If x = (4−15)(4 - \sqrt{15})(4−15), find the values of (x2+1x2)\Big(x^2 + \dfrac{1}{x^2}\Big)(x2+x21).
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Given,
x = (4−15)(4 - \sqrt{15})(4−15)
∴1x=1(4−15)\therefore \dfrac{1}{x} = \dfrac{1}{(4 - \sqrt{15})}∴x1=(4−15)1
Rationalizing,
⇒1x=1(4−15)×(4+15)(4+15)=4+1542−(15)2=4+1516−15=4+151=4+15.\Rightarrow \dfrac{1}{x} = \dfrac{1}{(4 - \sqrt{15})} \times \dfrac{(4 + \sqrt{15})}{(4 + \sqrt{15})} \\[1em] = \dfrac{4 + \sqrt{15}}{4^2 - (\sqrt{15})^2} \\[1em] = \dfrac{4 + \sqrt{15}}{16 - 15} \\[1em] = \dfrac{4 + \sqrt{15}}{1} \\[1em] = 4 + \sqrt{15}.⇒x1=(4−15)1×(4+15)(4+15)=42−(15)24+15=16−154+15=14+15=4+15.
By formula,
x2+1x2=(x+1x)2−2x^2 + \dfrac{1}{x^2} = \Big(x + \dfrac{1}{x}\Big)^2 - 2x2+x21=(x+x1)2−2
Substituting values we get :
⇒x2+1x2=(4−15+4+15)2−2=82−2=64−2=62.\Rightarrow x^2 + \dfrac{1}{x^2} = (4 - \sqrt{15} + 4 + \sqrt{15})^2 - 2 \\[1em] = 8^2 - 2 \\[1em] = 64 - 2 \\[1em] = 62.⇒x2+x21=(4−15+4+15)2−2=82−2=64−2=62.
Hence, x2+1x2x^2 + \dfrac{1}{x^2}x2+x21 = 62.
Answered By
Show that : 1(3−8)+1(7−6)+1(5−2)−1(8−7)−1(6−5)=5\dfrac{1}{(3 - \sqrt{8})} + \dfrac{1}{(\sqrt{7} - \sqrt{6})}+ \dfrac{1}{(\sqrt{5} - 2)} - \dfrac{1}{(\sqrt{8} - \sqrt{7})} - \dfrac{1}{(\sqrt{6} - \sqrt{5})} = 5(3−8)1+(7−6)1+(5−2)1−(8−7)1−(6−5)1=5
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