GATE CSE 1987 | Question: 1-xxii

Question

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

• A. All complex roots
• B. At least one real root
• C. Four pairs of imaginary roots
• D. None of the above

Comments

(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.

Answer 1

Using Descartes' rule$^{1}$of sign,which is used to determine the number of real zeros of a polynomial function.

1. https://www.mathplanet.com/education/algebra-2/polynomial-functions/descartes-rule-of-sign

Finding no of +ve real no (ignore the coefficient of polynomial):

Let f(x) be $7x^{7} + 14x^{6} + 12x^{5} + 3x^{4} + 12x^{3} + 10x^{2} + 5x +7 =0$.

As there is no sign change in f(x), it has no positive real roots.

Replace x by -x in f(x) to find negative real roots

$f(-x) = -7x^{7} + 14x^{6} - 12x^{5} + 3x^{4} - 12x^{3} + 10x^{2} - 5x +7 =0$.

As change of sign takes place in 5 pairs, it means that there are 5 negative real roots.

As the degree of polynomial is 7,max 7 roots are possible. Out of which 5 negative real roots are present and 2 imaginary root (As imaginary roots always exist as conjugate pairs).

Option B: Atleast 1 real roots

Link for reference: https://math.stackexchange.com/questions/745583/intuition-behind-descartes-rule-of-signs