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15 votes
15 votes

The equation $7x^{7}+14x^{6}+12x^{5}+3x^{4}+12x^{3}+10x^{2}+5x+7=0$ has

  1. All complex roots
  2. At least one real root
  3. Four pairs of imaginary roots
  4. None of the above
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1 comment

(B) is correct.

Imaginary roots always exist in pairs in these types of equations.

No of imaginary roots cannot be 8 (4 pairs) as the degree of the polynomial is 7. (max no of roots = degree of the polynomial). So (C) is wrong.

No of imaginary roots can be 6 and number of real roots has to be at least once.

If no of imaginary roots is 4. No of real roots = 3

If no of imaginary roots is 2. No of real roots = 5

If no of imaginary roots is 0. No of real roots = 7

So even without finding the roots, we can say the number of real roots is at least 1.
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5 Answers

33 votes
33 votes
Best answer

Since the polynomial has the highest degree $7$. So there are $7$ roots possible for it

Now suppose if an imaginary number $a+bi$ is also the root of this polynomial then $a-bi$ will also be the root of this polynomial

That means there must be an even number of complex roots possible because they occur in pairs.

Now we will solve this question option wise

A) All complex root

This is not possible. The polynomial has $7$ roots and as I mention a polynomial should have an even number of complex roots and  $7$ is not even. So this option is wrong

B) At least one real root

This is possible. Since polynomial has $7$ roots and only an even number of the complex root is possible, that means this polynomial has max $6$ complex roots and Hence a minimum of one real root. So this option is correct

C) Four pairs of imaginary roots

$4$ pair means $8$ complex root. But this polynomial can have at most $7$ roots. So this option is also wrong

Hence answer should be (B).

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3 Comments

Which is that one real root?
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That means there must be even number of complex root possible becoz they occur in pair.

plz exlain this line

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0

suppose if an imaginary number a+bi is also root of this polynomial then a-bi will also be the root of this polynomial which makes 2 roots. suppose other root x+iy present then it makes x-iy to be a root. so totally 4 roots possible. so if complext roots present they must be even in number

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7 votes
7 votes

Imaginary or complex roots always occur in pairs

Since given polynomial is of degree 7, total roots must be 7.(Some of them might repeat, but still these 7 will count those repeated ones too!)

Now, we'll consider possibilities

Real roots Complex Possibility?
7 0 Yes
6 1 No
5 2 Yes
4 3 No
3 4 Yes
2 5 No
1 6 Yes
0 7 No

 

Now look at options

  1. All complex roots (Not possible from the table)!!
  2. At least one real root.. -Yes many cases are there.
  3. Four pairs of imaginary roots-Means total of 8 roots-surely not!!

Answer - B

 

1 vote
1 vote
here just try some value sto check the behaviour of th function let put x=0 f is +ive, lets try x=-1 value is -ive  , hence aatlest 1 real root exist between -1 and 0.
1 vote
1 vote

The Intermediate Value Theorem states that if $f$ is continuous in an interval (generally closed $[a, b]$) and $f(a)$ and $f(b)$ have differing signs, then there is a root of $f$ in $(a, b)$.

Due to this, polynomials (which are known to be continuous everywhere) of odd degree have atleast one real root.

https://mathinsight.org/intermediate_value_theorem_location_roots_refresher

http://www.sosmath.com/calculus/limcon/limcon06/limcon06.html

Answer:

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