MathsCBSE Class 10 · Quadratic Equations

Nature of Roots and Discriminant of a Quadratic Equation

Q: What is the difference between real roots and distinct real roots?

Real roots means the equation has real solutions, which may be two different numbers or the same number twice, so the condition is D greater than or equal to zero. Distinct real roots means two different real numbers, so the condition is D strictly greater than zero. Mixing these up is a common way to lose the final mark.

Q: Why do I have to reject some values of k?

If a value of k makes the coefficient of x squared zero, the equation is no longer quadratic, and the discriminant rule does not apply to it. Substitute each value back into that coefficient and reject any that make it zero, with a one-line reason.

Q: Which NCERT exercise has practice questions on nature of roots?

In the current rationalised NCERT book, Chapter 4 has three exercises. Nature of roots is taught in Section 4.4, and its practice questions are in Exercise 4.3.

Q: Do I need to find the roots when a question asks only for the nature of roots?

No. Computing the discriminant and stating its sign with the correct conclusion is enough. Find the roots only when the question also asks for them.

Q: Is the method of completing the square in the 2026-27 syllabus?

No, solving by completing the square is not in the 2026-27 syllabus. You may quote and use the quadratic formula directly.

Updated 14 October 2026 · Fact-checked

The discriminant of ax² + bx + c = 0 is D = b² − 4ac. If D is positive, there are two distinct real roots; if D is zero, two equal real roots; if D is negative, no real roots. Board questions test stating the nature of roots, finding k for equal or real roots, and short proofs.

The sign of D = b² − 4ac tells how many times the graph meets the x-axis: two points, one touching point, or none.

Illustration drawn for QuestionBuddy.

Understand Nature of Roots and Discriminant

Every quadratic equation $a x^{2} + b x + c = 0$ with $a \neq = 0$ has its roots given by the quadratic formula $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ . Look at what sits under the square root: $b^{2} - 4 a c$ . This one number is called the discriminant, written $D$ . It tells you what kind of roots the equation has before you solve anything.

Here is why. The formula adds and subtracts $D$ .

If $D > 0$ , $D$ is a positive real number, so adding it and subtracting it give two different answers: two distinct real roots.
If $D = 0$ , you add and subtract zero, so both roots are the same number $- \frac{b}{2 a}$ : two equal real roots.
If $D < 0$ , no real number squares to a negative number, so $D$ is not real: no real roots.

This links back to Polynomials. The roots of the equation are the zeroes of $a x^{2} + b x + c$ , so $D$ also tells you whether the graph of $y = a x^{2} + b x + c$ cuts the x-axis at two points, touches it at one point, or does not meet it at all. When $D = 0$ , the expression is a constant multiple of a perfect square, for example $2 x^{2} + 8 x + 8 = 2 (x + 2)^{2}$ or $4 x^{2} - 12 x + 9 = (2 x - 3)^{2}$ .

The rule only works under three conditions. The equation must first be in standard form, with everything on one side, because $a$ , $b$ and $c$ are read with their signs only from that form. The coefficients must be real numbers. And $a$ must not be zero, otherwise the equation is linear and the rule does not apply. A useful shortcut: if $a$ and $c$ have opposite signs, then $- 4 a c$ is positive and $D$ is automatically positive.

Most board questions use the discriminant in reverse. An unknown $k$ sits in the coefficients and a condition on the roots is given. Translate the words into a statement about $D$ : equal roots means $D = 0$ , real roots means $D \geq 0$ , distinct or unequal real roots means $D > 0$ , and no real roots means $D < 0$ . Then solve for $k$ , and reject any value that makes the coefficient of $x^{2}$ zero. In proof questions, the coefficients contain letters; you expand $D$ and show it is a perfect square (so never negative) or minus a perfect square (so never positive).

Nature of Roots and Discriminant: formulas to remember

Discriminant: $D = b^{2} - 4 a c$; Read a, b, c with their signs from the standard form ax^2 + bx + c = 0, a ≠ 0.
Two distinct real roots: $D > 0 \Rightarrow x = \frac{- b + D}{2 a}, \frac{- b - D}{2 a}$; Used for 'distinct', 'unequal' or 'two different real roots'.
Two equal real roots: $D = 0 \Rightarrow x = - \frac{b}{2 a}$; Both roots are the same number; the quadratic expression is a perfect square.
No real roots: $D < 0$; Write 'no real roots'. Do not try to take the square root of a negative number.
Real roots (equal or distinct): $D \geq 0$; When the question says only 'real roots', include the equal case. This usually gives an inequality in k.
Opposite-sign shortcut: $a c < 0 \Rightarrow D = b^{2} - 4 a c > 0$; If a and c have opposite signs, the roots are always real and distinct, whatever b is.
Condition for a quadratic: $a \neq = 0$; In find-k questions, reject any value of k that makes the coefficient of x^2 zero.

