1. Introduction to Quadratic Equations
A quadratic equation in one variable is an equation that can be written in the form \(ax^2 + bx + c = 0\), where the unknown appears with highest exponent 2. The word "quadratic" comes from quadratus, Latin for "square".
Why do they matter? Whenever two things multiply to a fixed product — the two dimensions of a garden, the speed and time of a journey, the number of rows and benches in a hall — the relationship often becomes quadratic. Squaring also appears when a ball is thrown: its height is a quadratic function of time.
One variable
A single letter, usually \(x\), carrying the square.
Highest power 2
After simplification the \(x^2\) term must survive.
Three coefficients
\[a x^2 + b x + c\] with \(a \neq 0\).
Two roots
Up to two real roots — equal, distinct or none.
2. What You Will Learn
- Recognise a quadratic equation and express it in standard form.
- Identify \(a\), \(b\) and \(c\), including equations not yet in standard form.
- Find roots — by substitution, by factorisation and by the quadratic formula.
- Compute the discriminant and predict the nature of the roots.
- Fix unknown coefficients when a condition on the roots is given.
- Form and solve quadratic equations from word problems.
- Answer competency and case-based questions with reasoning, not just calculation.
3. Identifying a Quadratic Equation
A quadratic equation contains a variable with highest power 2. When it is written out, collected and simplified, the term with \(x^2\) must still be present with a non-zero coefficient.
An equation of the form \(ax^2 + bx + c = 0\), where \(a \neq 0\), and \(a\), \(b\), \(c\) are real numbers, is a quadratic equation in one variable \(x\).
Not every equation that contains an \(x^2\) term is automatically quadratic — sometimes the squared terms cancel when both sides are simplified.
| Equation | Simplify / note | Quadratic? |
|---|---|---|
| \(2x^2 - 4x + 3 = 0\) | Already in standard form | Yes |
| \(x^2 = 25\) | \(x^2 - 25 = 0\), so \(b = 0\) | Yes |
| \(x(x + 3) = 0\) | \(x^2 + 3x = 0\), so \(c = 0\) | Yes |
| \(4x - 7 = 0\) | No \(x^2\) term at all | No — linear |
| \((x + 1)^2 = x^2 + 5\) | \(x^2 + 2x + 1 = x^2 + 5\) gives \(2x = 4\) | No — the \(x^2\) terms cancel |
| \(x^2(x - 1) = 0\) | Expands to \(x^3 - x^2 = 0\) | No — highest power is 3 |
| \((x - 2)(x + 2) = 0\) | \(x^2 - 4 = 0\) | Yes |
Always simplify both sides completely before deciding. A squared term that cancels does not make the equation quadratic.
4. Standard Form
Exams expect a quadratic to be written as
\[ax^2 + bx + c = 0, \qquad a \neq 0\]
where:
- \(a\) is the coefficient of \(x^2\) (must not be 0, otherwise there is no \(x^2\) term and the equation is no longer quadratic);
- \(b\) is the coefficient of \(x\);
- \(c\) is the constant term.
Either \(b\) or \(c\) may be 0 — the equation is still quadratic as long as \(a \neq 0\).
Write \(x(2x - 5) = 3\) in standard form. Expand: \(2x^2 - 5x = 3\). Bring everything to one side: \(2x^2 - 5x - 3 = 0\). Here \(a = 2\), \(b = -5\), \(c = -3\).
5. Identifying \(a\), \(b\) and \(c\)
Before applying any method, put the equation in standard form and read the coefficients with their signs. A common exam pitfall is misreading a negative coefficient.
| Equation (as given) | Standard form | \(a\) | \(b\) | \(c\) |
|---|---|---|---|---|
| \(3x^2 - 2x + 5 = 0\) | \(3x^2 - 2x + 5 = 0\) | 3 | -2 | 5 |
| \(x^2 = 7\) | \(x^2 - 7 = 0\) | 1 | 0 | -7 |
| \(4x - x^2 = -1\) | \(-x^2 + 4x + 1 = 0\) | -1 | 4 | 1 |
| \(2x(x - 3) = 0\) | \(2x^2 - 6x = 0\) | 2 | -6 | 0 |
| \((x + 1)^2 = 4\) | \(x^2 + 2x - 3 = 0\) | 1 | 2 | -3 |
| \(\dfrac{x^2}{2} + 3x = 1\) | \(\dfrac{1}{2}x^2 + 3x - 1 = 0\) | \(\dfrac{1}{2}\) | 3 | -1 |
Multiplying an equation by a non-zero constant does not change its roots, so \(x^2 + 6x - 2 = 0\) and \(2x^2 + 12x - 4 = 0\) have exactly the same solutions.
