Quadratic Equations takes the degree-2 polynomial ax² + bx + c from Chapter 2, sets it equal to zero, and finds the values of x that satisfy it. The chapter covers the standard form, turning word situations into equations, roots, solving by factorisation, the quadratic formula, and the discriminant b² − 4ac that decides whether the roots are distinct, equal or not real. Board papers test it through discriminant and equal-roots items, factorisation and formula questions, and word problems on area, age, speed and work where an impossible root has to be rejected.
Key Concepts
1. What Is a Quadratic Equation
In Chapter 2 you met the quadratic polynomial ax2 + bx + c, a polynomial of degree 2. When this polynomial is set equal to zero, the result is a quadratic equation.
A quadratic equation in the variable x is an equation of the form
ax2 + bx + c = 0, where a, b, c are real numbers and a ≠ 0
This arrangement, with the terms written in descending order of their degrees, is called the standard form of a quadratic equation. More generally, any equation p(x) = 0 where p(x) is a polynomial of degree 2 is a quadratic equation.
- 2x2 + x − 300 = 0 is quadratic with a = 2, b = 1, c = −300.
- 4x − 3x2 + 2 = 0 is quadratic. In standard form it reads −3x2 + 4x + 2 = 0, so a = −3, b = 4, c = 2.
- x2 − 9 = 0 (b = 0) and 5x2 + 2x = 0 (c = 0) are also quadratic.
Why a ≠ 0 matters. If a = 0, the x2 term disappears and what is left, bx + c = 0, is a linear equation. The coefficients b and c are allowed to be zero; a is never allowed to be zero.
Simplify First, Then Decide
An equation may look quadratic and turn out linear, or look cubic and turn out quadratic. Expand everything, bring all terms to one side and look at the highest power of x that survives. In NCERT Example 2, x(x + 1) + 8 = (x + 2)(x − 2) simplifies to x + 12 = 0 (linear, the x2 terms cancel), while (x + 2)3 = x3 − 4 simplifies to x2 + 2x + 2 = 0 (quadratic, the x3 terms cancel). The rule: the x2 term must survive and higher powers must vanish.
2. Representing Situations as Quadratic Equations
Many real problems about area, products of numbers, ages, cost and speed lead to quadratic equations. The skill tested here is translation: turning a paragraph into one equation in one unknown.
A four-step recipe
- Choose one unknown and say clearly what it stands for, with units: “Let the breadth of the hall be x metres.”
- Write every other quantity in terms of x: length = (2x + 1) m.
- Use the one fact that has not been used yet to form the equation: area = 300 m2, so x(2x + 1) = 300.
- Rearrange into standard form: 2x2 + x − 300 = 0.
This is the prayer hall problem from the start of the NCERT chapter: a hall of carpet area 300 m2 whose length is one metre more than twice its breadth.
Standard Situation Types
| Situation | Let | Equation formed |
|---|---|---|
| Two consecutive positive integers with product 306 | smaller = x | x(x + 1) = 306, so x2 + x − 306 = 0 |
| Rohan’s mother is 26 years older; product of their ages 3 years from now is 360 | Rohan’s age = x years | (x + 3)(x + 29) = 360, so x2 + 32x − 273 = 0 |
| Train covers 480 km; if speed were 8 km/h less it would take 3 hours more | speed = x km/h | 480/(x − 8) − 480/x = 3, so x2 − 8x − 1280 = 0 |
| John and Jivanti have 45 marbles; each loses 5; product of what is left is 124 | John’s marbles = x | (x − 5)(40 − x) = 124, so x2 − 45x + 324 = 0 |
The speed equation in detail. Time = distance/speed. At x km/h the train takes 480/x hours; at (x − 8) km/h it takes 480/(x − 8) hours, which is 3 hours more. So 480/(x − 8) − 480/x = 3. Taking the LCM on the left, 480[x − (x − 8)] / [x(x − 8)] = 3, that is 3840 = 3x(x − 8), or 3x2 − 24x − 3840 = 0. Dividing by 3 gives x2 − 8x − 1280 = 0.
When a question says only “represent the situation”, stop at the standard form. When it says “find”, carry on and solve.
3. Roots of a Quadratic Equation
A real number α is called a root of the quadratic equation ax2 + bx + c = 0 (a ≠ 0) if
aα2 + bα + c = 0
We also say that x = α is a solution of the equation, or that α satisfies the equation. For example, putting x = 1 in 2x2 − 3x + 1 gives 2 − 3 + 1 = 0, so 1 is a root of 2x2 − 3x + 1 = 0.
