Let f(x) = $x^2+ax+b,a,b \in R$. If $f(1)+f(2)+f(3)=0$, then the nature of the roots of the equation $f(x) =0$ is .....

Question

Let $f(x) = x^2+ax+b$ for $a,b \in \mathbb{R}$. If $f(1)+f(2)+f(3)=0$, then the nature of the roots of the equation $f(x) =0$ is

(A) real

(B) imaginary

(C) real and distinct

(D) equal roots

My attempts:

\begin{align} f(1) &= 1+a+b \\ f(2) &= 4+2a+b \\ f(3) &= 9+3a+b \\ f(1)+f(2)+f(3) &= 1+a+b+4+2a+b+9+3a+b \\ 0 &= 14+6a+3b \end{align}

now how can we take it further about the nature of the roots , whether the roots of $f(x)=0$ is imaginary or real , please help, thanks...

Answer 1

We have three real terms summing up to $0$.

They can't be all zero as a quadratic has at most two zeroes.

Hence at least one term is positive and at least one term is negative, hence the roots must be distinct real roots.

Answer 2

Brute force,

\begin{align} \Delta &= a^2-4b \\ &=a ^2-4\left( -\frac{14+6a}{3} \right) \\ &= a^2+8a \color{red}{+16} +\frac{56}{3} \color{red}{-16} \\ &= (a+4)^2+\frac{8}{3} \\ &> 0 \end{align}

Answer 3

Continuing with your method, $ \ 14 + 6a + 3b \ = \ 0 \ \Rightarrow \ b \ = \ -\left(2a + \frac{14}{3} \right) \ \ . $ The quadratic polynomial is then $ \ x^2 + ax - \left(2a + \frac{14}{3} \right) \ \ , $ which upon "completing the square" becomes $$ \left(x \ + \ \frac{a}{2} \right)^2 \ - \ \left(\frac{a^2}{4} + 2a + \frac{14}{3} \right) \ \ . $$ Multiplying the latter expression in parentheses through by $ \ 12 \ $ produces $ \ 3a^2 + 24a + 56 \ \ ; $ completing the square for this yields $ \ 3·(a + 4)^2 + 8 \ \ , $ so this expression is positive for all real $ \ a \ \ . $ Our quadratic polynomial is thus represented by an "upward-opening" parabola with its vertex "below" the $ \ x-$axis, so it has two $ \ x-$intercepts, corresponding to two real distinct roots for the quadratic equation. [This is equivalent to finding that the discriminant of the polynomial is positive.]

$$ \ \ $$

You could also approach the choices by "elimination".

$ \mathbf{(B) : \ \ }$ A quadratic polynomial with non-real zeroes (pure imaginary or complex) corresponds to a parabola which does not have $ \ x-$intercepts, so the values of $ \ f(x) \ $ can only be always positive or always negative. The sum of any three values of such a function could not sum to zero. $ \ \ \mathbf{[ \ X \ ]} $

$ \mathbf{(D) : \ \ }$ If the quadratic polynomial has two equal (real) zeroes, then the corresponding parabola has a single $ \ x-$intercept, and all other values of $ \ f(x) \ $ can only be always positive or always negative. At most then, one of the terms in $ \ f(1) + f(2) + f(3) \ $ could be zero, but the sum could not equal zero. $ \ \ \mathbf{[ \ X \ ]} $

$ \mathbf{(A) : \ \ }$ The inclusion of this choice actually makes this a poorly-written multiple-choice question, since it could be correct, but incompletely describes the character of the zeroes. $ \ \ \mathbf{[ \ ? \ ]} \ \ $ (A possible alternative "distractor" would be "one real, one imaginary".)

The better description of the zeroes is

$ \mathbf{(C) : \ \ }$ As explained in the other postings, the sum of the three function values could only equal zero if one has a sign opposite to the other two (as all three cannot have the same sign and all three cannot equal zero). This implies that the function can have both positive and negative values: for the corresponding parabola, it then must have two $ \ x-$intercepts, which is equivalent to saying that the quadratic polynomial has two real and distinct zeroes. $ \ \ \mathbf{[ \ ! \ ]} $