Quadratic Expressions

MathsQuadratic EquationsFor JEE aspirants

A quadratic expression $a x^{2} + b x + c$ with $a \neq = 0$ traces a parabola: opening upward if $a > 0$ and downward if $a < 0$ . Its sign for all real $x$ is decided by the leading coefficient $a$ and the discriminant $D = b^{2} - 4 a c$ , giving six standard cases from 'always positive' to 'always negative'. The vertex $x = - b / (2 a)$ gives the maximum or minimum value $\frac{4 a c - b ^{2}}{4 a}$ . This page covers the sign analysis, greatest and least values, and the powerful technique for finding the range of a rational expression in $x$ by reducing to a quadratic in $x$ with $y$ as a parameter. Three JEE-style solved examples included.

Key Formulas - Quick Reference

Standard quadratic expression: $f (x) = a x^{2} + b x + c$ with $a \neq = 0$
Vertex: $x = - \frac{b}{2 a}$
Extremum value: $f (- \frac{b}{2 a}) = \frac{4 a c - b ^{2}}{4 a} = - \frac{D}{4 a}$
If $a > 0$ : minimum value $= \frac{4 a c - b ^{2}}{4 a}$ ; no maximum
If $a < 0$ : maximum value $= \frac{4 a c - b ^{2}}{4 a}$ ; no minimum
Always positive ( $f (x) > 0 \forall x$ ): $a > 0$ and $D < 0$
Always negative ( $f (x) < 0 \forall x$ ): $a < 0$ and $D < 0$
Non-negative ( $f (x) \geq 0 \forall x$ ): $a > 0$ and $D \leq 0$
Perfect square: $D = 0$
Range of rational expression: set $y = P (x) / Q (x)$ and require the resulting quadratic in $x$ to have real solutions (discriminant $\geq 0$ )

1. The Quadratic Expression as a Parabola

The expression $f (x) = a x^{2} + b x + c$ (with $a, b, c$ real and $a \neq = 0$ ) is called a real quadratic expression in $x$ . Its graph is a parabola:

If $a > 0$ , the parabola opens upward: $f (x) \to + \infty$ as $∣ x ∣ \to \infty$ .
If $a < 0$ , the parabola opens downward: $f (x) \to - \infty$ as $∣ x ∣ \to \infty$ .

The vertex lies at $x = - b / (2 a)$ and gives the minimum value (if $a > 0$ ) or maximum value (if $a < 0$ ):

f (- \frac{b}{2 a}) = \frac{4 a c - b ^{2}}{4 a} = - \frac{D}{4 a}

2. Sign of the Quadratic Expression

The sign of $f (x)$ for all real $x$ depends on the sign of $a$ and the discriminant $D = b^{2} - 4 a c$ . Six standard sub-cases:

Case I: $a > 0$ (parabola opens upward)

Discriminant	Sign of $f (x)$	Geometric picture
$D < 0$	$f (x) > 0 \forall x \in R$	Parabola stays entirely above the $x$ -axis
$D = 0$	$f (x) \geq 0 \forall x \in R$	Parabola touches the $x$ -axis at exactly one point (double root)
$D > 0$	$f (x) > 0$ outside $[α, β]$ ; $f (x) < 0$ inside	Parabola cuts the $x$ -axis at two points $α, β$

Case II: $a < 0$ (parabola opens downward)

Discriminant	Sign of $f (x)$	Geometric picture
$D < 0$	$f (x) < 0 \forall x \in R$	Parabola stays entirely below the $x$ -axis
$D = 0$	$f (x) \leq 0 \forall x \in R$	Parabola touches the $x$ -axis at exactly one point
$D > 0$	$f (x) < 0$ outside $[α, β]$ ; $f (x) > 0$ inside	Parabola cuts the $x$ -axis at two points

Figure 1: All six sign configurations of the quadratic expression

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

, organised by sign of

a

(row) and sign of

D

(column).

3. Greatest and Least Values

When $a > 0$ (parabola opens upward)

The expression has no greatest value (it goes to $+ \infty$ ) and a least value of $\frac{4 a c - b ^{2}}{4 a}$ attained at $x = - \frac{b}{2 a}$ .

When $a < 0$ (parabola opens downward)

The expression has no least value (it goes to $- \infty$ ) and a greatest value of $\frac{4 a c - b ^{2}}{4 a}$ attained at $x = - \frac{b}{2 a}$ .

