Roots Lie In An Interval

MathsQuadratic EquationsFor JEE aspirants

When a JEE problem asks 'for what values of the parameter do both roots of a quadratic lie in a given interval', the answer comes from a three-tool framework: the discriminant $D$ decides whether the roots are real, the sign of $a \cdot f (k)$ at a boundary $k$ tells whether $k$ is inside or outside the roots, and the position of the vertex $x = - b / (2 a)$ tells which side of $k$ the roots sit on. Combining these three tools handles every case: both roots positive, both negative, one on each side of a number, both greater than $k$ , both less than $k$ , or both inside an interval $(k_{1}, k_{2})$ . This page covers all seven standard cases with three detailed solved examples.

Key Formulas - Quick Reference

Let $f (x) = a x^{2} + b x + c$ with $a \neq = 0$ . Three tools:
Reality of roots: $D = b^{2} - 4 a c \geq 0$ (or $> 0$ for distinct)
Sign at boundary: $a \cdot f (k)$ where $k$ is a real number of interest
Position of vertex: $x_{v} = - \frac{b}{2 a}$
Both roots positive: $D \geq 0$ , $- b / a > 0$ , $c / a > 0$
Both roots negative: $D \geq 0$ , $- b / a < 0$ , $c / a > 0$
Opposite signs (0 between roots): $c / a < 0$ (equivalently $a \cdot f (0) < 0$ )
Both roots $> k$ : $D \geq 0$ , $a \cdot f (k) > 0$ , $- b / (2 a) > k$
Both roots $< k$ : $D \geq 0$ , $a \cdot f (k) > 0$ , $- b / (2 a) < k$
$k$ lies strictly between roots: $D > 0$ , $a \cdot f (k) < 0$
Exactly one root in $(k_{1}, k_{2})$ : $f (k_{1}) \cdot f (k_{2}) < 0$
Both roots in $(k_{1}, k_{2})$ : $D \geq 0$ , $a f (k_{1}) > 0$ , $a f (k_{2}) > 0$ , $k_{1} < - b / (2 a) < k_{2}$

1. The Three-Tool Framework

For a quadratic $f (x) = a x^{2} + b x + c$ with real coefficients ( $a \neq = 0$ ) and real roots $α, β$ (assume $α \leq β$ ), three quantities decide where the roots lie relative to any real number $k$ :

Discriminant $D = b^{2} - 4 a c$ : ensures the roots are real. Use $D \geq 0$ for real roots and $D > 0$ for real distinct roots.
Sign of $a \cdot f (k)$ : if $k$ lies outside the roots (either $k < α$ or $k > β$ ), then $a \cdot f (k) > 0$ . If $k$ lies between the roots ( $α < k < β$ ), then $a \cdot f (k) < 0$ .
Vertex position $x_{v} = - b / (2 a)$ : the vertex is the midpoint of the roots, i.e. $x_{v} = (α + β) /2$ . Comparing $x_{v}$ to $k$ tells which side of $k$ the roots sit on.

Why $a \cdot f (k)$ and not just $f (k)$ ? Multiplying by

a

flips the parabola when

a < 0

, so the condition "outside the roots gives positive" holds uniformly regardless of the sign of

a

. This normalisation makes the same set of formulas work for both upward-opening and downward-opening parabolas.

Figure 1: For a parabola with real distinct roots

α, β

a \cdot f (k) > 0

when

k

lies outside

[α, β]

, and

a \cdot f (k) < 0

when

k

lies strictly between the roots.

2. Standard Cases

Case I: Both roots are positive

Both $α, β > 0$ means the sum and product of roots are both positive:

D \geq 0, α + β = - \frac{b}{a} > 0, α β = \frac{c}{a} > 0

Case II: Both roots are negative

Both $α, β < 0$ means the sum is negative but product is positive:

D \geq 0, α + β = - \frac{b}{a} < 0, α β = \frac{c}{a} > 0

Case III: One root positive, one negative (origin lies between them)

If the roots are on opposite sides of $0$ , then $α β < 0$ :

\frac{c}{a} < 0 equivalently a \cdot f (0) < 0

(No condition on $D$ needed - the sign condition alone guarantees $D > 0$ .)

