Let alpha_1, alpha_2, ldots, alpha_7 be the r | 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

$\alpha_1 \alpha_2-\alpha_3 \alpha_4 +\alpha_5 \alpha_6$

$=1-(-3)+5=9$

Let $\alpha_1, \alpha_2, \ldots, \alpha_7$ be the roots of the equation $x^7+$ $3 x^5-13 x^3-15 x=0$ and $\left|\alpha_1\right| \geq\left|\alpha_2\right| \geq \ldots \geq\left|\alpha_7\right|$. Then $\alpha_1 \alpha_2-\alpha_3 \alpha_4+\alpha_5 \alpha_6$ is equal to $..................$.

Similar Questions

Number of positive integral values of $'K'$ for which the equation $k = \left| {x + \left| {2x - 1} \right|} \right| - \left| {x - \left| {2x - 1} \right|} \right|$ has exactly three real solutions, is

The number of integral values of $m$ for which the quadratic expression, $(1 + 2m)x^2 -2(1+ 3m)x + 4(1 + m),$ $x\in R,$ is always positive, is

The number of solutions, of the equation $\mathrm{e}^{\sin x}-2 e^{-\sin x}=2$ is

If $x+\frac{1}{x}=a, x^2+\frac{1}{x^3}=b$, then $x^3+\frac{1}{x^2}$ is

Solution of the equation $\sqrt {x + 3 - 4\sqrt {x - 1} } + \sqrt {x + 8 - 6\sqrt {x - 1} } = 1$ is