MathsICSE Class 10 · Quadratic Equations in One Variable

Nature of Roots and the Discriminant: Rules and Solved Questions (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

For ax² + bx + c = 0, the discriminant D = b² − 4ac decides the roots without solving. If D > 0 there are two distinct real roots; if D = 0 two equal real roots, each −b/(2a); if D < 0 there are no real roots. Finding k for equal roots means solving D = 0.

The sign of D tells how many times the parabola meets the x-axis: twice, once at x = −b/2a, or never.

Illustration drawn for QuestionBuddy.

Understand Nature of Roots and the Discriminant

In the quadratic formula $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ , everything depends on the number under the square root. It is called the discriminant, because it discriminates between the kinds of roots:

$D = b^{2} - 4 a c .$

The CISCE syllabus lists the three cases exactly:

$D > 0$ : two distinct real roots, $\frac{- b + D}{2 a}$ and $\frac{- b - D}{2 a}$ . If $D$ is also a perfect square (and $a, b, c$ are rational), the roots are rational and the equation factorises.
$D = 0$ : two equal real roots, both equal to $- \frac{b}{2 a}$ . The quadratic is a perfect square, such as $x^{2} - 6 x + 9 = (x - 3)^{2}$ .
$D < 0$ : no real roots, because a negative number has no real square root.

Computing D carefully. Two traps cost marks in recent papers. In 2024 Q1(iii), the equation $p x^{2} - q x + r = 0$ has $b = - q$ , so $b^{2} = q^{2}$ (not $- q^{2}$ ); equal roots need $q^{2} = 4 p r$ . In 2025 Q1(i), $3 x^{2} + 7 x + 2 = 0$ has $b^{2} = (7)^{2} = 7$ (not 49), so $D = 7 - 24 < 0$ and there are no real roots. The examiners reported both slips.

Finding an unknown coefficient. A favourite question gives a coefficient as $k$ and states the nature of the roots:

equal roots: solve $D = 0$ ;
real roots: solve $D \geq 0$ ;
distinct real roots: $D > 0$ ;
no real roots: $D < 0$ .

For 'equal roots' you usually get a simple equation in $k$ , sometimes with two answers (for example $(k - 1)^{2} = 16$ gives $k = 5$ or $k = - 3$ ). If the question also asks for the roots, put each value of $k$ back and use $x = - \frac{b}{2 a}$ . If $k$ is the coefficient of $x^{2}$ , reject $k = 0$ , since the equation would no longer be quadratic.

Proving roots are real. To show an equation has real roots for all values of the letters, write $D$ as a sum of squares, which can never be negative. For example $(a - b)^{2} + 4 k^{2} \geq 0$ .

Using the discriminant in context. In word problems, $D < 0$ tells you a situation is impossible (a rectangle with that perimeter cannot have that area), and $D = 0$ often marks a largest or smallest possible value.

Nature of Roots and the Discriminant: formulas to remember

Discriminant: $D = b^{2} - 4 a c$; For $a x^{2} + b x + c = 0$ , $a \neq = 0$ .
Two distinct real roots: $D > 0$; Rational if $D$ is a perfect square (with rational coefficients).
Two equal real roots: $D = 0 \Rightarrow x = - \frac{b}{2 a}$; The quadratic is a perfect square.
No real roots: $D < 0$; No real number satisfies the equation.
Real roots (equal or distinct): $D \geq 0$; Used when the question says only 'real roots'.

How to solve Nature of Roots and the Discriminant questions

Every nature-of-roots question starts with a careful value of D.

1Write the equation in standard form and read $a$ , $b$ , $c$ with signs.
2Compute $b^{2}$ in brackets: $(- q)^{2} = q^{2}$ , $(7)^{2} = 7$ , $(26)^{2} = 24$ .
3Find $D = b^{2} - 4 a c$ and compare it with 0.
4State the nature in the exact words: 'two distinct real roots', 'two equal real roots', or 'no real roots'.
5For an unknown $k$ : set $D = 0$ , $D > 0$ , $D \geq 0$ or $D < 0$ as the question demands, and solve.
6If roots are asked after finding $k$ , use $x = - \frac{b}{2 a}$ for equal roots.
7Reject a value of $k$ that makes $a = 0$ .

