if a= 6 - rt35, find the value of a^3 + 1/a^2
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Answered by
1
Answer:
a = (6 - sqrt 35)
a + 1/a
= (6 - sqrt 35) + 1/(6 - sqrt 35)
= ((6 - sqrt 35)^2 + 1) / (6 - sqrt 35)
= (36 -12 sqrt 35 + 35 +1) / (6 - sqrt 35)
= (72 -12 sqrt 35) / (6 - sqrt 35)
= 12 * (6 - sqrt 35) / (6 - sqrt 35)
= 12
=> a + 1 / a = 12 --- (1)
a^2 + 1 / a^2
= (a + 1/a)^2 - 2
= 12^2 - 2 --- from Eq[1]
= 144 - 2
= 142
=> a^2 + 1/a^2 = 142
Step-by-step explanation:
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Answered by
7
Answer:
Assuming you mean a^2 - (1/a)^2, this is
(6 - √35)^2 + (1 / (6 - √35))^2
(36 - 12√35 + 35) + (1 / (36 - 12√35 + 35))
(71 - 12√35) + (1 / (71 - 12√35))
(71 - 12√35) + [ (71+12√35) / (71 - 12√35)(71+12√35) ]
(71 - 12√35) + [ (71+12√35) / (5041 - 5040) ]
(71 - 12√35) + (71+12√35)
142
hope it helps you. :D
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