Question

If a= 6 - √35,find the value of a²+1/a²

Answer 1

a=6-_/35
1/a=1/(6-_/35)=(6+_/35)(6^2-35)=6+_/35 (rationalised the denominator)
So, a^2+1/a^2=(6-_/35)^2(6+_/35)^2
=36+35-12_/35 +36+35+12_/35
=71+71=142...

Answer 2

Answer:

$a^{2}+\frac{1}{a^{2}}=142$

Step-by-step explanation:

$Given \\a=6-\sqrt{35}---(1)$

$\frac{1}{a}=\frac{1}{6-\sqrt{35}}\\=\frac{6+\sqrt{35}}{(6-\sqrt{35})(6+\sqrt{35})}\\=\frac{6+\sqrt{35}}{6^{2}-(\sqrt{35})^{2}}\\=\frac{6+\sqrt{35}}{36-35}\\=6+\sqrt{35}---(2)$

$a+\frac{1}{a}\\=6-\sqrt{35}+6+\sqrt{35}\\=12---(3)\\\: [From\: (1)\:and\:(2)]$

$Now,\\a^{2}+\frac{1}{a^{2}}\\=\big(a+\frac{1}{a}\big)^{2}-2\\=(12)^{2}-2\\=144-2\\=142\:[from \:(3)]$

Therefore,

$a^{2}+\frac{1}{a^{2}}=142$

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