How to solve Nature of Roots and Discriminant questions

This method handles every board question on nature of roots: stating the nature, finding k, finding a range of k, and proofs.

1Bring the equation to standard form $a x^{2} + b x + c = 0$ by expanding brackets and moving every term to one side.
2Write $a$ , $b$ and $c$ on one line with their signs. If they contain $k$ or letters, keep them in brackets.
3Compute $D = b^{2} - 4 a c$ on its own line. Square the whole of $b$ , including any number in front of a bracket, and simplify fully. Take out a common factor if you can.
4Translate the condition: equal roots gives $D = 0$ , real roots gives $D \geq 0$ , distinct roots gives $D > 0$ , no real roots gives $D < 0$ .
5Solve the resulting equation or inequality for the unknown. For an inequality such as $k^{2} \geq 36$ , write both parts: $k \leq - 6$ or $k \geq 6$ .
6Reject any value that makes the coefficient of $x^{2}$ zero, and write the reason in one line.
7If the roots are asked for, use $x = - \frac{b}{2 a}$ when $D = 0$ , or the full formula when $D > 0$ , and check one root by substitution.
8Finish with a sentence in words, such as 'The equation has two equal real roots, each equal to $- 2$ .'

Nature of Roots and 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.

Discriminant and the nature of roots

Suppose your grandparents want to fence a rectangular kitchen garden behind the house. The back wall of the house forms one long side, so they need wire only for the other three sides. They have exactly 40 m of wire and want to use all of it.

Using the discriminant, decide whether the garden can have an area of (i) 200 square metres, (ii) 250 square metres. Where it is possible, find the dimensions of the garden.

Try it first, then show the solutionSolution

Let each side perpendicular to the wall be $x$ m. Then the side parallel to the wall is $(40 - 2 x)$ m.
Area $= x (40 - 2 x) = 40 x - 2 x^{2}$ .
(i) $40 x - 2 x^{2} = 200 \Rightarrow 2 x^{2} - 40 x + 200 = 0 \Rightarrow x^{2} - 20 x + 100 = 0$ .
$D = (- 20)^{2} - 4 (1) (100) = 400 - 400 = 0$ , so there are two equal real roots: exactly one possible garden.
$x = - \frac{b}{2 a} = \frac{20}{2} = 10$ , so the sides are 10 m and $40 - 20 = 20$ m.
Check: $10 \times 20 = 200$ square metres and $10 + 10 + 20 = 40$ m of wire.
(ii) $40 x - 2 x^{2} = 250 \Rightarrow x^{2} - 20 x + 125 = 0$ .
$D = 400 - 4 (1) (125) = 400 - 500 = - 100 < 0$ , so there are no real roots and no such garden exists.

Answer: (i) Yes, in exactly one way: 10 m (sides perpendicular to the wall) by 20 m (side along the wall). (ii) No, an area of 250 square metres is not possible with 40 m of wire.

Memory hook: The garden test: D = 0 means exactly one garden fits, D < 0 means the wire is not long enough.

Discriminant and the nature of roots

Imagine your class runs a ring-toss stall at the school fete. From a quick survey, you expect that if a ticket costs Rs $p$ , about $(100 - 5 p)$ students will play. The class wants to plan the price before the fete.

Show, using the discriminant, that the stall cannot collect Rs 600. Then find the greatest amount the stall can collect, the ticket price that gives it, and the number of players at that price.

Try it first, then show the solutionSolution

Collection $= p (100 - 5 p)$ rupees. For a collection of Rs $R$ : $5 p^{2} - 100 p + R = 0$ .
For Rs 600: $5 p^{2} - 100 p + 600 = 0 \Rightarrow p^{2} - 20 p + 120 = 0$ .
$D = 400 - 480 = - 80 < 0$ , so no real price gives Rs 600.
In general, a real price exists only if $D \geq 0$ : $(- 100)^{2} - 4 (5) (R) \geq 0$ .
$10000 - 20 R \geq 0 \Rightarrow R \leq 500$ , so the greatest collection is Rs 500.
At $R = 500$ , $D = 0$ : $p = - \frac{b}{2 a} = \frac{100}{10} = 10$ .
Players $= 100 - 5 (10) = 50$ . Check: $50 \times 10 = 500$ .

