Q38 / 90 · JEE Main 2023 (12 Apr Shift 1) · Mathematics

Let α, β be the roots of the quadratic equation x2+6x+3=0. Then α23+β23+α14+β14α15+β15+α10+β10 is equal to

  • Correct answer
    A
    81
  • B
    9
  • C
    72
  • D
    729

Solution

To find the value of α23+β23+α14+β14α15+β15+α10+β10,

Let an=αn+βn

Hence,

α23+β23+α14+β14α15+β15+α10+β10=a23+a14a15+a10

Now, x2+6x+3=0 has roots α & β

So, x=-6±-62

⇒x=6-1±i2

⇒x=3-1±i2

Hence, α=3 ei3π4 and β=3ei5π4

Now, solving α23+β23+α14+β14α15+β15+α10+β10

=323ei23×3π4+ei23×5π4+314ei14×3π4+ei14×5π4315ei15×3π4+ei15×5π4+310ei10×3π4+ei×10×5π4

=3439ei23×3π4+ei23×5π4+ei14×3π4+ei14×5π435ei15×3π4+ei15×5π4+ei10×3π4+ei×10×5π4

=9×391+i-1+i2+0351+i-1+i2+0as ei21π2+ei35π2=0+isin21π2+0+isin35π2=i-i=0

=9×392i2352i2

=9×34

=81

Hence, the correct answer is option: 'A'.

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