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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. $1/65$
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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.

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6 Comments

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.
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2
Correct answer is -516/65.
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A is not a correct answer , your approach is wrong.
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why the

α1+α2+α3+α4+α5=−a ? it should be a only?
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@  brother the sum of roots will be “-a”

thus  the 5th root becomes -4/65  which gives  a= -516/65

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@rish1602 I am also getting this result but there is no such options given

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0

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