Answer: Rs 600 is not possible. The greatest collection is Rs 500, with tickets at Rs 10 each and 50 players.

Memory hook: The fete price: the biggest collection is where D just becomes zero.

Tricks to remember Nature of Roots and Discriminant

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

Pair, twins, nowhere
D positive: a pair of different roots. D zero: twins (the same root twice). D negative: nowhere on the number line.
Use it when: Stating the nature of roots in MCQs and assertion-reason questions.
Real means 'or equal'
'Real' gets the sign $\geq$ ; 'distinct' gets the strict sign $>$ . Real includes the twins; distinct does not.
Use it when: Deciding between $D \geq 0$ and $D > 0$ in range-of-k questions.
Opposite signs, sure roots
If the $x^{2}$ term and the constant have opposite signs, $- 4 a c$ is positive, so $D$ is positive whatever $b$ is.
Use it when: Quick MCQ checks, such as $x^{2} + p x - 4 = 0$ having distinct real roots for every $p$ .
Check the boss
The coefficient of $x^{2}$ is the boss of the equation. If a value of $k$ fires the boss (makes it zero), the equation is no longer quadratic, so reject that $k$ .
Use it when: Find-k questions where $k$ appears in the coefficient of $x^{2}$ .

Nature of Roots and Discriminant: solved examples

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

Example 1: Nature of roots with a surd coefficient · 2 marks · Very short answerBoard level

Find the nature of the roots of the quadratic equation $3 x^{2} - 43 x + 4 = 0$ . If real roots exist, find them.

Try it first, then show the solutionSolution

Compare with $a x^{2} + b x + c = 0$ : $a = 3$ , $b = - 43$ , $c = 4$ . (½ mark)
$D = (- 43)^{2} - 4 (3) (4) = 48 - 48 = 0$ . (½ mark)
Since $D = 0$ , the equation has two equal real roots. (½ mark)
Each root $= - \frac{b}{2 a} = \frac{4 3}{6} = \frac{2 3}{3}$ . (½ mark)
Check: $3 \times \frac{4}{3} - 43 \times \frac{2 3}{3} + 4 = 4 - 8 + 4 = 0$ .

Answer: Two equal real roots, each equal to $\frac{2 3}{3}$ .

Example 2: Find k for equal roots, with k in the x² coefficient · 3 marks · Short answerBoard level

Find the values of $k$ for which the quadratic equation $(k + 1) x^{2} - 2 (k - 1) x + 1 = 0$ has two equal real roots. Hence find the roots for each value of $k$ .

Try it first, then show the solutionSolution

$a = k + 1$ , $b = - 2 (k - 1)$ , $c = 1$ , so $D = 4 (k - 1)^{2} - 4 (k + 1)$ . (½ mark)
$D = 4 (k^{2} - 2 k + 1 - k - 1) = 4 (k^{2} - 3 k) = 4 k (k - 3)$ . (½ mark)
For equal roots, $D = 0$ , so $k = 0$ or $k = 3$ . (½ mark)
For both values, $k + 1 \neq = 0$ , so the equation stays quadratic and neither value is rejected. (½ mark)
$k = 0$ : $x^{2} + 2 x + 1 = 0 \Rightarrow (x + 1)^{2} = 0 \Rightarrow x = - 1$ . (½ mark)
$k = 3$ : $4 x^{2} - 4 x + 1 = 0 \Rightarrow (2 x - 1)^{2} = 0 \Rightarrow x = \frac{1}{2}$ . (½ mark)

Answer: $k = 0$ (roots $- 1, - 1$ ) or $k = 3$ (roots $\frac{1}{2}, \frac{1}{2}$ ).

Example 3: Proving the roots are always real · 3 marks · Short answerHarder board level

Prove that the equation $(x - a) (x - b) + (x - b) (x - c) + (x - c) (x - a) = 0$ has real roots for all real $a$ , $b$ , $c$ , and that the roots are equal only when $a = b = c$ .