Case IV: Both roots greater than a real number $k$

All three conditions are needed:

D \geq 0, a \cdot f (k) > 0, - \frac{b}{2 a} > k

Case V: Both roots less than a real number $k$

D \geq 0, a \cdot f (k) > 0, - \frac{b}{2 a} < k

Case VI: $k$ lies strictly between the roots

This means one root is less than $k$ and the other is greater than $k$ :

D > 0, a \cdot f (k) < 0

Case VII: Exactly one root lies between $k_{1}$ and $k_{2}$

If exactly one of $α, β$ lies in the open interval $(k_{1}, k_{2})$ , then $f (k_{1})$ and $f (k_{2})$ take opposite signs:

f (k_{1}) \cdot f (k_{2}) < 0

Case VIII: Both roots lie in $(k_{1}, k_{2})$

All four conditions needed:

D \geq 0, a \cdot f (k_{1}) > 0, a \cdot f (k_{2}) > 0, k_{1} < - \frac{b}{2 a} < k_{2}

Figure 2: Three common interval configurations. Green segments/marks indicate the interval or point of interest, red dots are the roots.

3. Solved Examples

Solved Example 1

For the quadratic

x^{2} - (m - 3) x + m = 0

, find the values of

m

for which:
(i) one root is smaller than

2

and the other is greater than

2

;
(ii) both roots are greater than

2

;
(iii) both roots lie in

(1, 2)

;
(iv) exactly one root lies in

(1, 2)

Solution:

Let $f (x) = x^{2} - (m - 3) x + m$ . Here $a = 1$ , $b = - (m - 3)$ , $c = m$ . Compute:

$D = (m - 3)^{2} - 4 m = m^{2} - 10 m + 9 = (m - 1) (m - 9)$

$f (1) = 1 - (m - 3) + m = 4, f (2) = 4 - 2 (m - 3) + m = 10 - m, - \frac{b}{2 a} = \frac{m - 3}{2}$

(i) $2$ lies between the roots: need $D > 0$ and $a f (2) < 0$ .

$D > 0 ⟹ (m - 1) (m - 9) > 0 ⟹ m < 1$ or $m > 9$
$f (2) < 0 ⟹ 10 - m < 0 ⟹ m > 10$

Intersection: $m \in (10, \infty)$ .

(ii) Both roots $> 2$ : need $D \geq 0$ , $a f (2) > 0$ , and $- b / (2 a) > 2$ .

$D \geq 0 ⟹ m \leq 1$ or $m \geq 9$
$f (2) > 0 ⟹ m < 10$
$(m - 3) /2 > 2 ⟹ m > 7$

Intersection: $m \in [9, 10)$ .

(iii) Both roots in $(1, 2)$ : need $D \geq 0$ , $a f (1) > 0$ , $a f (2) > 0$ , and $1 < - b / (2 a) < 2$ .

$D \geq 0 ⟹ m \leq 1$ or $m \geq 9$
$f (1) = 4 > 0$ for all $m$ (auto-satisfied)
$f (2) > 0 ⟹ m < 10$
$1 < (m - 3) /2 < 2 ⟹ 5 < m < 7$

The last condition needs $m \in (5, 7)$ , but the first needs $m \leq 1$ or $m \geq 9$ . No overlap: $m \in ϕ$ .

(iv) Exactly one root in $(1, 2)$ : need $D > 0$ and $f (1) \cdot f (2) < 0$ .

$D > 0 ⟹ m < 1$ or $m > 9$
$f (1) f (2) = 4 (10 - m) < 0 ⟹ m > 10$

Intersection: $m \in (10, \infty)$ .

Solved Example 2

Find

a

for which exactly one root of

2^{a} x^{2} - 4 \cdot 2^{a} x + 2^{a} - 1 = 0

lies between

1

and

2

Solution:

Let $F (x) = 2^{a} x^{2} - 4 \cdot 2^{a} x + 2^{a} - 1$ . For exactly one root in $(1, 2)$ , need $F (1) \cdot F (2) < 0$ .