6. Roots and Solutions
A root (or solution) of the quadratic equation \(ax^2 + bx + c = 0\) is a value of \(x\) that satisfies the equation. Putting the value into the left-hand side must give 0.
\(x = k\) is a root of \(ax^2 + bx + c = 0\) if and only if \(ak^2 + bk + c = 0\).
A quadratic equation can have at most two real roots. Depending on the coefficients it has:
- two distinct real roots, or
- two equal real roots (one distinct value repeated), or
- no real roots.
Which of these happens is decided by the discriminant, introduced in Section 10.
7. Verifying a Root
To check whether \(x = k\) is a root, substitute \(k\) into the left-hand side of the standard form. If the result is 0, it is a root.
Check whether \(x = -2\) and \(x = 3\) are roots of \(x^2 - x - 6 = 0\).
For \(x = -2\): \((-2)^2 - (-2) - 6 = 4 + 2 - 6 = 0\) ✓
For \(x = 3\): \((3)^2 - 3 - 6 = 9 - 3 - 6 = 0\) ✓
Both values are roots. Indeed \((x + 2)(x - 3) = x^2 - x - 6\).
Verification alone does not prove you have found all roots. It proves the values you tested are correct. Solving the equation systematically shows the full set.
8. Solving by Factorisation
The most reliable first method is factorisation: rewrite the quadratic as a product of two linear factors, then use the zero product rule.
If \(p \times q = 0\), then \(p = 0\) or \(q = 0\) (or both).
Solve \(x^2 - 5x + 6 = 0\).
Step 1: Standard form done: \(a = 1\), \(b = -5\), \(c = 6\).
Step 2: Need \(m + n = -5\), \(mn = 6 \implies m = -2, n = -3\).
Step 3–4: \(x^2 - 2x - 3x + 6 = x(x - 2) - 3(x - 2) = (x - 2)(x - 3)\).
Step 5: \((x - 2)(x - 3) = 0 \implies x = 2\) or \(x = 3\).
Verify: \(4 - 10 + 6 = 0\) ✓ and \(9 - 15 + 6 = 0\) ✓. So the roots are \(x = 2\) and \(x = 3\).
Solve \(2x^2 + 7x + 3 = 0\).
Step 2: \(a \times c = 2 \times 3 = 6\), and \(b = 7\). Find \(m + n = 7\), \(mn = 6 \implies m = 1, n = 6\).
Step 3–4: \(2x^2 + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)\).
Step 5: \((2x + 1)(x + 3) = 0 \implies x = -\dfrac{1}{2}\) or \(x = -3\).
Verify: \(2\left(\frac{1}{4}\right) - \frac{7}{2} + 3 = \frac{1}{2} - \frac{7}{2} + 3 = 0\) ✓; \(18 - 21 + 3 = 0\) ✓.
Solve \(x(x - 3) = 10\).
Step 1: \(x^2 - 3x - 10 = 0\), so \(a = 1, b = -3, c = -10\).
Step 2: \(m + n = -3\), \(mn = -10 \implies m = -5, n = 2\).
Step 3–4: \(x^2 - 5x + 2x - 10 = x(x - 5) + 2(x - 5) = (x - 5)(x + 2)\).
Step 5: \(x = 5\) or \(x = -2\).
Verify: \(5(2) = 10\) ✓; \((-2)(-5) = 10\) ✓.
It works quickly when the two numbers \(m\) and \(n\) are easy to spot, i.e. when the roots are rational. When the numbers are not obvious, switch to the quadratic formula.
9. Solving with the Quadratic Formula
Every quadratic equation \(ax^2 + bx + c = 0\) (with \(a \neq 0\)) can be solved by the quadratic formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Solve \(x^2 - 4x - 5 = 0\).
Here \(a = 1\), \(b = -4\), \(c = -5\). Discriminant \(D = (-4)^2 - 4(1)(-5) = 16 + 20 = 36\).
\[x = \frac{-(-4) \pm \sqrt{36}}{2(1)} = \frac{4 \pm 6}{2}\]
So \(x = \dfrac{4 + 6}{2} = 5\) and \(x = \dfrac{4 - 6}{2} = -1\).
Verify: \(25 - 20 - 5 = 0\) ✓; \(1 + 4 - 5 = 0\) ✓.
Solve \(2x^2 - 5x + 2 = 0\). Here \(a = 2\), \(b = -5\), \(c = 2\).
\(D = 25 - 16 = 9\).
\[x = \frac{5 \pm 3}{4}\]
So \(x = \dfrac{5 + 3}{4} = 2\) and \(x = \dfrac{5 - 3}{4} = \dfrac{1}{2}\).