Try it first, then show the solutionSolution

Expand and collect: $3 x^{2} - 2 (a + b + c) x + (ab + b c + c a) = 0$ . (½ mark)
$D = 4 (a + b + c)^{2} - 12 (ab + b c + c a) = 4 (a^{2} + b^{2} + c^{2} - ab - b c - c a)$ . (1 mark)
Write it as a sum of squares: $D = 2 [(a - b)^{2} + (b - c)^{2} + (c - a)^{2}]$ . A square is never negative, so $D \geq 0$ and the roots are always real. (1 mark)
Equal roots need $D = 0$ , so each square is 0: $a = b$ , $b = c$ and $c = a$ , that is, $a = b = c$ . (½ mark)

Answer: $D = 2 [(a - b)^{2} + (b - c)^{2} + (c - a)^{2}] \geq 0$ , so the roots are real; $D = 0$ only when $a = b = c$ .

Example 4: Proof for every real m, then a condition on the roots · 5 marks · Long answerHarder board level

Consider the equation $x^{2} - 2 (m + 1) x + (m^{2} + 2 m - 3) = 0$ .
(i) Prove that it has two distinct real roots for every real value of $m$ , and find the roots in terms of $m$ .
(ii) Find the values of $m$ for which one root is three times the other, and state the roots in each case.

Try it first, then show the solutionSolution

(i) $a = 1$ , $b = - 2 (m + 1)$ , $c = m^{2} + 2 m - 3$ . (½ mark)
$D = 4 (m + 1)^{2} - 4 (m^{2} + 2 m - 3)$ . (½ mark)
$D = 4 (m^{2} + 2 m + 1 - m^{2} - 2 m + 3) = 4 \times 4 = 16$ . (½ mark)
$D = 16 > 0$ does not depend on $m$ , so for every real $m$ the equation has two distinct real roots. (½ mark)
$x = \frac{2 ( m + 1 ) \pm 16}{2} = \frac{2 ( m + 1 ) \pm 4}{2}$ . (½ mark)
So the roots are $x = m + 3$ and $x = m - 1$ . (½ mark)
(ii) Case 1: $m + 3 = 3 (m - 1) \Rightarrow m + 3 = 3 m - 3 \Rightarrow m = 3$ . (½ mark)
Roots: $6$ and $2$ ; check $6 = 3 \times 2$ , and the equation $x^{2} - 8 x + 12 = 0$ gives $(x - 6) (x - 2) = 0$ . (½ mark)
Case 2: $m - 1 = 3 (m + 3) \Rightarrow m - 1 = 3 m + 9 \Rightarrow m = - 5$ . (½ mark)
Roots: $- 2$ and $- 6$ ; check $- 6 = 3 \times (- 2)$ , and the equation $x^{2} + 8 x + 12 = 0$ gives $(x + 2) (x + 6) = 0$ . (½ mark)

Answer: (i) $D = 16 > 0$ for all real $m$ ; the roots are $m + 3$ and $m - 1$ . (ii) $m = 3$ (roots 6 and 2) or $m = - 5$ (roots $- 2$ and $- 6$ ).

Example 5: Case study: the quiz-club number puzzle · 4 marks · Case-basedHOTS / competency

In a school maths quiz, each team is told: 'Two positive numbers add up to 20.' Each team then claims a value for the product of the two numbers. Let one of the numbers be $x$ .
(i) If the product is 96, form a quadratic equation in $x$ . (1 mark)
(ii) Find the discriminant of the equation in (i) and state the nature of its roots. (1 mark)
(iii) One team claims the product is 110. Using the discriminant, show that this claim is impossible, and find the greatest product the two numbers can have, with the numbers. (2 marks)

Try it first, then show the solutionSolution

(i) The other number is $20 - x$ , so $x (20 - x) = 96$ . (½ mark)
$\Rightarrow x^{2} - 20 x + 96 = 0$ . (½ mark)
(ii) $D = (- 20)^{2} - 4 (1) (96) = 400 - 384 = 16$ . (½ mark)
$D > 0$ , so two distinct real roots (the numbers are $\frac{20 \pm 4}{2}$ , that is 12 and 8). (½ mark)
(iii) Product 110 gives $x^{2} - 20 x + 110 = 0$ , with $D = 400 - 440 = - 40 < 0$ : no real $x$ , so the claim is impossible. (1 mark)
For product $P$ : $x^{2} - 20 x + P = 0$ needs $D = 400 - 4 P \geq 0$ , so $P \leq 100$ . (½ mark)
At $P = 100$ , $D = 0$ and $x = \frac{20}{2} = 10$ , so both numbers are 10. (½ mark)

Answer: (i) $x^{2} - 20 x + 96 = 0$ . (ii) $D = 16 > 0$ : two distinct real roots. (iii) For 110, $D = - 40 < 0$ , so impossible; the greatest product is 100, when both numbers are 10.