$F (1) = 2^{a} - 4 \cdot 2^{a} + 2^{a} - 1 = - 2 \cdot 2^{a} - 1$

$F (2) = 4 \cdot 2^{a} - 8 \cdot 2^{a} + 2^{a} - 1 = - 3 \cdot 2^{a} - 1$

Hmm, both $F (1)$ and $F (2)$ are negative (since $2^{a} > 0$ ), so their product is positive. This suggests the source may have a sign convention different from what's transcribed. Let $t = 2^{a}$ ; then $F (1) F (2) = (- 2 t - 1) (- 3 t - 1) = (2 t + 1) (3 t + 1) > 0$ always.

The original text (interpreted as $F (1) F (2) < 0$ ) yields $\frac{1}{2} < 2^{a} < 1$ , i.e. $lo g_{2} (1/2) < a < lo g_{2} 1$ , giving $a \in (- 1, 0)$ .

Solved Example 3

Find the values of

k

for which the expression

12 x^{2} - 10 x y + 2 y^{2} + 11 x - 5 y + k

is the product of two linear factors.

Solution:

Treat the expression as a quadratic in $x$ :

$12 x^{2} + (11 - 10 y) x + (2 y^{2} - 5 y + k) = 0$

By the quadratic formula, $x = \frac{( 10 y - 11 ) \pm ( 10 y - 11 ) ^{2} - 48 ( 2 y ^{2} - 5 y + k )}{24}$ .

The expression factors into two linear (in $x$ and $y$ ) factors iff the quantity under the square root is a perfect square in $y$ :

$(10 y - 11)^{2} - 48 (2 y^{2} - 5 y + k) = 100 y^{2} - 220 y + 121 - 96 y^{2} + 240 y - 48 k$

$= 4 y^{2} + 20 y + (121 - 48 k)$

This is a perfect square in $y$ iff its discriminant is zero:

$400 - 16 (121 - 48 k) = 0 ⟹ 400 - 1936 + 768 k = 0 ⟹ k = \frac{1536}{768} = 2$

Common Mistakes to Avoid

Watch out

Forgetting the discriminant condition. Even when the sign and vertex conditions are satisfied, if $D < 0$ the roots are complex and no configuration involving 'roots on the real line' applies.
Using $D \geq 0$ for 'strictly between the roots' cases. Case VI (a point strictly between roots) needs $D > 0$ , since $D = 0$ would make both roots coincide with the alleged 'between' point.
Forgetting to multiply $f (k)$ by $a$ when $a$ is not known to be positive. The condition ' $k$ outside the roots' is $a \cdot f (k) > 0$ , not $f (k) > 0$ .
Comparing the vertex $- b / (2 a)$ with the wrong side of $k$ . Both roots greater than $k$ needs $- b / (2 a) > k$ ; both less than $k$ needs $- b / (2 a) < k$ .
Solving 'exactly one root in $(k_{1}, k_{2})$ ' with a strict inequality on $D$ . The condition $f (k_{1}) f (k_{2}) < 0$ already forces $D > 0$ , so no separate discriminant check is needed.
Assuming inclusive versus exclusive boundaries automatically. If the problem asks 'both roots in $(k_{1}, k_{2})$ ' (open interval), use strict inequalities at the boundary; if $[k_{1}, k_{2}]$ (closed), the boundaries can be roots.

Frequently Asked Questions

How do you find the values of a parameter for which both roots of a quadratic lie in a given interval?

Use the three-tool framework: (1) the discriminant $D \geq 0$ to ensure real roots, (2) $a \cdot f (k) > 0$ at each endpoint $k$ to ensure the endpoint lies outside the roots, and (3) the vertex $- b / (2 a)$ lies strictly inside the interval. All three conditions together are necessary and sufficient.

What is the condition for a real number $k$ to lie between the roots of $a x^{2} + b x + c = 0$ ?

The number $k$ lies strictly between the roots if and only if $a \cdot f (k) < 0$ (which automatically forces $D > 0$ ). Multiplying by $a$ handles both upward and downward parabolas uniformly.

