JEE Main 2023 — Mathematics Algebra

Given:

$${x}^{7}+3{x}^{5}-13{x}^{3}-15x=0$$x7+3x5−13x3−15x=0

$$\Rightarrow x({x}^{6}+3{x}^{4}-13{x}^{2}-15)=0$$⇒x(x6+3x4−13x2−15)=0

So, $x=0$x=0 is one of the root.

Now,

$$({x}^{6}+3{x}^{4}-13{x}^{2}-15)=0$$(x6+3x4−13x2−15)=0

Put ${x}^{2}=t$x2=t, then we have

$${t}^{3}+3{t}^{2}-13t-15=0$$t3+3t2−13t−15=0

$$\Rightarrow (t-3)({t}^{2}+6t+5)=0$$⇒(t−3)(t2+6t+5)=0

$$\Rightarrow (t-3)(t+1)(t+5)=0$$⇒(t−3)(t+1)(t+5)=0

So, $t=3,t=-1,t=-5$t=3,t=−1,t=−5

Now we are getting

$${x}^{2}=3,{x}^{2}=-1,{x}^{2}=-5$$x2=3,x2=−1,x2=−5

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

From the given condition $|{\alpha }_{1}|\geq |{\alpha }_{2}|\geq \ldots .\geq |{\alpha }_{7}|$∣α1​∣≥∣α2​∣≥….≥∣α7​∣

We can clearly say that $|{\alpha }_{7}|=0$∣α7​∣=0 and

and $|{\alpha }_{6}|=\sqrt{5}=|{\alpha }_{5}|$∣α6​∣=5​=∣α5​∣

and $|{\alpha }_{4}|=\sqrt{3}=|{\alpha }_{3}|$∣α4​∣=3​=∣α3​∣ and

$$|{\alpha }_{2}|=1=|{\alpha }_{1}|$$∣α2​∣=1=∣α1​∣

So we can have,

$${\alpha }_{1}=\sqrt{5}i,{\alpha }_{2}=-\sqrt{5}i,{\alpha }_{3}=\sqrt{3}$$α1​=5​i,α2​=−5​i,α3​=3​

$${\alpha }_{4}=-\sqrt{3},{\alpha }_{5}=i,{\alpha }_{6}=-i$$α4​=−3​,α5​=i,α6​=−i

$${\alpha }_{7}=0$$α7​=0

Hence,

$${\alpha }_{1}{\alpha }_{2}-{\alpha }_{3}{\alpha }_{4}+{\alpha }_{5}{\alpha }_{6}$$α1​α2​−α3​α4​+α5​α6​

$$=5-(-3)+1=9$$=5−(−3)+1=9