Roots and zeroes are the same numbers. The zeroes of the quadratic polynomial ax2 + bx + c and the roots of the quadratic equation ax2 + bx + c = 0 are identical. Saying “1 is a root of 2x2 − 3x + 1 = 0” and “1 is a zero of 2x2 − 3x + 1” means the same thing.
At most two roots. A quadratic polynomial has at most two zeroes (Chapter 2), so a quadratic equation has at most two roots. It can have two different real roots, one real root repeated twice, or no real roots at all.
Using a Given Root
A very common short question gives one root and an unknown coefficient. Substitute the root and solve for the coefficient.
Illustration: If x = 2 is a root of x2 + kx − 6 = 0, find k and the other root.
Put x = 2: 4 + 2k − 6 = 0, so 2k = 2 and k = 1. The equation is x2 + x − 6 = 0 = (x + 3)(x − 2), so the other root is −3.
4. Solving by Factorisation
If the quadratic ax2 + bx + c can be written as a product of two linear factors, the roots come from setting each factor equal to zero. This works because of the zero product rule: if the product of two numbers is 0, at least one of them must be 0.
Splitting the Middle Term
You learnt this in Class 9. To factorise ax2 + bx + c:
- Multiply a and c.
- Find two numbers whose product is a × c and whose sum is b.
- Rewrite bx as the sum of two terms using those numbers.
- Group in pairs and take out the common factor from each pair.
- Take out the common bracket to get two linear factors.
Illustration (NCERT Example 3): 2x2 − 5x + 3 = 0.
a × c = 2 × 3 = 6 and b = −5. The pair −2 and −3 has product 6 and sum −5.
2x2 − 2x − 3x + 3 = 2x(x − 1) − 3(x − 1) = (2x − 3)(x − 1)
So 2x − 3 = 0 or x − 1 = 0, giving x = 3/2 or x = 1.
Repeated Roots
Sometimes both factors are the same. NCERT Example 5 solves 3x2 − 2√6 x + 2 = 0. Here a × c = 6, and the split is −√6 x − √6 x:
3x2 − √6 x − √6 x + 2 = √3 x(√3 x − √2) − √2(√3 x − √2) = (√3 x − √2)(√3 x − √2)
The factor √3 x − √2 = 0 gives x = √2/√3 = √(2/3). The root is repeated twice, once for each factor, so the roots are √(2/3), √(2/3).
A perfect square such as 100x2 − 20x + 1 = (10x − 1)2 behaves the same way: roots 1/10, 1/10.
Surds and fractions. For √2 x2 + 7x + 5√2 = 0, a × c = 10 and 5 + 2 = 7, so split 7x as 5x + 2x and write 2x as √2 × √2 x. With fractions, multiply through first: 2x2 − x + 1/8 = 0 becomes 16x2 − 8x + 1 = (4x − 1)2 = 0.
Never divide both sides by x. In x2 = 5x, dividing by x throws away the root x = 0. Write x2 − 5x = 0, so x(x − 5) = 0, giving x = 0 or x = 5.
5. The Quadratic Formula
Factorisation is fast when the numbers are friendly. The quadratic formula works for every quadratic equation that has real roots. The roots of ax2 + bx + c = 0 are
x = (−b ± √(b2 − 4ac)) / 2a, provided b2 − 4ac ≥ 0
The ± sign gives two roots: (−b + √(b2 − 4ac)) / 2a and (−b − √(b2 − 4ac)) / 2a.
History from NCERT: Brahmagupta (C.E. 598 to 665) gave a formula for ax2 + bx = c, and Sridharacharya (C.E. 1025) derived the quadratic formula by completing the square, as quoted by Bhaskara II.
Where the Formula Comes From
The NCERT chapter states the formula and asks you to apply it. The working below shows why it is true, using Sridharacharya’s idea of completing the square.
- Divide ax2 + bx + c = 0 by a: x2 + (b/a)x + c/a = 0.
- Add and subtract (b/2a)2: [x2 + (b/a)x + (b/2a)2] − (b/2a)2 + c/a = 0.
- The bracket is a perfect square: (x + b/2a)2 = b2/4a2 − c/a = (b2 − 4ac) / 4a2.
- If b2 − 4ac ≥ 0, take square roots: x + b/2a = ± √(b2 − 4ac) / 2a.
- So x = (−b ± √(b2 − 4ac)) / 2a.
Step 3 also shows why the formula needs b2 − 4ac ≥ 0: the left side is a square, which can never be negative.
Using the Formula Without Slips
- Write the equation in standard form. Everything must be on one side, equal to 0.
- Read a, b and c with their signs. In x2 + 7x − 60 = 0, c = −60, with its minus sign.
