Quadratic Equations

Quadratic Equations

Quadratic Equations usually carries 5 to 8 marks in the board paper: 6 in the 2026-27 sample paper, 8 in the main 2026 paper, and 5 in the 2025 paper and in the second 2026 paper.

A 5-mark word problem from this chapter has come in every one of these papers: speed and time, cost and quantity, areas and dimensions, numbers, right-triangle sides. There may also be a 1-mark MCQ on the nature of roots or equal roots, and sometimes a 2-mark discriminant question.

Where marks are usually lost:

• not rejecting a root that makes no sense (a negative speed, a negative number of rows, a cut larger than the sheet);
• sign slips in $b^2 - 4ac$, especially when $c$ is negative;
• dividing both sides by an expression in $x$ and losing a root;
• forgetting that $a \neq 0$ when finding $k$ for equal roots.

Revise in 5 Minutes

Standard form: $ax^2 + bx + c = 0$, with $a \neq 0$. Always simplify first: if the $x^2$ terms cancel, the equation is not quadratic.

Methods in the syllabus (completing the square is not)

• Factorisation: split $bx$ into two terms whose product is $ac \cdot x^2$, then take out common factors.
• Quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Discriminant $D = b^2 - 4ac$

A perfect-square $D$ (with rational $a$, $b$, $c$) means the roots are rational; it is not needed for real roots.

Word problems: set-ups to know

• Speed: time = distance ÷ speed; boat upstream $(u - v)$, downstream $(u + v)$.
• Fixed total shared: amount per person $= \frac{\text{total}}{n}$.
• Rectangle: perimeter gives $l + b$; area gives $lb$.
• Consecutive integers $x, x + 1$; consecutive odd or even $x, x + 2$.
• Right triangle: leg² + leg² = hypotenuse²; reject negative lengths.
• Work: if A alone takes $x$ hours and B alone $y$ hours, together they take $t$ hours where $\frac{1}{x} + \frac{1}{y} = \frac{1}{t}$.

"Is it possible?" Form the equation and look at $D$: if $D < 0$, it is not possible.

Traps

• Reject roots that make no sense, and say why in one line.
• Never divide by an expression containing $x$: bring all terms to one side and factorise.
• In "equal roots, find $k$", throw away any $k$ that makes $a = 0$.
• Watch the sign of $c$ when finding $D$.

How to use this page: try each question on paper first, then read the answer. The marks against each step show what an examiner looks for. The 1-mark MCQs and Assertion-Reason questions are in the quiz at the end, together with questions that test how well you understand the chapter; every quiz answer comes with its explanation.

Short Answer Questions (2 and 3 Marks)

Question 1 (2 marks)

Find the roots of the quadratic equation $\sqrt{5}x^2 + 9x + 4\sqrt{5} = 0$ by factorisation.

Answer.

1. $\sqrt{5}x^2 + 5x + 4x + 4\sqrt{5} = \sqrt{5}x(x + \sqrt{5}) + 4(x + \sqrt{5}) = (\sqrt{5}x + 4)(x + \sqrt{5})$ — 1 mark
2. Roots $x = -\frac{4}{\sqrt{5}}$ and $x = -\sqrt{5}$ — 1 mark

Question 2 (2 marks)

Show that the equation $2x^2 - 2(p + q)x + p^2 + q^2 = 0$ has equal roots when $p = q$, and no real roots when $p \neq q$.

Answer.

1. $D = 4(p + q)^2 - 8(p^2 + q^2) = -4p^2 + 8pq - 4q^2 = -4(p - q)^2$ — 1 mark
2. If $p = q$, $D = 0$: equal roots; if $p \neq q$, $D < 0$: no real roots — 1 mark

Question 3 (2 marks)

Find the nature of the roots of the equation $25x^2 - 20x + 4 = 0$. If the roots are real, find them.

Answer.