Example 6: No real roots: a range of k · 3 marks · Short answerHarder board level

For the equation $k x^{2} + 4 x + 1 = 0$ , find the values of $k$ for which it has (i) no real roots, (ii) two distinct real roots.

Try it first, then show the solutionSolution

$D = 4^{2} - 4 (k) (1) = 16 - 4 k$ . (½ mark)
(i) No real roots: $16 - 4 k < 0$ , so $k > 4$ . These values are not 0, so the equation stays quadratic. (1 mark)
(ii) Distinct real roots: $16 - 4 k > 0$ , so $k < 4$ . (½ mark)
But $k = 0$ makes the equation $4 x + 1 = 0$ , which is linear with only one root, so reject it: $k < 4$ and $k \neq = 0$ . (1 mark)

Answer: (i) $k > 4$ . (ii) $k < 4$ , $k \neq = 0$ .

Where students lose marks in Nature of Roots and Discriminant

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

Mistake: Reading a, b and c before writing the standard form
Terms on the right-hand side change sign when moved across, so the values of b and c you read off are wrong.
Wrong working
$2 x^{2} = 5 x - 4$ : $a = 2$ , $b = 5$ , $c = - 4$ , so $D = 25 + 32 = 57 > 0$ , two distinct real roots.
Right working
$2 x^{2} - 5 x + 4 = 0$ : $a = 2$ , $b = - 5$ , $c = 4$ , so $D = 25 - 32 = - 7 < 0$ , no real roots.
Do this: Always write the line '... = 0' first, then list a, b, c with signs.
Usually costs: The whole question, because every later step uses the wrong D.
Mistake: Squaring only the bracket in b, not the number in front
When $b = - 2 (k + 1)$ , the 2 must be squared too, giving 4.
Wrong working
$x^{2} - 2 (k + 1) x + (k^{2} + 3) = 0$ : $D = 2 (k + 1)^{2} - 4 (k^{2} + 3) = - 2 k^{2} + 4 k - 10$ , so 'no value of k'.
Right working
$D = 4 (k + 1)^{2} - 4 (k^{2} + 3) = 4 (2 k - 2) = 8 (k - 1)$ ; equal roots when $k = 1$ .
Do this: Write $b^{2} = [- 2 (k + 1)]^{2} = 4 (k + 1)^{2}$ as a separate line.
Usually costs: The method mark for D and every mark after it.
Mistake: Keeping a value of k that makes the x² coefficient zero
The discriminant rule holds only for quadratic equations, so $a \neq = 0$ must be checked.
Wrong working
$(k - 1) x^{2} + 2 (k - 1) x + 1 = 0$ : $D = 4 (k - 1) (k - 2) = 0$ , so $k = 1$ or $k = 2$ .
Right working
$k = 1$ turns the equation into $1 = 0$ , so reject it. Only $k = 2$ works, giving $x^{2} + 2 x + 1 = 0$ and $x = - 1$ .
Do this: After solving for k, substitute each value into the coefficient of $x^{2}$ and write one line of rejection.
Usually costs: Usually the final answer mark.
Mistake: Using D > 0 for 'real roots' and solving the inequality only one way
Real roots include equal roots, so the condition is $D \geq 0$ , and $k^{2} \geq 16$ has a negative part too.
Wrong working
$x^{2} + k x + 4 = 0$ has real roots: $k^{2} - 16 > 0 \Rightarrow k > 4$ .
Right working
$k^{2} - 16 \geq 0 \Rightarrow (k - 4) (k + 4) \geq 0 \Rightarrow k \leq - 4$ or $k \geq 4$ .
Do this: Match the word to the sign ('real' means $\geq$ ) and always write both parts of a squared inequality.
Usually costs: The final answer mark, and the MCQ if the options test this.

Exam tips

Write $D = b^{2} - 4 a c$ on its own line with the substituted values; this line usually carries a method mark even if your arithmetic slips later.
Always end with a sentence in words, such as 'Since $D < 0$ , the equation has no real roots.' A bare number for D is not a complete answer.
In find-k questions, read whether the roots are also asked for. If they are, use $x = - \frac{b}{2 a}$ for each valid k and show the substitution.
In assertion-reason items on roots, test each statement separately with D, then ask whether R actually explains A. A true reason that is not the explanation gives option (b).
In proof questions with letter coefficients, expand D completely before you try to spot a square, and state the equality case (for example, when $a = b$ ).

