Some Other Forms of Equations

MathsQuadratic EquationsFor JEE aspirants

Beyond quadratics, JEE Main and Advanced regularly test polynomial equations of higher degree - especially cubics and quartics. For any polynomial equation $a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0} = 0$ with real coefficients and $a_{n} \neq = 0$ , the sum, sum-of-products-in-pairs, and product of all $n$ roots are elegantly expressed in terms of the coefficients through Vieta's formulas. This page covers the general relation between roots and coefficients, applications to cubic and quartic equations, the factor theorem, intermediate-value results for real roots, the shape of the cubic curve, and four solved examples including a classical result on cubics with $b^{2} < 2 a c$ .

Key Formulas - Quick Reference

For $a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{0} = 0$ with roots $α_{1}, α_{2}, \dots, α_{n}$ :
Sum of roots: $\sum α_{i} = - \frac{a _{n - 1}}{a _{n}}$
Sum of products in pairs: $\sum_{i < j} α_{i} α_{j} = \frac{a _{n - 2}}{a _{n}}$
General elementary symmetric sum: $\sum α_{i_{1}} α_{i_{2}} \dots α_{i_{r}} = (- 1)^{r} \frac{a _{n - r}}{a _{n}}$
Product of all roots: $α_{1} α_{2} \dots α_{n} = (- 1)^{n} \frac{a _{0}}{a _{n}}$
Cubic $a x^{3} + b x^{2} + c x + d = 0$ with roots $α, β, γ$ : $α + β + γ = - b / a$ ; $α β + β γ + γ α = c / a$ ; $α β γ = - d / a$
Quartic $a x^{4} + b x^{3} + c x^{2} + d x + e = 0$ with roots $α, β, γ, δ$ : $\sum α = - b / a$ ; $\sum α β = c / a$ ; $\sum α β γ = - d / a$ ; $α β γ δ = e / a$
Number of complex (non-real) roots is always even when coefficients are real
If degree is odd, at least one real root exists

1. Polynomial Equation of Degree $n$

Consider the polynomial equation

$a_{n} x^{n} + a_{n - 1} x^{n - 1} + a_{n - 2} x^{n - 2} + \dots + a_{1} x + a_{0} = 0 (a_{n} \neq = 0)$

where $a_{0}, a_{1}, \dots, a_{n}$ are real. By the Fundamental Theorem of Algebra, this equation has exactly $n$ roots (real or complex), counted with multiplicity. If $α_{1}, α_{2}, \dots, α_{n}$ are the roots, we can factorise as

$a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{0} = a_{n} (x - α_{1}) (x - α_{2}) \dots (x - α_{n})$

Expanding the right side and comparing coefficients of like powers of $x$ gives Vieta's formulas.

2. Vieta's Formulas: Relation between Roots and Coefficients

Sum of roots: $α_{1} + α_{2} + \dots + α_{n} = - \frac{a _{n - 1}}{a _{n}}$
Sum of products taken two at a time: $\sum_{i < j} α_{i} α_{j} = \frac{a _{n - 2}}{a _{n}}$
General $r$ -th elementary symmetric sum: $\sum α_{i_{1}} α_{i_{2}} \dots α_{i_{r}} = (- 1)^{r} \frac{a _{n - r}}{a _{n}}$
Product of all roots: $α_{1} α_{2} \dots α_{n} = (- 1)^{n} \frac{a _{0}}{a _{n}}$

Special Case: Cubic Equation

For $a x^{3} + b x^{2} + c x + d = 0$ with roots $α, β, γ$ :

α + β + γ = - \frac{b}{a}, α β + β γ + γ α = \frac{c}{a}, α β γ = - \frac{d}{a}

Special Case: Quartic (Biquadratic) Equation

For $a x^{4} + b x^{3} + c x^{2} + d x + e = 0$ with roots $α, β, γ, δ$ :

