Let 1 , 2 , … , 7 be the roots of the equation x 7 + 3 x 5 - 13 x 3 - 15 x = 0 and 1 2 … 7 . Then 1 2 - 3 4 + 5 6 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 $α_{1}, α_{2}, \dots, α_{7}$ be the roots of the equation $x^{7} + 3 x^{5} - 13 x^{3} - 15 x = 0$ and $| α_{1} |$ $\geq$ $| α_{2} |$ $\geq \dots \geq$ $| α_{7} |$ . Then $α_{1} α_{2} - α_{3} α_{4} + α_{5} α_{6}$ is equal to ______. [2023]

Ans.
(9)

Given equation is, $x^{7} + 3 x^{5} - 13 x^{3} - 15 x = 0$

$\Rightarrow x (x^{6} + 3 x^{4} - 13 x^{2} - 15) = 0$

Here, $x = 0$ is one of the roots. Replacing $x^{2} = t$

So, $t^{3} + 3 t^{2} - 13 t - 15 = 0 \Rightarrow (t - 3) (t^{2} + 6 t + 5) = 0$

So, $t = 3, t = - 1, t = - 5$

Now, $x^{2} = 3, x^{2} = - 1, x^{2} = - 5$

$\Rightarrow x = \pm \sqrt{3}, x = \pm i, x = \pm \sqrt{5} i$

From the given condition, $| α_{1} |$ $\geq$ $| α_{2} |$ $\geq \dots \geq$ $| α_{7} |$

We can say that $| α_{7} |$ $= 0$ and $| α_{1} |$ $= \sqrt{5} =$ $| α_{2} |$

and $| α_{4} |$ $= \sqrt{3} =$ $| α_{3} |$ , and $| α_{5} |$ $= 1 =$ $| α_{6} |$

So, we have,

$α_{5} = i, α_{6} = - i, α_{3} = \sqrt{3}, α_{4} = - \sqrt{3}, α_{2} = \sqrt{5} i, α_{1} = - \sqrt{5} i$

So, $α_{1} α_{2} - α_{3} α_{4} + α_{5} α_{6} = 1 - (- 3) + 5 = 9$

Want Us to Email you About Special Offers & Updates?