Solve \(x^2 + 6x + 6 = 0\). Here \(a = 1\), \(b = 6\), \(c = 6\).
\(D = 36 - 24 = 12\).
\[x = \frac{-6 \pm \sqrt{12}}{2} = \frac{-6 \pm 2\sqrt{3}}{2} = -3 \pm \sqrt{3}\]
So the two roots are \(-3 + \sqrt{3}\) and \(-3 - \sqrt{3}\).
Do not drop the \(\pm\) symbol, and do not forget the factor \(2a\) in the denominator. Write \(b\) with its sign: if \(b = -4\), then \(-b = 4\).
10. The Discriminant
The expression under the square root in the formula has its own name and role.
For \(ax^2 + bx + c = 0\), the discriminant is \(D = b^2 - 4ac\).
Without solving the equation, the discriminant tells you what kind of roots to expect:
| Discriminant | Nature of roots |
|---|---|
| \(D > 0\) | Two distinct real roots |
| \(D = 0\) | Two equal real roots (one real root, repeated) |
| \(D < 0\) | No real roots |
\(2x^2 - 7x + 3 = 0\): \(D = 49 - 24 = 25 > 0\), so there are two distinct real roots. Indeed \(x = \dfrac{7 \pm 5}{4} = 3, \dfrac{1}{2}\).
\(x^2 - 6x + 9 = 0\): \(D = 36 - 36 = 0\), so there are two equal real roots: \(x = \dfrac{6}{2} = 3\) (repeated). Note \(x^2 - 6x + 9 = (x - 3)^2\).
\(3x^2 + 2x + 5 = 0\): \(D = 4 - 60 = -56 < 0\), so the equation has no real roots.
For what value of \(k\) does \(x^2 + 2x + k = 0\) have equal roots?
Equal roots \(\iff D = 0\), i.e. \(4 - 4k = 0 \implies k = 1\).
Suppose two equations have the same coefficients except one has \(c = 2\) and the other \(c = -2\). The one with \(c = -2\) has \(D = b^2 + 8 > 0\), so it always has two distinct real roots. You can answer such questions purely from the sign of \(D\).
11. Nature of Roots — Summary
The table below summarises everything about the nature of roots in one place.
| Value of \(D = b^2 - 4ac\) | Nature of roots | Example | Roots |
|---|---|---|---|
| \(D > 0\) and a perfect square | Two distinct rational real roots | \(x^2 - 5x + 6 = 0\), \(D = 1\) | \(2, 3\) |
| \(D > 0\) but not a perfect square | Two distinct irrational real roots | \(x^2 + 6x + 6 = 0\), \(D = 12\) | \(-3 \pm \sqrt{3}\) |
| \(D = 0\) | Two equal real roots (repeated) | \((x - 3)^2 = 0\) | \(3, 3\) |
| \(D < 0\) | No real roots | \(x^2 + 2 = 0\) | none (in reals) |
When a standard-form equation is a perfect square like \((x - 3)^2\), the two roots coincide. Exam questions sometimes word this as "equal roots" or "one root repeated twice".
12. Unknown Parameters \(k\), \(p\), \(m\)
A parameter is an unknown number inside the coefficients. The discriminant links it to the nature of roots, and substitution links it to a given root.
If \(x = 2\) is a root of \(x^2 + px - 6 = 0\), find \(p\).
Substitute \(x = 2\): \(4 + 2p - 6 = 0 \implies 2p = 2 \implies p = 1\).
So the equation is \(x^2 + x - 6 = 0 = (x + 3)(x - 2)\), confirming 2 is a root and giving the other root \(-3\).
For which \(k\) does \(kx^2 + 6x + 1 = 0\) have equal roots?
Equal roots \(\iff D = 0\): \(36 - 4k = 0 \implies k = 9\). The equation is \(9x^2 + 6x + 1 = (3x + 1)^2\), root \(x = -\dfrac{1}{3}\) (repeated).
For which \(p\) does \(px^2 - 2x + 1 = 0\) have two distinct real roots?
Need \(a \neq 0\) and \(D > 0\): \(4 - 4p > 0 \implies p < 1\), and \(p \neq 0\). So \(p < 1\), \(p \neq 0\).
For which \(m\) does \(2x^2 + mx + 8 = 0\) have no real roots?
Need \(D < 0\): \(m^2 - 64 < 0 \implies m^2 < 64 \implies -8 < m < 8\).
13. Application Word Problems
Word problems ask you to frame the equation first. The 4-step plan below works for every such problem.
The product of two consecutive positive integers is 306. Let the integers be \(x\) and \(x + 1\). Then \(x(x+1) = 306\), i.e. \(x^2 + x - 306 = 0\). Factorising: \((x + 18)(x - 17) = 0 \implies x = 17\) (we reject \(-18\) as negative). The integers are 17 and 18. Check: \(17 \times 18 = 306\) ✓.
