ધારો કે alpha_1, alpha_2, ldots, alpha_7 એ સમ | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

Given equation can be rearranged as

$x\left(x^6+3 x^4-13 x^2-15\right)=0$

clearly $x=0$ is one of the root and other part can be observed by rep

Vedclass · Accepted Answer

$9$

Vedclass · Answer

Given equation can be rearranged as

$x\left(x^6+3 x^4-13 x^2-15\right)=0$

clearly $x=0$ is one of the root and other part can be observed by replacing $x ^2= t$ from which we have $\quad t^3+3 t^2-13 t-15=0$

$\Rightarrow \quad( t -3)\left( t ^2+6 t +5\right)=0$

So, $\quad t=3, t=-1, t=-5$

Now we are getting $x ^2=3, x ^2=-1, x ^2=-5$

$\Rightarrow x=\pm \sqrt{3}, x=\pm i , x=\pm \sqrt{5} i$

From the given condition $\left|\alpha_1\right| \geq\left|\alpha_2\right| \geq \ldots \geq\left|\alpha_2\right|$

We can clearly say that $\left|\alpha_7\right|=0$ and and

$\left|\alpha_6\right|=\sqrt{5}=\left|\alpha_5\right|$

and $\quad\left|\alpha_4\right|=\sqrt{3}=\left|\alpha_3\right|$ and $\left|\alpha_2\right|=1=\left|\alpha_1\right|$

So we can have, $\alpha_1=\sqrt{5} i , \alpha_2=-\sqrt{5} i , \alpha_3=\sqrt{3} i$,

$\alpha_4=-\sqrt{3}, \alpha_5= i , \alpha_6=- i$

Hence

$\alpha_1 \alpha_2-\alpha_3 \alpha_4 +\alpha_5 \alpha_6$

$=1-(-3)+5=9$

ધારો કે $\alpha_1, \alpha_2, \ldots, \alpha_7$ એ સમીકરણ $x^7+3 x^5-13 x^3-15 x=0$ નાં બીજ છે અને $\left|a_1\right| \geq\left|\alpha_2\right| \geq \ldots \geq\left|\alpha_7\right|$ તો $\alpha_1 \alpha_2-\alpha_3 \alpha_4+\alpha_5 \alpha_6=......$

Similar Questions

જો $\alpha ,\beta$ એ સમીકરણ $x^2 -ax + b = 0$ ના ઉકેલો હોય અને $\alpha^n + \beta^n = V_n$, હોય તો

સમીકરણ ${x^2}\, + \,\left| {2x - 3} \right|\, - \,4\, = \,0,$ ના ઉકેલો નો સરવાળો ...... થાય.

સમીકરણ $x^2 - |x| - 6 = 0$ ના વાસ્તવિક બીજનો ગુણાકાર = .......

સમીકરણ ${e^{\sin x}} - {e^{ - \sin x}} - 4$ $ = 0$ ના વાસ્તવિક બીજની સંખ્યા મેળવો.