α + β + γ + δ = - \frac{b}{a}

α β + α γ + α δ + β γ + β δ + γ δ = \frac{c}{a}

α β γ + α β δ + α γ δ + β γ δ = - \frac{d}{a}

α β γ δ = \frac{e}{a}

3. Some Important Results

A polynomial equation of degree $n$ has exactly $n$ roots (real or complex), counted with multiplicity.
If all coefficients are real, then complex (non-real) roots occur in conjugate pairs. Hence the number of complex roots is always even.
If the degree is odd, the number of real roots is also odd; in particular, at least one real root exists.
Factor Theorem: $α$ is a root of $f (x) = 0$ iff $(x - α)$ is a factor of $f (x)$ .
Intermediate value result: If $f (x) = 0$ is a polynomial equation and $f (p), f (q)$ have opposite signs, then $f (x) = 0$ has at least one real root in $(p, q)$ (in fact, an odd number of them, counted with multiplicity).
If $α$ is a repeated root of $f (x) = 0$ with multiplicity $r$ , then $f (α) = f^{'} (α) = f^{''} (α) = \dots = f^{(r - 1)} (α) = 0$ and $f^{(r)} (α) \neq = 0$ .

4. Shape of the Cubic Curve

Consider $f (x) = a x^{3} + b x^{2} + c x + d$ with $a > 0$ . Key features of its graph:

As $x \to \infty$ , $f (x) \to \infty$ ; as $x \to - \infty$ , $f (x) \to - \infty$ (since the $x^{3}$ term dominates).
Because $f$ is continuous and takes both large positive and large negative values, it crosses the $x$ -axis at least once. So every cubic has at least one real root.
The derivative $f^{'} (x) = 3 a x^{2} + 2 b x + c$ may have two, one, or zero real roots.
If $f^{'} (x) = 0$ has two distinct real roots $p > q$ , then $f (q)$ is a local maximum and $f (p)$ is a local minimum.
If $f^{'} (x) = 0$ has a repeated root $r$ , then $(r, f (r))$ is a point of inflection with horizontal tangent.
If $f^{'} (x) = 0$ has no real roots, the cubic is monotonic (no local extrema).
$f^{''} (x) = 6 a x + 2 b$ vanishes at $x = - b / (3 a)$ ; every cubic has a point of inflection there.

f^{'} (x) = 0

has 2 distinct roots local max and min

f^{'} (x) = 0

has a repeated root horizontal inflection point

f^{'} (x) = 0

has no real roots strictly monotonic

Figure 1: Three shapes of the cubic

f (x) = a x^{3} + b x^{2} + c x + d

(with

a > 0

) determined by the discriminant of

f^{'} (x) = 3 a x^{2} + 2 b x + c

5. Solved Examples

Solved Example 1

b^{2} < 2 a c

, prove that

a x^{3} + b x^{2} + c x + d = 0

has exactly one real root.

Solution 1 (using Vieta's):

Let $α, β, γ$ be the roots. By Vieta's formulas:

$α + β + γ = - \frac{b}{a}, α β + β γ + γ α = \frac{c}{a}$

Then

$α^{2} + β^{2} + γ^{2} = (α + β + γ)^{2} - 2 (α β + β γ + γ α) = \frac{b ^{2}}{a ^{2}} - \frac{2 c}{a} = \frac{b ^{2} - 2 a c}{a ^{2}}$

Since $b^{2} < 2 a c$ , we get $α^{2} + β^{2} + γ^{2} < 0$ , which is impossible if all three roots are real. So at least one root is non-real. But complex roots of a real-coefficient polynomial come in conjugate pairs, so exactly two roots are non-real and exactly one is real.

Solution 2 (using calculus):

Suppose $f (x) = a x^{3} + b x^{2} + c x + d$ has three real roots. Then between any two consecutive roots, $f^{'} (x) = 3 a x^{2} + 2 b x + c$ has a zero (Rolle's theorem), so $f^{'} (x) = 0$ has two real roots.

Discriminant of $f^{'}$ : $(2 b)^{2} - 4 \cdot 3 a \cdot c = 4 b^{2} - 12 a c = 4 (b^{2} - 3 a c)$ .

Given $b^{2} < 2 a c < 3 a c$ (assuming $a c > 0$ ; the case $a c \leq 0$ is handled separately), so the discriminant is negative, contradiction. Hence not all roots are real.