A garden is 4 m longer than it is wide. Let breadth be \(x\) m; length is \(x + 4\) m. Area \(= x(x+4) = 96\), so \(x^2 + 4x - 96 = 0\). Factorising: \((x + 12)(x - 8) = 0 \implies x = 8\) (reject \(-12\)). The garden is 8 m by 12 m. Check: \(8 \times 12 = 96\) ✓.
A train travels 200 km at a uniform speed. If the speed had been 10 km/h more, the journey would have taken 1 hour less. Let the speed be \(x\) km/h. Time at speed \(x\) is \(\dfrac{200}{x}\), at \(x+10\) it is \(\dfrac{200}{x+10}\). Given \(\dfrac{200}{x} - \dfrac{200}{x+10} = 1\). Multiplying by \(x(x+10)\): \(200(x+10) - 200x = x(x+10)\), so \(x^2 + 10x - 2000 = 0\). Factorising: \((x + 50)(x - 40) = 0 \implies x = 40\) km/h (reject \(-50\)). Speed = 40 km/h. Check: \(\frac{200}{40} = 5\) h, \(\frac{200}{50} = 4\) h, difference 1 h ✓.
The hypotenuse of a right triangle is 13 cm and the difference between the other two sides is 7 cm. Let the shorter side be \(x\) cm; the other is \(x + 7\) cm. By Pythagoras: \(x^2 + (x+7)^2 = 13^2\). Expanding: \(x^2 + x^2 + 14x + 49 = 169\), so \(2x^2 + 14x - 120 = 0\), i.e. \(x^2 + 7x - 60 = 0\). Factorising: \((x + 12)(x - 5) = 0 \implies x = 5\). Sides are 5 cm and 12 cm. Check: \(5^2 + 12^2 = 25 + 144 = 169 = 13^2\) ✓.
A ball thrown upward has height \(h = 25t - 5t^2\) metres after \(t\) seconds. On the ground, \(h = 0\): \(25t - 5t^2 = 0 \implies 5t(5 - t) = 0\). So \(t = 0\) (when thrown) or \(t = 5\). The ball returns to the ground after 5 seconds. Check: \(25(5) - 5(25) = 125 - 125 = 0\) ✓.
The sum of the squares of two consecutive positive integers is 365. Let them be \(n\) and \(n+1\). Then \(n^2 + (n+1)^2 = 365\), so \(n^2 + n^2 + 2n + 1 = 365\), giving \(2n^2 + 2n - 364 = 0\), i.e. \(n^2 + n - 182 = 0\). Factorising: \((n + 14)(n - 13) = 0 \implies n = 13\). The integers are 13 and 14. Check: \(169 + 196 = 365\) ✓.
A rectangular park has area 240 m². Its length is 4 m more than twice its breadth. Let breadth be \(b\) m; length is \(2b + 4\) m. Then \(b(2b+4) = 240\), so \(2b^2 + 4b - 240 = 0\), i.e. \(b^2 + 2b - 120 = 0\). Factorising: \((b + 12)(b - 10) = 0 \implies b = 10\). Breadth is 10 m, length \(2(10)+4 = 24\) m. Check: \(10 \times 24 = 240\) ✓.
A rectangular garden has perimeter 64 m and area 240 m². Let length be \(l\) m and breadth \(b\) m. Perimeter \(2(l+b) = 64 \implies l + b = 32\), and \(lb = 240\). So \(l(32 - l) = 240 \implies 32l - l^2 = 240 \implies l^2 - 32l + 240 = 0\). Factorising: \((l - 20)(l - 12) = 0 \implies l = 20\) (or \(12\)). The garden is 20 m by 12 m. If fencing costs ₹12 per metre, cost \(= 64 \times 12 =\) ₹768. Check: \(2(32) = 64\), \(20 \times 12 = 240\) ✓.
14. Competency and Reasoning Questions
Competency questions test whether you understand why a method works, not just whether you can run it. They often hide a trap or ask for a choice of method.
From the following, pick the equation that is not quadratic: \(x^2 + 2x + 1 = 0\), \((x+1)(x-1) = 0\), \(x(x+3) = x^2 + 7\).
Reasoning. The third: \(x(x+3) = x^2 + 7\) gives \(x^2 + 3x = x^2 + 7\), so \(3x = 7\) — the \(x^2\) terms cancel. Seeing this without expanding wastes ten seconds; expanding still proves it.
Solve \(x^2 - 49 = 0\) in the fastest way.
Reasoning. Differences of squares: \(x^2 - 49 = (x - 7)(x + 7) = 0\), so \(x = \pm 7\). The quadratic formula would also work but is slower here.