Nature of Roots and the Discriminant in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

D < 0 means impossible

Suppose you want to fence a rectangular plot with exactly 40 m of fencing and enclose an area of 120 m².

Show that this is impossible. What is the largest area you could enclose?

Try it first, then show the solutionSolution

If one side is $x$ , the other is $20 - x$ : $x (20 - x) = 120 \Rightarrow x^{2} - 20 x + 120 = 0$ .
$D = 400 - 480 = - 80 < 0$ : no real $x$ , so impossible.
For area $A$ : $x^{2} - 20 x + A = 0$ needs $D = 400 - 4 A \geq 0$ , so $A \leq 100$ ; equality ( $D = 0$ ) gives $x = 10$ , a 10 m square.

Answer: Impossible; the largest area is 100 m², a 10 m by 10 m square.

Memory hook: A negative discriminant is maths saying 'cannot be done'.

D = 0 marks the highest point

Imagine a ball thrown upwards whose height after $t$ seconds is $h = 20 t - 5 t^{2}$ metres.

Can the ball reach 25 m? What is the greatest height it reaches?

Try it first, then show the solutionSolution

$20 t - 5 t^{2} = 25 \Rightarrow t^{2} - 4 t + 5 = 0$ ; $D = 16 - 20 = - 4 < 0$ , so no.
For height $H$ : $5 t^{2} - 20 t + H = 0$ , $D = 400 - 20 H \geq 0 \Rightarrow H \leq 20$ .
At $H = 20$ , $D = 0$ and $t = 2$ s (one moment only: the top).

Answer: No; the greatest height is 20 m, reached at $t = 2$ s.

Memory hook: Equal roots: the ball touches that height once, at the very top.

Tricks to remember Nature of Roots and the Discriminant

Short memory aids. Say each one aloud once, then use it on a question the same day.

Plus, zero, minus
D positive: two roots; D zero: one repeated root; D negative: none (real).
Use it when: Stating the nature instantly.
Square in brackets
Always write $b^{2}$ as $(b)^{2}$ : $(- q)^{2} = q^{2}$ , $(7)^{2} = 7$ .
Use it when: Avoiding the 2024 and 2025 MCQ traps.
Equal roots → minus b over 2a
When $D = 0$ the ± part vanishes, leaving $- \frac{b}{2 a}$ .
Use it when: Finding the repeated root quickly.
Sum of squares is never negative
To prove roots are always real, make $D$ look like $(\dots)^{2} + (\dots)^{2}$ .
Use it when: Proof-type questions.

Nature of Roots and the Discriminant: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Recognising no real roots · 1 mark · MCQBoard level

The equation $2 x^{2} - 4 x + 3 = 0$ has:
(a) two distinct real roots
(b) two equal real roots
(c) no real roots
(d) more than two real roots

Try it first, then show the solutionSolution

$D = (- 4)^{2} - 4 (2) (3) = 16 - 24 = - 8 < 0$ .

Answer: (c) no real roots

Example 2: Nature of roots of two equations · 3 marks · Short answerBoard level

Without solving, state the nature of the roots of (i) $x^{2} - 6 x + 9 = 0$ , (ii) $2 x^{2} + 5 x - 3 = 0$ . Where the roots are equal, find them.

Try it first, then show the solutionSolution

(i) $D = 36 - 36 = 0$ : two equal real roots, $x = \frac{6}{2} = 3$ (each).
(ii) $D = 25 + 24 = 49 > 0$ : two distinct real roots; since 49 is a perfect square they are rational.

Answer: (i) Equal real roots, $x = 3, 3$ . (ii) Two distinct real (rational) roots.

Example 3: Finding k for equal roots · 3 marks · Short answerHarder board level

Find the value of $k$ for which $k x^{2} - 12 x + 9 = 0$ has equal roots, and find the roots.