Completing the square gives the same result:

a x^{2} + b x + c = a (x + \frac{b}{2 a})^{2} + \frac{4 a c - b ^{2}}{4 a}

. Since the squared term is always

\geq 0

, the extremum equals the constant

\frac{4 a c - b ^{2}}{4 a}

, achieved when

x = - b / (2 a)

4. Range of a Rational Expression in $x$

To find the range of a rational expression $\frac{P ( x )}{Q ( x )}$ where $P, Q$ are quadratics in $x$ (or one is linear), use this four-step reduction:

Set the rational expression equal to $y$ and clear denominators: $P (x) = y \cdot Q (x)$ .
Rearrange as a quadratic equation in $x$ with $y$ as a parameter.
For real $x$ to exist, the discriminant of that quadratic (in $x$ ) must be $\geq 0$ . This gives an inequality in $y$ .
Solve the inequality in $y$ ; the resulting range is the range of the original expression.

5. Solved Examples

Solved Example 1

Find the maximum and minimum values of

\frac{x ^{2} - x + 1}{x ^{2} + x + 1}

for real

x

Solution:

Set $y = \frac{x ^{2} - x + 1}{x ^{2} + x + 1}$ . Cross-multiplying:

$y (x^{2} + x + 1) = x^{2} - x + 1$

$⟹ (y - 1) x^{2} + (y + 1) x + (y - 1) = 0$

For real $x$ , discriminant $\geq 0$ :

$(y + 1)^{2} - 4 (y - 1)^{2} \geq 0$

Factor as difference of squares:

$[(y + 1) - 2 (y - 1)] [(y + 1) + 2 (y - 1)] \geq 0$

$⟹ (3 - y) (3 y - 1) \geq 0 ⟹ (y - 3) (3 y - 1) \leq 0$

$⟹ \frac{1}{3} \leq y \leq 3$

So $y_{m i n} = \frac{1}{3}, y_{m a x} = 3$ .

Aside: the case $y = 1$ makes the quadratic degenerate to a linear equation $2 x = 0$ , giving $x = 0$ , which is valid. So $y = 1$ is attained too, consistent with $1 \in [1/3, 3]$ .

Solved Example 2

min {x^{2} + (a - b) x + (1 - a - b)} > max {- x^{2} + (a + b) x - (1 + a + b)}

, prove that

a^{2} + b^{2} < 4

Solution:

Complete the square in each expression. For the first (leading coefficient $+ 1 > 0$ , minimum exists):

$x^{2} + (a - b) x + (1 - a - b) = (x + \frac{a - b}{2})^{2} + (1 - a - b) - \frac{( a - b ) ^{2}}{4}$

Minimum value $= 1 - a - b - \frac{( a - b ) ^{2}}{4}$ .

For the second (leading coefficient $- 1 < 0$ , maximum exists):

$- x^{2} + (a + b) x - (1 + a + b) = - (x - \frac{a + b}{2})^{2} + \frac{( a + b ) ^{2}}{4} - (1 + a + b)$

Maximum value $= \frac{( a + b ) ^{2}}{4} - 1 - a - b$ .

The given inequality becomes:

$1 - a - b - \frac{( a - b ) ^{2}}{4} > \frac{( a + b ) ^{2}}{4} - 1 - a - b$

$⟹ 2 > \frac{( a - b ) ^{2} + ( a + b ) ^{2}}{4} = \frac{2 a ^{2} + 2 b ^{2}}{4} = \frac{a ^{2} + b ^{2}}{2}$

$⟹ a^{2} + b^{2} < 4$

Solved Example 3

a x^{2} - b x + 5 = 0

does not have two distinct real roots, find the minimum value of

5 a + b

Solution:

Since the equation does not have two distinct real roots, either the roots are complex ( $D < 0$ ) or they are real and equal ( $D = 0$ ). In either case, $f (x) = a x^{2} - b x + 5$ does not change sign, so it is either always $\geq 0$ or always $\leq 0$ .

Note $f (0) = 5 > 0$ , so $f (x) \geq 0$ for all real $x$ . In particular:

$f (- 5) \geq 0 ⟹ 25 a + 5 b + 5 \geq 0 ⟹ 5 a + b \geq - 1$

Hence the minimum value of $5 a + b$ is $- 1$ , achieved when equality holds in $f (- 5) \geq 0$ , i.e. when $- 5$ is a repeated root.

Common Mistakes to Avoid

Watch out

Confusing the max/min value of the expression with the location of the extremum. The value is $\frac{4 a c - b ^{2}}{4 a}$ ; the location is $x = - b / (2 a)$ .
Assuming a quadratic with $a > 0$ has both a maximum and minimum. It has only a minimum (at the vertex); it has no maximum since $f (x) \to + \infty$ .
Using $D < 0$ alone to conclude 'always positive'. You also need $a > 0$ . If $a < 0$ and $D < 0$ , the expression is always negative.
In the range-of-rational-expression technique, forgetting to check when the coefficient of $x^{2}$ becomes zero (i.e. $y$ takes a value that makes the equation linear rather than quadratic). Handle that value separately.
In the range technique, using $D > 0$ instead of $D \geq 0$ . The equality case corresponds to a repeated real root of $x$ , which is still a valid real $x$ , so $D \geq 0$ is correct.
Assuming $a x^{2} + b x + c > 0$ has 'no roots' means the discriminant is zero. It means $D < 0$ (strict inequality). $D = 0$ gives one repeated real root, so the expression touches zero.