For what value of \(k\) does \((k-1)x^2 + 2x - 8 = 0\) have two distinct real roots?
Reasoning. Coarse conditions: \((k-1) \neq 0 \implies k \neq 1\), and \(D = 4 + 32(k-1) > 0 \implies 32k - 28 > 0 \implies k > \frac{7}{8}\). Together, \(k > \frac{7}{8}\) with \(k \neq 1\).
A student solves \(x^2 - 5x + 6 = 0\), factors as \((x - 2)(x - 3)\), then writes roots \(-2, -3\). Where is the error?
Reasoning. The factorisation is right, but the roots are wrong. \((x - 2) = 0\) gives \(x = +2\), not \(-2\); \((x - 3) = 0\) gives \(x = +3\). Correct roots: 2 and 3. Substitution confirms: \(4 - 10 + 6 = 0\), \(9 - 15 + 6 = 0\).
Can \(x^2 + 1 = 0\) be solved in the real numbers?
Reasoning. Rewriting gives \(x^2 = -1\). A square of a real number is never negative, so there is no real solution. Using \(D = 0 - 4 = -4 < 0\) gives the same answer immediately: no real roots.
One root of \(x^2 - 9x + 20 = 0\) is 4. Find the other root without dividing polynomials.
Reasoning. The linear factor for root 4 is \((x - 4)\). Factorising the whole quadratic: \(x^2 - 9x + 20 = (x - 4)(x - 5)\), so the other root is 5. Verify: \(25 - 45 + 20 = 0\) ✓.
15. Case-Based Examples
Case-based questions describe a situation and then ask 2–3 related sub-questions. Answer the sub-questions using the case data only, and show all working.
Case Study 1 — School Ground
A school ground is 12 m longer than it is wide, and its area is 405 m². The school plans to fence the ground all around.
Question 1
Taking the breadth as \(x\) m, form the quadratic equation for the area of the ground.
Question 2
Solve the equation to find the breadth and length of the ground.
Question 3
Find the cost of fencing the ground at ₹45 per metre.
Solution. (1) Length \(= x + 12\), area \(= x(x + 12)\). So the equation is \(x^2 + 12x - 405 = 0\).
(2) Factorising: \((x + 27)(x - 15) = 0\), so \(x = 15\) (reject \(-27\)). Breadth = 15 m, Length = 27 m. Check \(15 \times 27 = 405\) ✓.
(3) Perimeter \(= 2(15 + 27) = 84\) m. Cost \(= 84 \times 45 =\) ₹3780.
Case Study 2 — Study Tour
A class plans a tour. A coach charges a total of ₹2400 for the trip. If 4 more students join, the cost per student decreases by ₹20.
Question 1
Let the number of students be \(n\). Write the equation relating the cost per student before and after 4 more students join.
Question 2
Find the number of students on the trip.
Question 3
Find the cost per student originally and after the extra students join, and verify the difference.
Solution. (1) Original cost per student \(= \dfrac{2400}{n}\); after \(n+4\) students it is \(\dfrac{2400}{n+4}\). Given \(\dfrac{2400}{n} - \dfrac{2400}{n+4} = 20\). Multiplying by \(n(n+4)\): \(2400(n+4) - 2400n = 20n(n+4)\), so \(9600 = 20n^2 + 80n\), i.e. \(n^2 + 4n - 480 = 0\).
(2) Factorising: \((n + 24)(n - 20) = 0\), so \(n = 20\) (reject \(-24\)). 20 students.
(3) Original: \(\dfrac{2400}{20} = \) ₹120 each; with 24 students: \(\dfrac{2400}{24} = \) ₹100 each. Difference \(= 20\) ✓.
Case Study 3 — Consecutive Odd Numbers
The product of two consecutive odd positive integers is 323.
Question 1
Let the smaller integer be \(x\). Form the quadratic equation.
Question 2
Find the two integers.
Question 3
Verify by substitution that both integers satisfy the product condition.
Solution. (1) Consecutive odd integers are \(x\) and \(x + 2\). Product: \(x(x+2) = 323\), so \(x^2 + 2x - 323 = 0\).
(2) Factorising: \((x + 19)(x - 17) = 0\), so \(x = 17\) (reject \(-19\)). Integers are 17 and 19.
(3) \(x = 17\): \(17(19) = 323\) ✓. The factorisation itself guarantees both factors satisfy the equation.
Case Study 4 — Fair Tickets
A fair charges ₹\(x\) per ticket. The number of visitors that evening is \((400 - 2x)\), so the revenue is \(R(x) = x(400 - 2x)\).