1. $D = (-20)^2 - 4 \times 25 \times 4 = 400 - 400 = 0$, so the roots are real and equal — 1 mark
2. Each root $= -\frac{b}{2a} = \frac{20}{50} = \frac{2}{5}$ — 1 mark

Question 4 (2 marks)

Solve $2x^2 - 4x - 3 = 0$ using the quadratic formula.

Answer.

1. $D = 16 + 24 = 40$, $\sqrt{D} = 2\sqrt{10}$ — 1 mark
2. $x = \frac{4 \pm 2\sqrt{10}}{4} = \frac{2 \pm \sqrt{10}}{2}$ — 1 mark

Question 5 (2 marks)

Rahul solved $x(x - 3) = 2(x - 3)$ by dividing both sides by $(x - 3)$ and wrote the answer $x = 2$. Is his answer complete? Find all the roots.

Answer.

Model answer:

No, Rahul's answer is not complete.

Dividing both sides by $(x - 3)$ is allowed only when $x - 3 \neq 0$. By doing this he has thrown away the case $x = 3$, and $x = 3$ does satisfy the equation: $3 \times 0 = 2 \times 0$.

The right way is to bring everything to one side: $x(x - 3) - 2(x - 3) = 0$, that is, $(x - 3)(x - 2) = 0$.

So the roots are $x = 2$ and $x = 3$. A quadratic equation can have two roots, so never divide by an expression that contains $x$.

Marking scheme:

1. No; dividing by $(x - 3)$ assumes $x \neq 3$, so the root $x = 3$ is lost — 1 mark
2. $x(x - 3) - 2(x - 3) = 0$ gives $(x - 3)(x - 2) = 0$; roots 2 and 3 — 1 mark

Question 6 (3 marks)

Solve for $x$: $\frac{4}{x + 1} - \frac{2}{x + 4} = 1$, where $x \neq -1, -4$.

Answer.

1. $4(x + 4) - 2(x + 1) = (x + 1)(x + 4)$, so $2x + 14 = x^2 + 5x + 4$ — 1 mark
2. $x^2 + 3x - 10 = 0$, that is, $(x + 5)(x - 2) = 0$ — 1 mark
3. $x = 2$ or $x = -5$ (neither is an excluded value) — 1 mark

Question 7 (3 marks)

The difference of two natural numbers is 5 and the sum of their squares is 157. Find the numbers.

Answer.

1. Let the smaller number be $x$; the larger is $x + 5$ and $x^2 + (x + 5)^2 = 157$ — 1 mark
2. $2x^2 + 10x - 132 = 0$, so $x^2 + 5x - 66 = 0$ and $(x + 11)(x - 6) = 0$ — 1 mark
3. $x = -11$ is not a natural number, so $x = 6$; the numbers are 6 and 11 — 1 mark

Question 8 (3 marks)

If $3$ and $-\frac{1}{2}$ are the roots of the quadratic equation $ax^2 - 5x + b = 0$, find the values of $a$ and $b$.

Answer.

1. $x = 3$: $9a - 15 + b = 0$ — 1 mark
2. $x = -\frac{1}{2}$: $\frac{a}{4} + \frac{5}{2} + b = 0$, that is, $a + 10 + 4b = 0$ — 1 mark
3. $b = 15 - 9a$ gives $a + 10 + 60 - 36a = 0$, so $a = 2$ and $b = -3$ — 1 mark

Question 9 (3 marks)

Solve for $x$ using the quadratic formula: $x^2 - 2ax + a^2 - 9b^2 = 0$.

Answer.

1. Here $A = 1$, $B = -2a$, $C = a^2 - 9b^2$ — 1 mark
2. $D = 4a^2 - 4(a^2 - 9b^2) = 36b^2$, so $\sqrt{D} = 6b$ — 1 mark
3. $x = \frac{2a \pm 6b}{2}$: $x = a + 3b$ or $x = a - 3b$ — 1 mark

Question 10 (3 marks)

The perimeter of a rectangular plot is 62 m and its area is 228 m². Find the length and breadth of the plot.

Answer.

