# Problems on solving an equation.

I came across this question while studying the theory of equations. I also have the answer plus explanation given. But I could not understand them. Please Help.

Let $(a_1, a_2, a_3, a_4, a_5)$ denote a rearrangement of $(3, -5, 7, 4, -9)$, then the equation $$a_1\cdot x^4 + a_2\cdot x^3 + a_3\cdot x^2 + a_4\cdot x + a_5 = 0$$ has,

a) at least two real roots

b) all four real roots

c) only imaginary roots

d) none of these.

ans. a)

explanation. $x=1$ is always a root of the equation.

As $a_1, a_2,...,a_5$ are rearrangement of $3,-5,...,-9$ is it necessary to check whether all the possible equations have atleast two real roots or what? A short cut or concept behind the solution? Please help.

Since all the coefficients are real, any complex roots must come in conjugate pairs.

Thus, we have the following possibilities for a degree 4 polynomial:

1. 4 real roots and 0 non-real roots (zero complex conjugate pairs)
2. 2 real roots and 2 non-real roots (one complex conjugate pair)
3. 0 real roots and 4 non-real roots (two complex conjugate pairs)

The sum of the coefficients is $0$, so as you stated, $x = 1$ is always a root. This rules out the possibility of having 0 real roots and 4 non-real roots. So for each arrangement, there are either exactly 2 real roots or exactly 4 real roots. Thus, every arrangement yields a polynomial with at least 2 real roots.

If you need to rule out the other answer choices, it suffices to check that the polynomial $3x^4-5x^3+7x^2-9x+4$ has 2 real roots and 2 non-real roots.

• Helped a lot.. Thank you so much. – Ahana Jul 11 '15 at 8:00