Question 1
Express \(R(x)\) in standard form \(ax^2 + bx + c\) and write \(a\), \(b\), \(c\).
Question 2
For which ticket prices is the revenue zero?
Question 3
Find the revenue when the ticket price is ₹50.
Solution. (1) \(R(x) = -2x^2 + 400x\), so \(a = -2\), \(b = 400\), \(c = 0\).
(2) \(x(400 - 2x) = 0 \implies x = 0\) (free entry, no revenue) or \(x = 200\). Revenue is zero at ₹0 and ₹200.
(3) \(R(50) = 50(400 - 100) = 50 \times 300 = \) ₹15,000.
Case Study 5 — Lawn and Path
A rectangular garden is 40 m by 30 m. A path of uniform width \(x\) m is cut around the inside edge, leaving a lawn of area 600 m² in the middle.
Question 1
Write the dimensions of the lawn in terms of \(x\), and form the quadratic equation.
Question 2
Solve to find the width of the path.
Question 3
The equation has two solutions. Why is one of them rejected?
Solution. (1) Lawn dimensions: \((40 - 2x)\) and \((30 - 2x)\). Area: \((40-2x)(30-2x) = 600\). Expanding: \(1200 - 80x - 60x + 4x^2 = 600\), so \(4x^2 - 140x + 600 = 0\), i.e. \(x^2 - 35x + 150 = 0\).
(2) Factorising: \((x - 30)(x - 5) = 0 \implies x = 5\) (reject \(30\)). Path width = 5 m.
(3) \(x = 30\) is rejected because a path of 30 m on each side would exceed half of the 30 m breadth, giving a negative lawn dimension. Only \(x = 5\) satisfies the geometry. Check: \((40-10)(30-10) = 30 \times 20 = 600\) ✓.
16. Important Formulae
\(ax^2 + bx + c = 0\), where \(a \neq 0\).
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
\[D = b^2 - 4ac\]
\(D > 0\): two distinct real roots; \(D = 0\): two equal real roots; \(D < 0\): no real roots.
Identify \(a\), \(b\), \(c\) correctly before applying the formula. Write the equation in standard form first, keep the signs of \(b\) and \(c\), and never drop the \(2a\) in the denominator.
17. Common Examination Mistakes
| Common mistake | Correct method |
|---|---|
| Not writing the equation in standard form before reading \(a\), \(b\), \(c\) | Simplify to \(ax^2 + bx + c = 0\) first, then read the coefficients with their signs |
| Stating \(a = 0\) or ignoring that condition | If \(a = 0\), the equation is linear, not quadratic |
| Dropping the \(\pm\) sign in the formula | Always compute both roots: \(x = \dfrac{-b + \sqrt{D}}{2a}\) and \(x = \dfrac{-b - \sqrt{D}}{2a}\) |
| Wrong sign substitution, e.g. using \(b = -4\) as \(b = +4\) | Substitute with signs: \(-b\) means the negative of \(b\), so \(b = -4\) gives \(-b = 4\) |
| Splitting the middle term using product \(c\) instead of \(ac\) | For \(ax^2 + bx + c\), find two numbers with product \(ac\) and sum \(b\) |
| Saying "no solution" for \(x^2 - 9 = 0\) | It has two real roots: \(x = \pm 3\). \(D \geq 0\) means real roots exist |
| Keeping a negative root as the answer in a word problem | Discard roots that make no sense (negative length, time, count), but state why you reject them |
| Forgetting to verify the found roots | Substitute each root back into the equation and check it gives 0 |
| Confusing "equal roots" with "no roots" | Equal roots still exist (they are real); write the repeated root once or twice as asked |
| Not simplifying, e.g. leaving \(\dfrac{-6 \pm \sqrt{48}}{2}\) | Simplify radicals: \(\sqrt{48} = 4\sqrt{3}\), then reduce the fraction |
For \(2x^2 - 7x + 3 = 0\), the middle term split needs numbers with product \(2 \times 3 = 6\) and sum \(-7\), so they are \(-1\) and \(-6\): \(2x^2 - x - 6x + 3 = (2x - 1)(x - 3)\). Correct roots: \(x = \dfrac{1}{2}\) and \(x = 3\).
18. CBSE Exam Tips
- Read whether the question wants real roots, integer roots or roots "as decimals" — this decides how you present the answer.
- Prefer factorisation when the numbers split easily; otherwise go straight to the quadratic formula. Do not waste time hunting for factors when \(a\), \(b\), \(c\) are large.
- Write the standard form before touching the formula, and show the substitution step — it carries marks even if the arithmetic slips.
- Compute the discriminant \(D\) before solving; it tells you whether roots exist and whether to expect a neat or radical form.
