A polynomial equation in x of degree n always has (1) n distinct roots (2) n real roots (3) n complex roots (4

Exercise 3.7 · Q3

x

(1)

n

distinct roots

(2)

n

real roots

(3)

n

complex roots

(4) at most one root.

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Concept understanding — Relations between Roots and Coefficients (Vieta's Formulas)

If $\alpha,\beta,\gamma$ are the roots of the monic cubic $x^3+px^2+qx+r=0$ , then $f(x)\equiv(x-\alpha)(x-\beta)(x-\gamma)$ . Expanding the right side and matching coefficients with $x^3+px^2+qx+r$ gives

S_1=\alpha+\beta+\gamma=-p,\qquad S_2=\alpha\beta+\beta\gamma+\gamma\alpha=q,\qquad S_3=\alpha\beta\gamma=-r.

For the biquadratic $x^4+px^3+qx^2+rx+s=0$ with roots $\alpha,\beta,\gamma,\delta$ , expanding $(x-\alpha)(x-\beta)(x-\gamma)(x-\delta)$ in the same way gives

S_1=\sum\alpha=-p,\quad S_2=\sum\alpha\beta=q,\quad S_3=\sum\alpha\beta\gamma=-r,\quad S_4=\alpha\beta\gamma\delta=s.

In general, for the monic degree- $n$ equation $x^n+p_1x^{n-1}+p_2x^{n-2}+\cdots+p_n=0$ with roots $\alpha_1,\ldots,\alpha_n$ ,

S_k=\sum(\text{products of the roots taken }k\text{ at a time})=(-1)^kp_k,\qquad k=1,\ldots,n.

These relations let any symmetric function of the roots -- one whose value is unchanged by permuting $\alpha,\beta,\gamma,\ldots$ among themselves -- be evaluated directly from the coefficients, without ever solving the equation. A symmetric function is first rewritten in terms of $S_1,S_2,S_3,\ldots$ : for example $\sum\alpha^2=S_1^2-2S_2$ , and for a cubic $\displaystyle\sum\frac1{\alpha^2}=\Big(\sum\frac1\alpha\Big)^2-2\sum\frac1{\alpha\beta}=\Big(\frac{S_2}{S_3}\Big)^2-2\frac{S_1}{S_3}=\frac{S_2^2-2S_1S_3}{S_3^2}$ .

Step 1. Rule out " $n$ distinct roots". A repeated root (e.g. $x=0$ with multiplicity $2$ in $x^2=0$ ) means the roots need not all be distinct.

Step 2. Rule out " $n$ real roots". E.g. $x^2+1=0$ has degree $2$ but $0$ real roots (both roots are non-real).

Step 3. Rule out "at most one root". A degree- $n$ equation can have up to $n$ roots, far more than one for $n>1$ .

Step 4. Confirm " $n$ complex roots". The Fundamental Theorem of Algebra guarantees exactly $n$ roots in $\mathbb C$ (the complex numbers, which include the reals), counted with multiplicity — always true, regardless of how many are real or repeated.

✓Final answer

n

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2024Set ANNUAL1 markMCQ
$\alpha, \beta$
(a) $\dfrac{q}{r}$
(b) $-\dfrac{q}{r}$
(c) $-\dfrac{q}{p}$
(d) $-\dfrac{p}{r}$
›Reveal solutionSolution
Writes the sum of reciprocals as (sum of pairwise products)/(product of roots) and substitutes Vieta's formulas for a cubic.

For $x^3+px^2+qx+r=0$ with roots $\alpha,\beta,\gamma$ : by Vieta's formulas, $\alpha\beta+\beta\gamma+\gamma\alpha=q$ and $\alpha\beta\gamma=-r$ .

$\displaystyle\sum\dfrac1\alpha=\dfrac1\alpha+\dfrac1\beta+\dfrac1\gamma=\dfrac{\beta\gamma+\alpha\gamma+\alpha\beta}{\alpha\beta\gamma}$ .

Substitute: $=\dfrac{q}{-r}=-\dfrac qr$ .

✓Final answer
$-\dfrac{q}{r}$
CBSE 2022Set ANNUAL1 markMCQ
$\alpha, \beta$
(a) $\dfrac{q}{r}$
(b) $-\dfrac{q}{r}$
(c) $-\dfrac{q}{p}$
(d) $-\dfrac{p}{r}$
›Reveal solutionSolution
Using Vieta's formulas for the cubic $x^3+px^2+qx+r$ , $\displaystyle\sum\dfrac{1}{\alpha}=-\dfrac{q}{r}$ .

For $x^3+px^2+qx+r=0$ with roots $\alpha,\beta,\gamma$ , Vieta's formulas give: $\alpha+\beta+\gamma=-p$ , $\alpha\beta+\beta\gamma+\gamma\alpha=q$ , $\alpha\beta\gamma=-r$ .

We want $\displaystyle\sum\dfrac{1}{\alpha}=\dfrac{1}{\alpha}+\dfrac{1}{\beta}+\dfrac{1}{\gamma}$ .

Combining over a common denominator, $\dfrac{1}{\alpha}+\dfrac{1}{\beta}+\dfrac{1}{\gamma}=\dfrac{\beta\gamma+\alpha\gamma+\alpha\beta}{\alpha\beta\gamma}$ .

Substituting the Vieta values: numerator $=q$ , denominator $=-r$ .

So $\displaystyle\sum\dfrac{1}{\alpha}=\dfrac{q}{-r}=-\dfrac{q}{r}$ .

✓Final answer
$\displaystyle\sum\dfrac{1}{\alpha}=-\dfrac{q}{r}$

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