- Work out b2 − 4ac on its own line first.
- Simplify the square root (√12 = 2√3, √289 = 17).
- Divide the whole numerator by 2a, then simplify each root.
Illustration (from NCERT Example 8): x2 + 7x − 60 = 0. Here a = 1, b = 7, c = −60.
b2 − 4ac = 49 − 4(1)(−60) = 49 + 240 = 289, and √289 = 17.
x = (−7 ± 17) / 2, so x = 10/2 = 5 or x = −24/2 = −12.
Illustration with a surd answer: 2x2 − 6x + 3 = 0. Here a = 2, b = −6, c = 3.
b2 − 4ac = 36 − 24 = 12, and √12 = 2√3.
x = (6 ± 2√3) / 4 = (3 ± √3) / 2. This equation cannot be factorised over whole numbers, which is exactly when the formula earns its place.
Choosing a method: if the question names a method, use it, since the method is marked. Otherwise factorise when a split is quick, and use the formula for large constants (such as −1280) or when no split appears within about 30 seconds.
6. Discriminant and Nature of Roots
The expression under the square root in the quadratic formula decides everything about the roots. It is called the discriminant:
D = b2 − 4ac
It “discriminates” between the three possible kinds of roots.
| Value of D | Nature of roots | The roots |
|---|---|---|
| D > 0 | Two distinct real roots | (−b + √D)/2a and (−b − √D)/2a |
| D = 0 | Two equal real roots (coincident roots) | −b/2a and −b/2a |
| D < 0 | No real roots | No real number has a negative square |
When D = 0, the formula gives x = (−b ± 0)/2a, so both roots equal −b/2a. When D < 0, √D is the square root of a negative number, and no real number squares to give a negative result, so the equation has no real roots.
Useful extra: with rational a, b, c, a perfect-square D (such as 289) gives rational roots; a positive non-square D (such as 12) gives surd roots.
Worked Illustrations from NCERT
3x2 − 2x + 1/3 = 0: D = (−2)2 − 4 × 3 × (1/3) = 4 − 4 = 0. Two equal real roots, each −b/2a = 2/6 = 1/3.
Finding an Unknown for Equal Roots
“Find k so that the equation has two equal roots” means: write D in terms of k, set D = 0, and solve.
2x2 + kx + 3 = 0: D = k2 − 4 × 2 × 3 = k2 − 24. Setting D = 0 gives k2 = 24, so k = ±2√6.
For “real roots” the condition is D ≥ 0. For “real and distinct roots” it is D > 0. For “no real roots” it is D < 0. Read the wording carefully before writing the condition.
Is the Situation Possible
The discriminant also answers practical questions. Form the equation from the situation. If D ≥ 0 the situation is possible and you go on to find the values; if D < 0 it is impossible.
NCERT Example 8, the pole in a circular park. A park is a circle of diameter 13 m, with two gates A and B at opposite ends of a diameter. A pole P is to stand on the boundary so that the difference of its distances from the two gates is 7 m. Let BP = x m, so AP = (x + 7) m. Since AB is a diameter, the angle APB in the semicircle is 90°, so by Pythagoras theorem
(x + 7)2 + x2 = 132, which gives 2x2 + 14x + 49 − 169 = 0, that is x2 + 7x − 60 = 0.
D = 49 + 240 = 289 > 0, so the pole can be placed. The roots are 5 and −12 (worked in Concept 5). A distance cannot be negative, so x = 5. The pole stands 5 m from gate B and 12 m from gate A. Check: 52 + 122 = 25 + 144 = 169 = 132.
7. Solving Word Problems and Rejecting Roots
The equation gives two roots; the situation usually accepts only one. After solving, test each root against the story.
- Lengths, ages, speeds, times, counts: must be positive. Reject negative roots.
- Counts of objects or people: must be whole numbers. Reject fractions.
- A quantity used in a subtraction: check that every derived quantity is positive too. If a tap takes x hours and the other takes (x − 10) hours, x = 3.75 gives a negative time for the second tap, so it is rejected even though 3.75 itself is positive.
- Both roots may be valid. The marbles equation x2 − 45x + 324 = (x − 9)(x − 36) = 0 gives 36 and 9: John has 36 and Jivanti 9, or the other way round. The toys equation gives 25 or 30 toys, and both cost Rs 750 in total. State both when both fit.
Relations to keep ready: time = distance/speed; upstream speed = u − v and downstream speed = u + v for a boat of speed u in a stream of speed v; a job done alone in x hours means 1/x of it is done in one hour; ages n years ago are x − n; consecutive integers are x and x + 1.