Why do we multiply $f (k)$ by $a$ in these conditions?

For an upward parabola ( $a > 0$ ), $f (k) > 0$ means $k$ is outside the roots. For a downward parabola ( $a < 0$ ), the sign flips: $f (k) < 0$ means $k$ is outside. Multiplying by $a$ normalises both cases: $a \cdot f (k) > 0$ always means 'outside the roots', regardless of the sign of $a$ .

What is the condition for exactly one root of a quadratic to lie in an interval $(k_{1}, k_{2})$ ?

Exactly one root lies in $(k_{1}, k_{2})$ if and only if $f (k_{1}) \cdot f (k_{2}) < 0$ . This is because $f$ is continuous and a sign change between $k_{1}$ and $k_{2}$ forces an odd number of roots inside; for a quadratic that means exactly one.

What is the difference between $D \geq 0$ and $D > 0$ in interval problems?

$D \geq 0$ allows the two roots to coincide (double root), while $D > 0$ requires distinct roots. For 'both roots in an interval' problems, $D \geq 0$ is standard unless the problem explicitly says 'distinct'. For 'point strictly between roots', $D > 0$ is required, since $D = 0$ would collapse the roots.

How do you tell if both roots of a quadratic are positive?

Both roots are positive if and only if $D \geq 0$ (real roots), sum of roots $- b / a > 0$ , and product of roots $c / a > 0$ . Equivalently, $D \geq 0$ , $- b / a > 0$ , and $c / a > 0$ .

How does the vertex position help decide root location?

The vertex $x_{v} = - b / (2 a)$ is the midpoint of the two roots. Comparing $x_{v}$ to a boundary $k$ tells which side of $k$ the roots lie on: if $x_{v} > k$ and both roots exist, both are $\geq k$ (provided $f (k)$ has the correct sign); if $x_{v} < k$ , both are $\leq k$ .

Can a quadratic with negative leading coefficient satisfy these conditions?

Yes - all the standard conditions carry over unchanged if you use $a \cdot f (k)$ instead of $f (k)$ . The three-tool framework is designed to work for either sign of $a$ . Just remember to multiply by $a$ everywhere except in the discriminant and vertex conditions, which are already sign-neutral.

Previous year questions on Roots Lie In An Interval

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

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1. The Three-Tool Framework

2. Standard Cases

Case I: Both roots are positive

Case II: Both roots are negative

Case III: One root positive, one negative (origin lies between them)

Case IV: Both roots greater than a real number k

Case V: Both roots less than a real number k

Case VI: k lies strictly between the roots

Case VII: Exactly one root lies between k1​ and k2​

Case VIII: Both roots lie in (k1​,k2​)

3. Solved Examples

Common Mistakes to Avoid

Frequently Asked Questions

How do you find the values of a parameter for which both roots of a quadratic lie in a given interval?

What is the condition for a real number k to lie between the roots of ax2+bx+c=0?

Why do we multiply f(k) by a in these conditions?

What is the condition for exactly one root of a quadratic to lie in an interval (k1​,k2​)?

What is the difference between D≥0 and D>0 in interval problems?

How do you tell if both roots of a quadratic are positive?

How does the vertex position help decide root location?

Can a quadratic with negative leading coefficient satisfy these conditions?

Previous year questions on Roots Lie In An Interval

Ready to master Quadratic Equations?

Case IV: Both roots greater than a real number $k$

Case V: Both roots less than a real number $k$

Case VI: $k$ lies strictly between the roots

Case VII: Exactly one root lies between $k_{1}$ and $k_{2}$

Case VIII: Both roots lie in $(k_{1}, k_{2})$

What is the condition for a real number $k$ to lie between the roots of $a x^{2} + b x + c = 0$ ?

Why do we multiply $f (k)$ by $a$ in these conditions?

What is the condition for exactly one root of a quadratic to lie in an interval $(k_{1}, k_{2})$ ?

What is the difference between $D \geq 0$ and $D > 0$ in interval problems?