1. Length + breadth $= 31$ m; if the breadth is $x$ m, the length is $(31 - x)$ m and $x(31 - x) = 228$ — 1 mark
2. $x^2 - 31x + 228 = 0$, so $(x - 12)(x - 19) = 0$ — 1 mark
3. $x = 12$ or $19$; the plot is 19 m long and 12 m broad — 1 mark

Long Answer and Case-Based Questions

Question 11 (5 marks)

A ferry on the Brahmaputra at Guwahati goes 24 km upstream and comes back to its starting point in 6 hours in all. The stream flows at 3 km/h. Find the speed of the ferry in still water.

Answer.

Model answer:

Let the speed of the ferry in still water be $x$ km/h. Then its speed upstream is $(x - 3)$ km/h and downstream is $(x + 3)$ km/h.

Time = distance ÷ speed, so

$\frac{24}{x - 3} + \frac{24}{x + 3} = 6$.

Multiplying both sides by $(x - 3)(x + 3) = x^2 - 9$: $24(x + 3) + 24(x - 3) = 6(x^2 - 9)$, that is, $48x = 6x^2 - 54$.

Dividing by 6: $x^2 - 8x - 9 = 0$. Splitting the middle term: $x^2 - 9x + x - 9 = (x - 9)(x + 1) = 0$.

So $x = 9$ or $x = -1$. A speed cannot be negative, so the speed of the ferry in still water is 9 km/h.

Check: upstream time $= \frac{24}{6} = 4$ hours and downstream time $= \frac{24}{12} = 2$ hours, which add up to 6 hours.

Marking scheme:

1. Let the speed in still water be $x$ km/h: upstream $x - 3$, downstream $x + 3$ — 1 mark
2. $\frac{24}{x - 3} + \frac{24}{x + 3} = 6$ — 1 mark
3. $24(x + 3) + 24(x - 3) = 6(x^2 - 9)$, so $48x = 6x^2 - 54$ — 1 mark
4. $x^2 - 8x - 9 = 0$, that is, $(x - 9)(x + 1) = 0$ — 1 mark
5. Speed cannot be negative, so the ferry moves at 9 km/h in still water (check: $4 + 2 = 6$ hours) — 1 mark

Question 12 (5 marks)

A goods train covers the first 200 km of its journey at a uniform speed. For the next 280 km it runs 20 km/h faster. The whole journey of 480 km takes 8 hours. Find the speed of the train for the first part and the time it takes for each part.

Answer.

Model answer:

Let the speed for the first 200 km be $x$ km/h. Then the speed for the next 280 km is $(x + 20)$ km/h.

Total time: $\frac{200}{x} + \frac{280}{x + 20} = 8$.

Multiplying by $x(x + 20)$: $200(x + 20) + 280x = 8x(x + 20)$, that is, $480x + 4000 = 8x^2 + 160x$.

So $8x^2 - 320x - 4000 = 0$. Dividing by 8: $x^2 - 40x - 500 = 0$.

Splitting the middle term: $x^2 - 50x + 10x - 500 = x(x - 50) + 10(x - 50) = (x - 50)(x + 10) = 0$.

So $x = 50$ or $x = -10$. Speed cannot be negative, so the train runs at 50 km/h for the first part and 70 km/h for the second.

Time for the first part $= \frac{200}{50} = 4$ hours and for the second $= \frac{280}{70} = 4$ hours, which add up to 8 hours.

Marking scheme:

1. Let the first speed be $x$ km/h: $\frac{200}{x} + \frac{280}{x + 20} = 8$ — 1 mark
2. $200(x + 20) + 280x = 8x(x + 20)$, so $480x + 4000 = 8x^2 + 160x$ — 1 mark
3. $8x^2 - 320x - 4000 = 0$, that is, $x^2 - 40x - 500 = 0$ — 1 mark
4. $(x - 50)(x + 10) = 0$; $x = -10$ is rejected, so the speed is 50 km/h — 1 mark
5. Times: $\frac{200}{50} = 4$ hours and $\frac{280}{70} = 4$ hours — 1 mark

Question 13 (5 marks)

For a Diwali craft fair, the students of a school in Kolkata make open boxes from rectangular cardboard sheets 40 cm long and 30 cm wide. From each corner of a sheet a square of side $x$ cm is cut off and the sides are folded up, as shown in the figure. If the base of the box must have an area of 600 cm², find $x$ and the volume of the box.