Always finish with a check and a sentence. Substitute the answer back into the original story as well as the equation, and write the answer with units: “The speed of the train is 40 km/h.”
Formula and Theorem Sheet
| Result | Statement | When to use it |
|---|---|---|
| Standard form | ax2 + bx + c = 0, a, b, c real, a ≠ 0 | Before reading a, b, c or applying any formula |
| Root | α is a root if aα2 + bα + c = 0 | Verifying roots; finding k from a given root |
| Splitting the middle term | Numbers p, q with p + q = b and pq = ac | Factorising ax2 + bx + c |
| Quadratic formula | x = (−b ± √(b2 − 4ac)) / 2a, when b2 − 4ac ≥ 0 | Any quadratic with real roots |
| Discriminant | D = b2 − 4ac | Nature of roots without solving |
| D > 0 | Two distinct real roots | “Real and unequal”, “possible” |
| D = 0 | Two equal real roots, each −b/2a | “Equal roots”, “coincident roots” |
| D < 0 | No real roots | “Not possible” situations |
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Important Definitions
| Term | Definition |
|---|---|
| Quadratic equation | An equation of the form ax2 + bx + c = 0, where a, b, c are real numbers and a ≠ 0; equivalently p(x) = 0 where p(x) has degree 2 |
| Standard form | ax2 + bx + c = 0, a ≠ 0, with the terms written in descending order of their degrees |
| Root (solution) | A real number α such that aα2 + bα + c = 0; we say α satisfies the equation |
| Quadratic formula | x = (−b ± √(b2 − 4ac)) / 2a, giving the roots of ax2 + bx + c = 0 when b2 − 4ac ≥ 0 |
| Discriminant | The expression b2 − 4ac, which determines whether ax2 + bx + c = 0 has real roots |
| Distinct real roots | Two different real roots; occur when b2 − 4ac > 0 |
| Equal (coincident) roots | Two real roots with the same value −b/2a; occur when b2 − 4ac = 0 |
| No real roots | The case b2 − 4ac < 0, when no real number satisfies the equation |
Solved Examples (NCERT-Based)
Example 1: Check whether these are quadratic equations (NCERT Exercise 4.1)
(i) (x − 2)(x + 1) = (x − 1)(x + 3) (ii) (x + 2)3 = 2x(x2 − 1) (iii) x3 − 4x2 − x + 1 = (x − 2)3
Solution:
(i) LHS = x2 − x − 2; RHS = x2 + 2x − 3. So x2 − x − 2 = x2 + 2x − 3, giving −3x + 1 = 0. The x2 terms cancel. Not quadratic.
(ii) LHS = x3 + 6x2 + 12x + 8; RHS = 2x3 − 2x. So x3 − 6x2 − 14x − 8 = 0, which has degree 3. Not quadratic.
(iii) RHS = x3 − 6x2 + 12x − 8. So x3 − 4x2 − x + 1 = x3 − 6x2 + 12x − 8, giving 2x2 − 13x + 9 = 0. The x3 terms cancel. Quadratic.
Example 2: Rohan’s age (NCERT Exercise 4.1)
Rohan’s mother is 26 years older than him. The product of their ages 3 years from now will be 360. Find Rohan’s present age.
Solution: Let Rohan’s age be x years; his mother’s age is x + 26. In 3 years: x + 3 and x + 29.
(x + 3)(x + 29) = 360 ⇒ x2 + 32x + 87 = 360 ⇒ x2 + 32x − 273 = 0.
Product −273, sum 32: the pair 39 and −7. So (x + 39)(x − 7) = 0, giving x = 7 or x = −39.
Age cannot be negative, so x = 7. Check: in 3 years the ages are 10 and 36, and 10 × 36 = 360. Rohan is 7 years old.
Example 3: Roots by factorisation (NCERT Exercise 4.2)
Find the roots of (i) x2 − 3x − 10 = 0 and (ii) 2x2 + x − 6 = 0.
Solution:
(i) Product −10, sum −3: −5 and 2. x2 − 5x + 2x − 10 = x(x − 5) + 2(x − 5) = (x − 5)(x + 2). Roots: 5 and −2.
(ii) a × c = −12, sum 1: 4 and −3. 2x2 + 4x − 3x − 6 = 2x(x + 2) − 3(x + 2) = (2x − 3)(x + 2). Roots: 3/2 and −2.
Example 4: Equal roots from a perfect square (NCERT Exercise 4.2)
Solve (i) 2x2 − x + 1/8 = 0 and (ii) 100x2 − 20x + 1 = 0.