Frequently Asked Questions

What is the difference between a quadratic equation and a quadratic expression?

A quadratic expression is $a x^{2} + b x + c$ - a polynomial in $x$ of degree 2. A quadratic equation is $a x^{2} + b x + c = 0$ - setting the expression equal to zero and asking for the values of $x$ that satisfy it. The expression is a function of $x$ ; the equation has specific roots.

How do you find the maximum or minimum value of a quadratic expression?

The extremum is at the vertex $x = - b / (2 a)$ , with value $\frac{4 a c - b ^{2}}{4 a} = - \frac{D}{4 a}$ . If $a > 0$ this is the minimum (no maximum); if $a < 0$ this is the maximum (no minimum). You can also derive this by completing the square.

When is $a x^{2} + b x + c$ positive for all real $x$ ?

$a x^{2} + b x + c > 0$ for all real $x$ if and only if $a > 0$ and $D = b^{2} - 4 a c < 0$ . Geometrically, the parabola opens upward and does not touch the $x$ -axis.

When is $a x^{2} + b x + c$ non-negative for all real $x$ ?

$a x^{2} + b x + c \geq 0$ for all real $x$ if and only if $a > 0$ and $D = b^{2} - 4 a c \leq 0$ . The equality $D = 0$ corresponds to the expression touching zero at exactly one point (a repeated root).

How do you find the range of a rational expression like $\frac{x ^{2} + 1}{x ^{2} + x + 1}$ ?

Set the expression equal to $y$ and clear denominators to get a quadratic in $x$ with $y$ as parameter. For $x$ to be real, the discriminant of that quadratic (in $x$ ) must be $\geq 0$ . Solve the resulting inequality in $y$ - the solution set is the range.

Why does completing the square give the vertex formula?

Writing $a x^{2} + b x + c = a (x + b / (2 a))^{2} + (4 a c - b^{2}) / (4 a)$ shows the expression is a shifted-and-scaled version of $x^{2}$ . The squared term is minimised when $x + b / (2 a) = 0$ , i.e. $x = - b / (2 a)$ , giving the constant $(4 a c - b^{2}) / (4 a)$ as the extremum.

When is a quadratic expression a perfect square?

$a x^{2} + b x + c$ is a perfect square (the square of a linear polynomial) if and only if $D = b^{2} - 4 a c = 0$ . In that case $a x^{2} + b x + c = a (x + b / (2 a))^{2}$ .

How can a quadratic expression in two variables be factored into linear factors?

For $A x^{2} + B x y + C y^{2} + D x + E y + F$ to factor into two linear factors in $x$ and $y$ , treat it as a quadratic in $x$ with $y$ as a parameter. Its discriminant (in $x$ ) must be a perfect square in $y$ , which itself requires that discriminant's own discriminant (in $y$ ) to be zero. This gives an algebraic condition on the coefficients.

Previous year questions on Quadratic Expressions

1 question from past papers, each with a step-by-step solution.

JEE Main 2026 Apr 2 Shift 1, Mathematics Q23

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1. The Quadratic Expression as a Parabola

2. Sign of the Quadratic Expression

Case I: a>0 (parabola opens upward)

Case II: a<0 (parabola opens downward)

3. Greatest and Least Values

When a>0 (parabola opens upward)

When a<0 (parabola opens downward)

4. Range of a Rational Expression in x

5. Solved Examples

Common Mistakes to Avoid

Frequently Asked Questions

What is the difference between a quadratic equation and a quadratic expression?

How do you find the maximum or minimum value of a quadratic expression?

When is ax2+bx+c positive for all real x?

When is ax2+bx+c non-negative for all real x?

How do you find the range of a rational expression like x2+x+1x2+1​?

Why does completing the square give the vertex formula?

When is a quadratic expression a perfect square?

How can a quadratic expression in two variables be factored into linear factors?

Previous year questions on Quadratic Expressions

Ready to master Quadratic Equations?

Case I: $a > 0$ (parabola opens upward)

Case II: $a < 0$ (parabola opens downward)

When $a > 0$ (parabola opens upward)

When $a < 0$ (parabola opens downward)

4. Range of a Rational Expression in $x$

When is $a x^{2} + b x + c$ positive for all real $x$ ?

When is $a x^{2} + b x + c$ non-negative for all real $x$ ?

How do you find the range of a rational expression like $\frac{x ^{2} + 1}{x ^{2} + x + 1}$ ?