Try it first, then show the solutionSolution

$D = 144 - 36 k = 0 \Rightarrow k = 4$ (and $k \neq = 0$ ✓).
Equation: $4 x^{2} - 12 x + 9 = 0 = (2 x - 3)^{2}$ .
Roots: $x = \frac{12}{8} = \frac{3}{2}$ , $\frac{3}{2}$ .

Answer: $k = 4$ ; roots $\frac{3}{2}, \frac{3}{2}$ .

Example 4: Two values of k · 4 marks · Short answerHarder board level

Find the values of $k$ for which $x^{2} + (k - 1) x + 4 = 0$ has equal roots. For each value, find the roots.

Try it first, then show the solutionSolution

$D = (k - 1)^{2} - 16 = 0 \Rightarrow (k - 1)^{2} = 16 \Rightarrow k - 1 = \pm 4$ .
$k = 5$ or $k = - 3$ .
$k = 5$ : $x^{2} + 4 x + 4 = 0$ , $x = - 2, - 2$ .
$k = - 3$ : $x^{2} - 4 x + 4 = 0$ , $x = 2, 2$ .

Answer: $k = 5$ (roots $- 2, - 2$ ) or $k = - 3$ (roots $2, 2$ ).

Example 5: Range of k for distinct roots · 4 marks · Short answerHOTS / competency

Find the values of $k$ for which $x^{2} - 4 x + k = 0$ has (a) two distinct real roots, (b) no real roots. Hence write the greatest integer $k$ for which the roots are real and distinct.

Try it first, then show the solutionSolution

$D = 16 - 4 k$ .
(a) $16 - 4 k > 0 \Rightarrow k < 4$ .
(b) $16 - 4 k < 0 \Rightarrow k > 4$ .
Greatest integer with $k < 4$ is 3 (check: $D = 4 > 0$ ).

Answer: (a) $k < 4$ (b) $k > 4$ ; greatest integer $k = 3$ .

Example 6: Proving the roots are always real · 4 marks · Long answerHOTS / competency

Prove that the roots of $(x - a) (x - b) = k^{2}$ are real for all real values of $a$ , $b$ and $k$ .

Try it first, then show the solutionSolution

Expand: $x^{2} - (a + b) x + ab - k^{2} = 0$ .
$D = (a + b)^{2} - 4 (ab - k^{2}) = a^{2} + 2 ab + b^{2} - 4 ab + 4 k^{2}$ .
$D = (a - b)^{2} + 4 k^{2}$ .
A sum of squares is never negative, so $D \geq 0$ and the roots are always real.

Answer: $D = (a - b)^{2} + 4 k^{2} \geq 0$ , so the roots are real.

Where students lose marks in Nature of Roots and the Discriminant

The slips examiners see most often, and what to write instead.

Mistake: Squaring a negative coefficient as negative
$(- q)^{2} = q^{2}$ , which is positive.
Wrong working
$p x^{2} - q x + r = 0$ : $D = - q^{2} - 4 p r$ .
Right working
$D = (- q)^{2} - 4 p r = q^{2} - 4 p r$ .
Do this: Put $b$ in brackets before squaring.
Usually costs: The MCQ in 2024 Q1(iii).
Mistake: Squaring a surd wrongly
$(7)^{2} = 7$ , not 49.
Wrong working
$3 x^{2} + 7 x + 2 = 0$ : $D = 49 - 24 = 25 > 0$ .
Right working
$D = 7 - 24 = - 17 < 0$ : no real roots.
Do this: $(n)^{2} = n$ .
Usually costs: The MCQ in 2025 Q1(i).
Mistake: Giving only one value of k
$(k - 1)^{2} = 16$ has two solutions.
Wrong working
$k - 1 = 4 \Rightarrow k = 5$ .
Right working
$k - 1 = \pm 4 \Rightarrow k = 5$ or $- 3$ .
Do this: Take ± when you take a square root.
Usually costs: Half the marks.
Mistake: Keeping k = 0 when k multiplies x²
Then the equation is not quadratic.
Wrong working
$k x^{2} + k x + 1 = 0$ , equal roots: $k^{2} - 4 k = 0 \Rightarrow k = 0$ or $4$ .
Right working
$k = 4$ only ( $k = 0$ rejected).
Do this: Check $a \neq = 0$ for each value.
Usually costs: The final mark.

