search
Log In
0 votes
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. $1/65$
in Quantitative Aptitude
recategorized by
199 views

1 Answer

0 votes

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
Correct answer is -516/65.
0
A is not a correct answer , your approach is wrong.
0
why the

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

Related questions

0 votes
1 answer
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
asked Sep 23, 2019 in Quantitative Aptitude Arjun 352 views
0 votes
1 answer
2
329 views
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 $0$ $1$ $2$ $3$
asked Sep 23, 2019 in Quantitative Aptitude Arjun 329 views
0 votes
1 answer
3
193 views
The number of real solutions of the equations $(9/10)^x = -3+x-x^2$ is $2$ $0$ $1$ none of the above
asked Sep 23, 2019 in Quantitative Aptitude Arjun 193 views
1 vote
3 answers
4
432 views
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
asked Sep 23, 2019 in Quantitative Aptitude Arjun 432 views
...