if a=6-√35, find the value of a²+1/a²
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a=6-√35
1/a= 1/6-√35 = 1/6-√35 ×6+√35/6+√35
= 6+√35/(6)^2 - (√35)^2
= 6+√35/36-35 = 6+√35
now, (6-√35)^2 + (6+√35)^2
= {(6)^2 + (√35)^2 - 2(6)(√35)} + {(6)^2 + (√35)^2 + 2(6)(√35)
= (36+35-12√35) + (36+35+12√35)
= 71-12√35+71+12√35
= 142
1/a= 1/6-√35 = 1/6-√35 ×6+√35/6+√35
= 6+√35/(6)^2 - (√35)^2
= 6+√35/36-35 = 6+√35
now, (6-√35)^2 + (6+√35)^2
= {(6)^2 + (√35)^2 - 2(6)(√35)} + {(6)^2 + (√35)^2 + 2(6)(√35)
= (36+35-12√35) + (36+35+12√35)
= 71-12√35+71+12√35
= 142
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