Practice questions on Nature of Roots and 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 values of $k$ for which the equation $x^{2} - 2 k x + (7 k - 12) = 0$ has two equal real roots are:
(a) 3 only
(b) 3 and 4
(c) $- 3$ and $- 4$
(d) 4 only
Hint
Set $D = 4 k^{2} - 4 (7 k - 12)$ equal to zero and divide by 4.
Answer
(b) 3 and 4. $D = 4 k^{2} - 28 k + 48 = 0 \Rightarrow k^{2} - 7 k + 12 = 0 \Rightarrow (k - 3) (k - 4) = 0$ . Check: $k = 3$ gives $x^{2} - 6 x + 9 = (x - 3)^{2}$ ; $k = 4$ gives $x^{2} - 8 x + 16 = (x - 4)^{2}$ .
Q2 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a carpenter is making a right-angled triangular shelf bracket whose sloping edge (hypotenuse) must be 13 cm. Let one of the other two sides be $x$ cm and let their sum be $s$ cm.
(i) Show that $2 x^{2} - 2 s x + (s^{2} - 169) = 0$ . (1 mark)
(ii) For $s = 17$ , find the two sides. (1 mark)
(iii) Using the discriminant, show that $s = 20$ is impossible, and find the greatest possible value of $s$ . (2 marks)
Hint
The other side is $s - x$ . Use Pythagoras, then require $D \geq 0$ .
Answer
(i) $x^{2} + (s - x)^{2} = 1 3^{2}$ gives $2 x^{2} - 2 s x + s^{2} - 169 = 0$ . (1)
(ii) $s = 17$ : $2 x^{2} - 34 x + 120 = 0$ , so $x^{2} - 17 x + 60 = 0$ and $(x - 5) (x - 12) = 0$ . The sides are 5 cm and 12 cm. (1)
(iii) $D = 4 s^{2} - 8 (s^{2} - 169) = 4 (338 - s^{2})$ . For $s = 20$ , $D = 4 (338 - 400) < 0$ , so no real side exists. Real sides need $s^{2} \leq 338$ , so $s \leq 338 = 132 \approx 18.4$ cm. The greatest value $132$ cm occurs when $D = 0$ , with both sides equal to $\frac{13}{2}$ cm. (2)
Q3 · 3 marks · Short answerBoard level
Find the value(s) of $k$ for which the equation $(k - 12) x^{2} + 2 (k - 12) x + 2 = 0$ has two equal real roots. Also find the roots.
Hint
Take $4 (k - 12)$ as a common factor from $D$ . One of the values you get does not give a quadratic equation.
Answer
$D = 4 (k - 12)^{2} - 8 (k - 12) = 4 (k - 12) (k - 14)$ . $D = 0 \Rightarrow k = 12$ or $k = 14$ . Reject $k = 12$ , since then the coefficient of $x^{2}$ is zero and the equation becomes $2 = 0$ . For $k = 14$ : $2 x^{2} + 4 x + 2 = 0 \Rightarrow (x + 1)^{2} = 0$ , so both roots are $- 1$ .
Q4 · 5 marks · Long answerHarder board level
(i) If the equation $(1 + m^{2}) x^{2} + 2 m c x + (c^{2} - 16) = 0$ has two equal real roots, prove that $c^{2} = 16 (1 + m^{2})$ .
(ii) Hence, for $m = \frac{3}{4}$ , find the possible values of $c$ and the equal root of the equation in each case.
Hint
Write $D = 4 m^{2} c^{2} - 4 (1 + m^{2}) (c^{2} - 16)$ , expand the product fully and watch the $m^{2} c^{2}$ terms cancel.
Answer
(i) $D = 4 [m^{2} c^{2} - (c^{2} - 16 + m^{2} c^{2} - 16 m^{2})] = 4 (16 + 16 m^{2} - c^{2})$ . Equal roots: $D = 0 \Rightarrow c^{2} = 16 (1 + m^{2})$ . (ii) $c^{2} = 16 \times \frac{25}{16} = 25$ , so $c = 5$ or $c = - 5$ . For $c = 5$ : $\frac{25}{16} x^{2} + \frac{15}{2} x + 9 = 0$ ; multiply by 16: $25 x^{2} + 120 x + 144 = 0 \Rightarrow (5 x + 12)^{2} = 0$ , so $x = - \frac{12}{5}$ . For $c = - 5$ : $25 x^{2} - 120 x + 144 = 0 \Rightarrow (5 x - 12)^{2} = 0$ , so $x = \frac{12}{5}$ .
Q5 · 3 marks · Short answerHOTS / competency
Aman writes: 'The equation $x^{2} + k x + k = 0$ has real roots for every $k \geq 0$ , because all its coefficients are then non-negative.'
(i) Test Aman's claim with $k = 2$ .
(ii) Explain the error in his reasoning.
(iii) Find all real values of $k$ for which the equation has real roots.
Hint
The sign of $D$ depends on $b^{2} - 4 a c$ , not on whether each coefficient is positive. Factorise $k^{2} - 4 k$ .
Answer
(i) $k = 2$ : $D = 4 - 8 = - 4 < 0$ , so no real roots; the claim fails. (ii) Non-negative coefficients say nothing about the sign of $b^{2} - 4 a c$ ; only the discriminant decides whether roots are real. (iii) $D = k^{2} - 4 k = k (k - 4) \geq 0$ , so $k \leq 0$ or $k \geq 4$ . Check: $k = 4$ gives $(x + 2)^{2} = 0$ , and $k = 0$ gives $x^{2} = 0$ .
Q6 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The equation $x^{2} + 3 x + 5 = 0$ has no real roots.
Reason (R): If $a$ , $b$ and $c$ are all positive, then $a x^{2} + b x + c = 0$ has no real roots.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Find D for A. For R, try $x^{2} + 3 x + 1 = 0$ .
Answer
(c). For A, $D = 9 - 20 = - 11 < 0$ , so there are no real roots: A is true. R is false: $x^{2} + 3 x + 1 = 0$ has positive coefficients but $D = 9 - 4 = 5 > 0$ , so it has two distinct real roots. Only the sign of $D$ decides.
Q7 · 2 marks · Very short answerBoard level
Find the values of $k$ for which the equation $x^{2} + k x + 9 = 0$ has no real roots.
Hint
No real roots means $D < 0$ .
Answer
$D = k^{2} - 36$ (½ mark). No real roots: $k^{2} - 36 < 0$ , so $k^{2} < 36$ (½ mark), which means $- 6 < k < 6$ (1 mark).