- In word problems, state what your variable represents (e.g. "let breadth = \(x\) m") and label the final answer with units.
- Always reject and explain impossible roots such as negative lengths, times or counts.
- Verify every root by substitution if time permits — it catches most sign errors.
- For case-based sections, read the whole case once before answering any sub-question.
19. Quick Revision
| Concept | Key point |
|---|---|
| Quadratic equation | \(ax^2 + bx + c = 0\), \(a \neq 0\) |
| Root | A value of \(x\) satisfying the equation |
| Maximum number of real roots | Two (equal or distinct) |
| Zero product rule | \(p \cdot q = 0 \implies p = 0\) or \(q = 0\) |
| Quadratic formula | \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) |
| Discriminant | \(D = b^2 - 4ac\) |
| \(D > 0\) | Two distinct real roots |
| \(D = 0\) | Two equal real roots |
| \(D < 0\) | No real roots |
| Word problems | Frame the equation, solve, reject impossible roots, verify |
20. Self-Practice Questions
Thirty original questions written in the CBSE exam style. Answers and solutions are in Section 21.
Conceptual
Q1.
Write \(3x^2 + 5x - 2x^2 + 7\) in standard form and state \(a\), \(b\), \(c\).
Q2.
Which of the following is not a quadratic equation? \(x^2 = 4\); \(x(x+2) = 0\); \((x+1)^2 = x^2 + 3\).
Q3.
In \(ax^2 + bx + c = 0\), why must \(a\) not be zero?
Q4.
What information does the discriminant \(D = b^2 - 4ac\) give without solving the equation?
Q5.
Does \(x^2 + 4 = 0\) have real roots? Justify.
Q6.
Verify by substitution that 2 and 3 are roots of \(x^2 - 5x + 6 = 0\).
Standard — solve
Q7.
Solve \(x^2 - 7x + 12 = 0\).
Q8.
Solve \(x^2 + 6x + 8 = 0\).
Q9.
Solve \(x^2 - 4 = 0\).
Q10.
Solve \(x^2 + 7x + 10 = 0\).
Q11.
Solve \(6x^2 + 5x - 6 = 0\).
Q12.
Solve \(2x^2 - 8x + 8 = 0\).
Q13.
Solve \(4x^2 - 4x + 1 = 0\).
Q14.
Solve \(x^2 + 5x + 6 = 0\).
Using the quadratic formula
Q15.
Solve \(9x^2 - 16 = 0\).
Q16.
Solve \(x^2 - 3x - 4 = 0\) using the quadratic formula.
Q17.
Solve \(2x^2 + x - 6 = 0\).
Q18.
Find the nature of the roots of \(x^2 + 6x + 9 = 0\) and then solve it.
Q19.
Solve \(3x^2 - 5x + 2 = 0\).
Q20.
Solve \(2x^2 - 7x + 3 = 0\).
Application based
Q21.
The product of two consecutive positive integers is 240. Form a quadratic equation and find the integers.
Q22.
A rectangular ground has breadth \(x\) m. Its length is 3 m more than the breadth and the area is 108 m². Find the dimensions.
Q23.
A ball, thrown upward, is at height \(h = 20t - 5t^2\) metres after \(t\) seconds. When does it return to the ground?
Q24.
The sum of the squares of two consecutive positive integers is 61. Find the integers.
Q25.
A train travels 180 km. If its speed were 15 km/h more, the journey would take 1 hour less. Find the speed of the train.
Competency / Reasoning
Q26.
A student claims the roots of \(x^2 - 7x + 12 = 0\) are \(-3\) and \(-4\). Is the claim correct? Explain.
Q27.
For what value of \(k\) does \(kx^2 + 2x + 1 = 0\) have equal roots?
Q28.
For the equation \(x^2 - 2x + c = 0\), the discriminant is \(D = 4 - 4c\). For which values of \(c\) does it have two distinct real roots?
Case based
Q29 — Garden fencing.
A rectangular garden has length 4 m more than its breadth, and an area of 96 m². (i) Taking breadth \(= x\), form the quadratic equation. (ii) Find the dimensions of the garden. (iii) Find the cost of fencing its boundary at ₹25 per metre.
Q30 — Fair stall.
At a school fair a stall charges ₹\(x\) per ride. The number of visitors is \((300 - 3x)\), so the revenue is \(R = x(300 - 3x)\). (i) Express \(R\) as \(ax^2 + bx + c\). (ii) At which ticket prices is the revenue zero? (iii) Find the revenue when the ticket price is ₹20.
21. Answers and Solutions
A1.
\(x^2 + 5x + 7 = 0\), with \(a = 1\), \(b = 5\), \(c = 7\).
A2.
