# ISI2015-MMA-12

199 views

Consider the polynomial  $x^5+ax^4+bx^3+cx^2+dx+4$ where $a,b,c,d$ are real numbers. If $(1+2i)$ and $(3-2i)$ are two two roots of this polynomial then the value of $a$ is

1. $-524/65$
2. $524/65$
3. $-1/65$
4. $1/65$

recategorized

Answer: $A$

Let the roots be $\alpha_1, \alpha_2, \alpha_3, \alpha_4,$ and $\alpha_5$

Now,

We know that the roots of the coefficients are the real numbers and two imaginary roots are given.
Therefore with this hint, we know other two roots as well, which are nothing but just the conjugate of these two imaginary roots. So, total $4$ roots are known out of $5$

Now, for any polynomial.

$$(x+\alpha_1)(x + \alpha_2)(x+\alpha_3)(x+\alpha_4)(x+\alpha_5) = 0$$

$$\therefore \; x^5+ x^4(\alpha_1 + \alpha_2+\alpha_3+\alpha_4+\alpha_5)+x^3(\alpha_1\alpha_2+\alpha_2\alpha_3+\dots) + x^2(\alpha_1.\alpha_2.\alpha_3 + \alpha_2.\alpha_3.\alpha_4+\dots) + x^1(\alpha_1\alpha_2\alpha_3\alpha_4 + \alpha_2\alpha_3\alpha_4\alpha_5+\dots) + x^0(\alpha_1\alpha_2\alpha_3\alpha_4\alpha_5) = 0$$

So, product of roots is given by:

$$\alpha_1\alpha_2\alpha_3\alpha_4\alpha_5 = 4 \implies (1+2\iota)(1-2\iota)(3-2\iota)(3+2\iota) \color{blue}{\alpha_5} = 4$$

$$\implies \color{blue}{\alpha_5} = \frac{4}{65}$$

Now, Sum of of roots is given by:

$$\alpha_1 + \alpha_2 + \alpha_3 + \alpha_4 + \alpha_5 = -a$$

$$\implies (1+2\iota)+(1-2i)+(3+2\iota)+(3-2\iota)+\frac{4}{65} = \color{red} {-a}$$

$$\implies \color {red}{a} = \color {red}{-\frac{524}{65}}$$

$\therefore \;A$ is the correct answer.

edited by
2
If you have taken α1,α2,α3,α4, and α5 as roots then equation will be:

(x-α1)(x-α2)(x-α3)(x-α4)(x-α5)=0

and product of roots will be = -4.

Also we know if we have a equation of the form:

$a_0x^{n} + a_{1}x^{n-1}+a_2x^{n-2}+a_3x^{n-3} + ....... + a_{n-1}x+a_n = 0$

then sum of roots taken 'p roots' at a time is given by:

$\sum x_1x_2x_3...x_p = (-1)^{p}a_p/a_0$

So here product of the roots will be $= (-1)^{5}4/1 = -4$    using which finally we will get a = -516/65

Correct me if I am wrong.
0
0
0
why the

α1+α2+α3+α4+α5=−a ? it should be a only?

## Related questions

1
352 views
If $a,b$ are positive real variables whose sum is a constant $\lambda$, then the minimum value of $\sqrt{(1+1/a)(1+1/b)}$ is $\lambda \: – 1/\lambda$ $\lambda + 2/\lambda$ $\lambda+1/\lambda$ None of the above
Consider the following system of equivalences of integers, $x \equiv 2 \text{ mod } 15$ $x \equiv 4 \text{ mod } 21$ The number of solutions in $x$, where $1 \leq x \leq 315$, to the above system of equivalences is $0$ $1$ $2$ $3$
The number of real solutions of the equations $(9/10)^x = -3+x-x^2$ is $2$ $0$ $1$ none of the above
Let $x$ be a positive real number. Then $x^2+\pi ^2 + x^{2 \pi} > x \pi+ (\pi + x) x^{\pi}$ $x^{\pi}+\pi^x > x^{2 \pi} + \pi ^{2x}$ $\pi x +(\pi+x)x^{\pi} > x^2+\pi ^2 + x^{2 \pi}$ none of the above