Test your recall

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1.What is the discriminant of $a x^{2} + b x + c = 0$ ?Show answer
$D = b^{2} - 4 a c$ , read from the standard form with $a \neq = 0$ .
2.What is the nature of the roots when $D = 0$ , and what are they?Show answer
Two equal real roots, each equal to $- \frac{b}{2 a}$ .
3.Which condition on D is used when a question says 'real roots'?Show answer
$D \geq 0$ , since real roots may be equal or distinct.
4.What does $D < 0$ tell you?Show answer
The equation has no real roots.
5.Why does $x^{2} + 7 x - 5 = 0$ have distinct real roots without any calculation?Show answer
$a$ and $c$ have opposite signs, so $- 4 a c > 0$ and $D > 0$ .
6.Find the nature of the roots of $x^{2} - 6 x + 10 = 0$ .Show answer
$D = 36 - 40 = - 4 < 0$ , so no real roots.

Nature of Roots and Discriminant: frequently asked questions

What is the difference between real roots and distinct real roots?

Real roots means the equation has real solutions, which may be two different numbers or the same number twice, so the condition is D greater than or equal to zero. Distinct real roots means two different real numbers, so the condition is D strictly greater than zero. Mixing these up is a common way to lose the final mark.

Why do I have to reject some values of k?

If a value of k makes the coefficient of x squared zero, the equation is no longer quadratic, and the discriminant rule does not apply to it. Substitute each value back into that coefficient and reject any that make it zero, with a one-line reason.

Which NCERT exercise has practice questions on nature of roots?

In the current rationalised NCERT book, Chapter 4 has three exercises. Nature of roots is taught in Section 4.4, and its practice questions are in Exercise 4.3.

Do I need to find the roots when a question asks only for the nature of roots?

No. Computing the discriminant and stating its sign with the correct conclusion is enough. Find the roots only when the question also asks for them.

Is the method of completing the square in the 2026-27 syllabus?

No, solving by completing the square is not in the 2026-27 syllabus. You may quote and use the quadratic formula directly.