Solved Example 2

α, β, γ

are the roots of

x^{3} - 3 x + 1 = 0

, find the value of

α^{3} + β^{3} + γ^{3}

Solution:

By Vieta's for $x^{3} + 0 \cdot x^{2} - 3 x + 1 = 0$ : $α + β + γ = 0$ , $α β + β γ + γ α = - 3$ , $α β γ = - 1$ .

Use the identity $α^{3} + β^{3} + γ^{3} - 3 α β γ = (α + β + γ) (α^{2} + β^{2} + γ^{2} - α β - β γ - γ α)$ .

Since $α + β + γ = 0$ , the right side is $0$ , giving

$α^{3} + β^{3} + γ^{3} = 3 α β γ = 3 (- 1) = - 3$

Solved Example 3

If two roots of the cubic

x^{3} - 6 x^{2} + 11 x - 6 = 0

are equal, find them. [Handle by parameterisation.]

Solution:

Suppose the roots are $α, α, β$ . By Vieta's:

$2 α + β = 6, α^{2} + 2 α β = 11, α^{2} β = 6$

From the first: $β = 6 - 2 α$ . Substitute into the second:

$α^{2} + 2 α (6 - 2 α) = 11 ⟹ - 3 α^{2} + 12 α - 11 = 0 ⟹ 3 α^{2} - 12 α + 11 = 0$

Discriminant $= 144 - 132 = 12 > 0$ , but roots are irrational: $α = 2 \pm \frac{1}{3}$ . Checking against the third equation reveals no consistent triple, so the cubic actually does not have a repeated root. Direct factorisation confirms: $x^{3} - 6 x^{2} + 11 x - 6 = (x - 1) (x - 2) (x - 3)$ , roots $1, 2, 3$ - all distinct.

Key takeaway: when Vieta's gives no consistent solution to the "two equal roots" ansatz, the polynomial has all distinct roots.

Solved Example 4

α, β, γ, δ

are the roots of

x^{4} - 4 x^{3} + 6 x^{2} - 4 x + 1 = 0

, find

α + β + γ + δ

and

α β γ δ

Solution:

Direct application of Vieta's for a quartic $a x^{4} + b x^{3} + c x^{2} + d x + e = 0$ :

$α + β + γ + δ = - \frac{b}{a} = - \frac{- 4}{1} = 4$

$α β γ δ = \frac{e}{a} = \frac{1}{1} = 1$

Observation: The polynomial is $(x - 1)^{4}$ , so all four roots equal $1$ : sum $= 4$ , product $= 1$ . Vieta's gives the answer without knowing the factorisation.

Common Mistakes to Avoid

Watch out

Forgetting the alternating sign pattern in Vieta's. The $r$ -th elementary symmetric sum has factor $(- 1)^{r}$ ; do not drop the sign.
Assuming a cubic with real coefficients can have exactly two real roots. Non-real roots come in conjugate pairs, so a real-coefficient cubic has either 1 or 3 real roots - never 2.
Applying Vieta's formulas without first normalising to monic form is fine, but forgetting to divide the appropriate coefficient by $a_{n}$ is a common error. Always keep the leading coefficient explicit.
Confusing $\sum α^{2}$ with $(\sum α)^{2}$ . The correct identity is $\sum α^{2} = (\sum α)^{2} - 2 \sum_{i < j} α_{i} α_{j}$ .
Assuming a repeated root of $f (x) = 0$ satisfies only $f (α) = 0$ . If the multiplicity is $r$ , then all derivatives up to order $r - 1$ vanish at $α$ .
Believing every polynomial equation of degree $n$ has $n$ distinct roots. It has $n$ roots counted with multiplicity; some may coincide.

Frequently Asked Questions

What are Vieta's formulas for a polynomial of degree $n$ ?

For $a_{n} x^{n} + \dots + a_{0} = 0$ with roots $α_{1}, \dots, α_{n}$ : the sum of roots equals $- a_{n - 1} / a_{n}$ , the sum of products taken two at a time equals $a_{n - 2} / a_{n}$ , and in general the $r$ -th elementary symmetric sum equals $(- 1)^{r} a_{n - r} / a_{n}$ . The product of all roots is $(- 1)^{n} a_{0} / a_{n}$ .

