x )^2=(√(5))^2⇒ x^2+1/x^2+2=5 (i) ∴ x^2+1/x^2=5 2=3 Again squaring, (x^2+1/x^2 )^2=(3)^2 ⇒ x^4+1/x^4+2=9 ⇒ x^4+1/x^4=9 2 ...

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Byju's Answer

Standard IX

Mathematics

Algebraic Identities

If x+1/x=√5, ...

Question

If $x + \frac{1}{x} = \sqrt{5}$ , find the values of $x^{2} + \frac{1}{x^{2}} a n d x^{4} + \frac{1}{x^{4}}$ .

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Solution

$x + \frac{1}{x} = \sqrt{5}$

Squaring both sides,

$∵ {(x + \frac{1}{x})}^{2} = (\sqrt{5})^{2} \Rightarrow x^{2} + \frac{1}{x^{2}} + 2 = 5 (i) ∴ x^{2} + \frac{1}{x^{2}} = 5 - 2 = 3 Again squaring, {(x^{2} + \frac{1}{x^{2}})}^{2} = (3)^{2} \Rightarrow x^{4} + \frac{1}{x^{4}} + 2 = 9 \Rightarrow x^{4} + \frac{1}{x^{4}} = 9 - 2 = 7 ∴ x^{2} + \frac{1}{x^{2}} = 3, a n d x^{4} + \frac{1}{x^{4}} = 7$

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If x+1x=√5, find the values of x2+1x2 and x4+1x4.

x+1x=√5 Squaring both sides, ∵ (x+1x)2=(√5)2⇒x2+1x2+2=5(i) ∴ x2+1x2=5−2=3Again squaring,(x2+1x2)2=(3)2⇒ x4+1x4+2=9⇒ x4+1x4=9−2=7∴x2+1x2=3,and x4+1x4=7

If $x + \frac{1}{x} = \sqrt{5}$ , find the values of $x^{2} + \frac{1}{x^{2}} a n d x^{4} + \frac{1}{x^{4}}$ .