Exam tips

Use the exact syllabus wording: 'two distinct real roots', 'two equal real roots', 'no real roots'.
Show $D$ as a number before stating the nature; the number earns the method mark.
In 2024 Q1(iii) the examiners reported candidates who chose $p^{2} = 4 q r$ ; the correct condition was $q^{2} = 4 p r$ .
In 2025 Q1(i) they reported candidates who took $b^{2}$ as 49 instead of 7.

Practice questions on Nature of Roots and the Discriminant

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The equation $x^{2} + 5 x + 2 = 0$ has:
(a) two distinct real roots
(b) two equal real roots
(c) no real roots
(d) one real root only
Hint
$(5)^{2} = 5$ .
Answer
(c). $D = 5 - 8 = - 3 < 0$ .
Q2 · 3 marks · Short answerBoard level
State the nature of the roots of $9 x^{2} - 30 x + 25 = 0$ and find them if they are real.
Hint
Compute $D$ .
Answer
$D = 900 - 900 = 0$ : equal real roots, $x = \frac{30}{18} = \frac{5}{3}$ (each).
Q3 · 1 mark · MCQHarder board level
The roots of $p x^{2} + q x + r = 0$ ( $p \neq = 0$ ) are real and distinct if:
(a) $q^{2} = 4 p r$
(b) $q^{2} < 4 p r$
(c) $q^{2} > 4 p r$
(d) $p^{2} > 4 q r$
Hint
$a = p$ , $b = q$ , $c = r$ .
Answer
(c)
Q4 · 3 marks · Short answerHarder board level
Find the values of $k$ for which $2 x^{2} + k x + 8 = 0$ has equal roots.
Hint
$k^{2} - 64 = 0$ .
Answer
$k = 8$ or $k = - 8$ .
Q5 · 3 marks · Short answerHarder board level
For what values of $k$ does $x^{2} + k x + 9 = 0$ have no real roots?
Hint
$k^{2} - 36 < 0$ .
Answer
$- 6 < k < 6$ .
Q6 · 4 marks · Short answerHOTS / competency
If $x = 2$ is a root of $x^{2} + p x - 6 = 0$ , find $p$ . Hence state the nature of the roots of $x^{2} + p x + p = 0$ .
Hint
Substitute $x = 2$ first.
Answer
$4 + 2 p - 6 = 0 \Rightarrow p = 1$ . Then $x^{2} + x + 1 = 0$ : $D = 1 - 4 = - 3 < 0$ , no real roots.
Q7 · 3 marks · Short answerHarder board level
Show that $3 x^{2} - 26 x + 2 = 0$ has equal roots, and find them.
Hint
$(26)^{2} = 24$ .
Answer
$D = 24 - 24 = 0$ ; root $= \frac{2 6}{6} = \frac{6}{3}$ (twice).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.D for $x^{2} + 3 x + 5 = 0$ ?Show answer
$9 - 20 = - 11$ : no real roots.
2.If $D = 0$ , what are the roots?Show answer
Both equal to $- \frac{b}{2 a}$ .
3.Equal roots of $x^{2} + k x + 16 = 0$ : k?Show answer
$k = \pm 8$ .
4.D for $x^{2} - 5 x + 6 = 0$ and the nature?Show answer
$25 - 24 = 1 > 0$ : two distinct real (rational) roots.
5.Condition for real roots (equal or not)?Show answer
$D \geq 0$ .

Nature of Roots and the Discriminant: frequently asked questions

What is the discriminant of a quadratic equation?

For ax² + bx + c = 0, it is D = b² − 4ac. Its sign tells you whether the roots are real and distinct, real and equal, or not real.

How do I find k so that a quadratic has equal roots?

Write the discriminant in terms of k and set it equal to zero, then solve for k. Check that no value makes the coefficient of x² zero.

Can a quadratic equation have only one root?

When D = 0 it has two equal roots, which is the same number counted twice. ICSE asks you to describe this as 'two equal real roots'.