How many real roots does a cubic equation have?

A cubic with real coefficients has either exactly one real root (with two complex conjugate roots) or three real roots (counting multiplicities). It can never have exactly two real roots, because non-real roots must come in conjugate pairs.

What is the factor theorem?

The factor theorem states that $x = α$ is a root of the polynomial equation $f (x) = 0$ if and only if $(x - α)$ is a factor of $f (x)$ . This lets you factor a polynomial once you know one of its roots.

How do you find $α^{2} + β^{2} + γ^{2}$ when you know Vieta's sums for a cubic?

Use the identity $α^{2} + β^{2} + γ^{2} = (α + β + γ)^{2} - 2 (α β + β γ + γ α)$ . Substitute the values from Vieta's formulas: $- b / a$ for the sum and $c / a$ for the sum of pairwise products.

When does a polynomial equation have a repeated root?

$α$ is a repeated root of $f (x) = 0$ with multiplicity $r$ if and only if $f (α) = f^{'} (α) = \dots = f^{(r - 1)} (α) = 0$ but $f^{(r)} (α) \neq = 0$ . To check for a double root, solve the system $f (α) = 0$ and $f^{'} (α) = 0$ simultaneously.

What is the intermediate value result for real roots?

If $f (x)$ is a polynomial and $f (p)$ and $f (q)$ have opposite signs, then $f (x) = 0$ has at least one real root in $(p, q)$ - in fact an odd number of them, counted with multiplicity. If $f (p)$ and $f (q)$ have the same sign, the equation has either no roots or an even number of roots in $(p, q)$ .

Why do complex roots of a real polynomial come in conjugate pairs?

If $f (x)$ has real coefficients and $f (z) = 0$ for some complex $z$ , then taking complex conjugates gives $\overline{f (z)} = f (\overset{z}{ˉ}) = 0$ , since conjugating the coefficients (which are real) leaves them unchanged. So $\overset{z}{ˉ}$ is also a root.

How is the graph of a cubic function shaped?

The cubic $f (x) = a x^{3} + b x^{2} + c x + d$ (with $a > 0$ ) rises from $- \infty$ to $+ \infty$ . Its exact shape depends on $f^{'} (x) = 3 a x^{2} + 2 b x + c$ : if the discriminant of $f^{'}$ is positive, $f$ has a local max and min; if zero, a horizontal inflection point; if negative, $f$ is strictly increasing. Every cubic has a point of inflection at $x = - b / (3 a)$ .

Does every polynomial equation of odd degree have a real root?

Yes. Since complex roots of a real-coefficient polynomial come in conjugate pairs, the number of complex roots is always even. If the total number of roots is odd (i.e. the degree is odd), then the number of real roots is also odd, and in particular at least one.

Previous year questions on Some Other Forms of Equations

5 questions from past papers, each with a step-by-step solution.

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1. Polynomial Equation of Degree n

2. Vieta's Formulas: Relation between Roots and Coefficients

Special Case: Cubic Equation

Special Case: Quartic (Biquadratic) Equation

3. Some Important Results

4. Shape of the Cubic Curve

5. Solved Examples

Common Mistakes to Avoid

Frequently Asked Questions

What are Vieta's formulas for a polynomial of degree n?

How many real roots does a cubic equation have?

What is the factor theorem?

How do you find α2+β2+γ2 when you know Vieta's sums for a cubic?

When does a polynomial equation have a repeated root?

What is the intermediate value result for real roots?

Why do complex roots of a real polynomial come in conjugate pairs?

How is the graph of a cubic function shaped?

Does every polynomial equation of odd degree have a real root?

Previous year questions on Some Other Forms of Equations

Ready to master Quadratic Equations?

1. Polynomial Equation of Degree $n$

What are Vieta's formulas for a polynomial of degree $n$ ?

How do you find $α^{2} + β^{2} + γ^{2}$ when you know Vieta's sums for a cubic?