\((x+1)^2 = x^2 + 3\) is not quadratic: expanding gives \(x^2 + 2x + 1 = x^2 + 3 \Rightarrow 2x = 2\), a linear equation.
A3.
If \(a = 0\), the \(x^2\) term disappears and the equation becomes linear (\(bx + c = 0\)), so it is no longer quadratic.
A4.
It tells whether the roots are real and whether they are equal: \(D > 0\) two distinct real, \(D = 0\) two equal real, \(D < 0\) no real roots.
A5.
No. \(D = 0 - 16 = -16 < 0\), so there are no real roots.
A6.
For \(x = 2\): \(4 - 10 + 6 = 0\) ✓. For \(x = 3\): \(9 - 15 + 6 = 0\) ✓.
A7.
\((x-3)(x-4) = 0 \Rightarrow x = 3\) or \(x = 4\).
A8.
\((x+2)(x+4) = 0 \Rightarrow x = -2\) or \(x = -4\).
A9.
\((x-2)(x+2) = 0 \Rightarrow x = \pm 2\).
A10.
\((x+2)(x+5) = 0 \Rightarrow x = -2\) or \(x = -5\).
A11.
\((3x-2)(2x+3) = 0 \Rightarrow x = \dfrac{2}{3}\) or \(x = -\dfrac{3}{2}\).
A12.
\(2(x-2)^2 = 0 \Rightarrow x = 2\) (repeated).
A13.
\((2x-1)^2 = 0 \Rightarrow x = \dfrac{1}{2}\) (repeated).
A14.
\((x+2)(x+3) = 0 \Rightarrow x = -2\) or \(x = -3\).
A15.
\((3x-4)(3x+4) = 0 \Rightarrow x = \pm \dfrac{4}{3}\).
A16.
\(D = 9 + 16 = 25\); \(x = \dfrac{3 \pm 5}{2} \Rightarrow x = 4\) or \(x = -1\).
A17.
\((2x-3)(x+2) = 0 \Rightarrow x = \dfrac{3}{2}\) or \(x = -2\).
A18.
\(D = 36 - 36 = 0\): two equal real roots. \((x+3)^2 = 0 \Rightarrow x = -3\) (repeated).
A19.
\((3x-2)(x-1) = 0 \Rightarrow x = \dfrac{2}{3}\) or \(x = 1\).
A20.
\((2x-1)(x-3) = 0 \Rightarrow x = \dfrac{1}{2}\) or \(x = 3\).
A21.
\(x(x+1) = 240 \Rightarrow x^2 + x - 240 = 0 \Rightarrow (x+16)(x-15) = 0\). Positive \(x = 15\); the integers are 15 and 16.
A22.
\(x(x+3) = 108 \Rightarrow x^2 + 3x - 108 = 0 \Rightarrow (x+12)(x-9) = 0 \Rightarrow x = 9\). Dimensions: breadth 9 m, length 12 m.
A23.
\(20t - 5t^2 = 0 \Rightarrow 5t(4-t) = 0 \Rightarrow t = 0\) or \(t = 4\). It returns to the ground after 4 seconds.
A24.
\(n^2 + (n+1)^2 = 61 \Rightarrow n^2 + n - 30 = 0 \Rightarrow (n+6)(n-5) = 0 \Rightarrow n = 5\). The integers are 5 and 6.
A25.
\(\dfrac{180}{x} - \dfrac{180}{x+15} = 1 \Rightarrow x^2 + 15x - 2700 = 0 \Rightarrow (x+60)(x-45) = 0 \Rightarrow x = 45\). Speed = 45 km/h.
A26.
Incorrect. \((x-3)(x-4) = 0\) gives \(x = +3\) and \(x = +4\). The claimed roots \(-3, -4\) would make the product of factors positive and the signs of the middle terms flip.
A27.
Equal roots \(\iff D = 0\): \(4 - 4k = 0 \Rightarrow k = 1\).
A28.
Two distinct real roots \(\iff D > 0\): \(4 - 4c > 0 \Rightarrow c < 1\), and the equation is quadratic for any real \(c\).
A29.
(i) \(x(x+4) = 96\), i.e. \(x^2 + 4x - 96 = 0\). (ii) \((x+12)(x-8) = 0 \Rightarrow x = 8\): breadth 8 m, length 12 m. (iii) Perimeter \(= 2(8+12) = 40\) m; cost \(= 40 \times 25 =\) ₹1000.
A30.
(i) \(R = -3x^2 + 300x\), so \(a = -3\), \(b = 300\), \(c = 0\). (ii) \(x(300 - 3x) = 0 \Rightarrow x = 0\) or \(x = 100\). (iii) \(R(20) = 20(300 - 60) = 20 \times 240 =\) ₹4800.