Let be the roots of the quadratic equation x 2 + 6 x + 3 = 0 . Then 23 + 23 + 14 + 14 15 + 15 + 10 + 10 is equal to [2023]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
Let $α, β$ be the roots of the quadratic equation $x^{2} + \sqrt{6} x + 3 = 0 .$ Then $\frac{α^{23} + β^{23} + α^{14} + β^{14}}{α^{15} + β^{15} + α^{10} + β^{10}}$ is equal to [2023]

1 729

2 9

3 81

4 72

Ans.
(3)

$Given, x^{2} + \sqrt{6} x + 3 = 0 ...(i)$

On solving eqn. (i), we get its roots

$α, β = \frac{- \sqrt{6} \pm \sqrt{6 - 12}}{2} = \frac{- \sqrt{6} \pm \sqrt{6} i}{2}$

$α, β = \sqrt{3} e^{\pm 3 π i / 4}$

So, required expression becomes:

$= \frac{{(\sqrt{3})}^{23} (2 \cos \frac{69 π}{4}) + {(\sqrt{3})}^{14} (2 \cos \frac{42 π}{4})}{{(\sqrt{3})}^{15} (2 \cos \frac{45 π}{4}) + {(\sqrt{3})}^{10} (2 \cos \frac{30 π}{4})}$

$= \frac{{(\sqrt{3})}^{23} \times 2 (- 1) + 0}{{(\sqrt{3})}^{15} \times 2 (- 1) + 0}$

$= {(\sqrt{3})}^{8} = 81$

Want Us to Email you About Special Offers & Updates?