Answer.

1. Base of the box: $(40 - 2x)$ cm by $(30 - 2x)$ cm; $(40 - 2x)(30 - 2x) = 600$ — 1 mark
2. $1200 - 140x + 4x^2 = 600$, so $x^2 - 35x + 150 = 0$ — 1 mark
3. $(x - 5)(x - 30) = 0$, so $x = 5$ or $x = 30$ — 1 mark
4. $x = 30$ is impossible (two 30 cm cuts need a sheet wider than 60 cm), so $x = 5$ — 1 mark
5. Base 30 cm by 20 cm, height 5 cm; volume $= 600 \times 5 = 3000$ cm³ — 1 mark

Question 14 (4 marks)

A carpenter in Jodhpur makes a wooden corner shelf in the shape of a right triangle, as shown in the figure. If the shortest side is $x$ cm, the other side containing the right angle is $(x + 1)$ cm and the longest side is $(x + 9)$ cm.

(i) Show that $x^2 - 16x - 80 = 0$. (1 mark)

Answer.

1. Pythagoras: $x^2 + (x + 1)^2 = (x + 9)^2$ gives $2x^2 + 2x + 1 = x^2 + 18x + 81$, that is, $x^2 - 16x - 80 = 0$ — 1 mark

(ii) Find the discriminant of this equation and state the nature of its roots. (1 mark)

Answer.

1. $D = 256 + 320 = 576 = 24^2 > 0$: two distinct real (rational) roots — 1 mark

(iii) Find the lengths of the three sides of the shelf. (2 marks)

Answer.

1. $x = \frac{16 \pm 24}{2}$, so $x = 20$ or $x = -4$; a length cannot be negative, so $x = 20$ — 1 mark
2. Sides 20 cm, 21 cm and 29 cm — 1 mark

OR

(iii) For a smaller shelf, the three sides of the right triangle are $x$ cm, $(x + 1)$ cm and $(x + 2)$ cm. Find them. (2 marks)

Answer.

1. $x^2 + (x + 1)^2 = (x + 2)^2$ gives $x^2 - 2x - 3 = 0$, that is, $(x - 3)(x + 1) = 0$ — 1 mark
2. $x = 3$ ($x = -1$ rejected); sides 3 cm, 4 cm and 5 cm — 1 mark

Question 15 (4 marks)

At a science fair in Bhopal, the science club launches a small water rocket straight up. Its height above the ground after $t$ seconds is $h = 20t - 5t^2$ metres.

(i) Write the quadratic equation for the times at which the rocket is 15 m above the ground, in standard form. (1 mark)

Answer.

1. $20t - 5t^2 = 15$, that is, $5t^2 - 20t + 15 = 0$ or $t^2 - 4t + 3 = 0$ — 1 mark

(ii) At what times is the rocket 15 m above the ground? (1 mark)

Answer.

1. $(t - 1)(t - 3) = 0$: after 1 s (going up) and after 3 s (coming down) — 1 mark

(iii) Can the rocket reach a height of 25 m? Justify your answer using the discriminant. (2 marks)

Answer.

1. $20t - 5t^2 = 25$ gives $t^2 - 4t + 5 = 0$ — 1 mark
2. $D = 16 - 20 = -4 < 0$: no real $t$, so the rocket never reaches 25 m — 1 mark

OR

(iii) After how many seconds does the rocket come back to the ground? (2 marks)

Answer.

1. $h = 0$: $20t - 5t^2 = 0$, that is, $5t(4 - t) = 0$ — 1 mark
2. $t = 0$ is the launch, so it lands after 4 s — 1 mark