Solution:
(i) Multiply by 8: 16x2 − 8x + 1 = 0. Product 16, sum −8: −4 and −4. 16x2 − 4x − 4x + 1 = 4x(4x − 1) − 1(4x − 1) = (4x − 1)2. Roots: 1/4, 1/4.
(ii) Product 100, sum −20: −10 and −10. 100x2 − 10x − 10x + 1 = 10x(10x − 1) − 1(10x − 1) = (10x − 1)2. Roots: 1/10, 1/10.
Example 5: The right triangle (NCERT Exercise 4.2)
The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Solution: Let the base be x cm; the altitude is (x − 7) cm. By Pythagoras theorem:
x2 + (x − 7)2 = 132 ⇒ x2 + x2 − 14x + 49 = 169 ⇒ 2x2 − 14x − 120 = 0 ⇒ x2 − 7x − 60 = 0.
Product −60, sum −7: −12 and 5. (x − 12)(x + 5) = 0, so x = 12 or x = −5.
A side cannot be negative, so x = 12. Base 12 cm, altitude 5 cm. Check: 122 + 52 = 144 + 25 = 169 = 132.
Example 6: Pottery articles (NCERT Exercise 4.2)
On a particular day the cost of production of each pottery article (in rupees) was 3 more than twice the number of articles produced. The total cost that day was Rs 90. Find the number of articles and the cost of each.
Solution: Let x articles be produced; cost of each = (2x + 3) rupees.
x(2x + 3) = 90 ⇒ 2x2 + 3x − 90 = 0. Here a = 2, b = 3, c = −90.
D = 32 − 4(2)(−90) = 9 + 720 = 729, and √729 = 27.
x = (−3 ± 27) / 4, so x = 24/4 = 6 or x = −30/4 = −7.5.
The number of articles must be a positive whole number, so x = 6. Cost of each = 2(6) + 3 = 15. 6 articles at Rs 15 each (check: 6 × 15 = 90).
Example 7: Nature of roots, and roots if real (NCERT Exercise 4.3)
(i) 2x2 − 3x + 5 = 0 (ii) 3x2 − 4√3 x + 4 = 0 (iii) 2x2 − 6x + 3 = 0
Solution:
(i) D = (−3)2 − 4(2)(5) = 9 − 40 = −31 < 0. No real roots.
(ii) D = (−4√3)2 − 4(3)(4) = 48 − 48 = 0. Two equal real roots, each −b/2a = 4√3/6 = 2√3/3 = 2/√3.
(iii) D = (−6)2 − 4(2)(3) = 36 − 24 = 12 > 0. Two distinct real roots: x = (6 ± √12)/4 = (6 ± 2√3)/4 = (3 ± √3)/2.
Example 8: Value of k for equal roots (NCERT Exercise 4.3)
Find k so that (i) 2x2 + kx + 3 = 0 and (ii) kx(x − 2) + 6 = 0 have two equal roots.
Solution:
(i) a = 2, b = k, c = 3. D = k2 − 24 = 0, so k2 = 24 and k = ±2√6.
(ii) Expand: kx2 − 2kx + 6 = 0, so a = k, b = −2k, c = 6. D = 4k2 − 24k = 4k(k − 6) = 0, so k = 0 or k = 6. If k = 0 the equation becomes 6 = 0, which is not a quadratic equation at all (a = k must be non-zero). So k = 6. Check: 6x2 − 12x + 6 = 6(x − 1)2, equal roots 1, 1.
Example 9: The friends’ ages, an impossible situation (NCERT Exercise 4.3)
The sum of the ages of two friends is 20 years. Four years ago, the product of their ages was 48. Is this possible? If so, find their present ages.
Solution: Let one friend be x years old; the other is 20 − x. Four years ago: x − 4 and 16 − x.
(x − 4)(16 − x) = 48 ⇒ 16x − x2 − 64 + 4x = 48 ⇒ −x2 + 20x − 112 = 0 ⇒ x2 − 20x + 112 = 0.
D = (−20)2 − 4(1)(112) = 400 − 448 = −48 < 0.
No real roots, so the situation is not possible.
Example 10: The rectangular park (NCERT Exercise 4.3)
Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth.
Solution: Length + breadth = 80/2 = 40. Let the length be x m; breadth = (40 − x) m. x(40 − x) = 400 ⇒ x2 − 40x + 400 = 0.
D = 1600 − 1600 = 0, so it is possible, with equal roots x = 40/2 = 20. Length 20 m, breadth 20 m: the park is a square. (For the mango grove in the same exercise, length twice breadth and area 800 m2, 2x2 = 800 gives 20 m by 40 m.)
Example 11: Speed of the train (NCERT Exercise 4.1, solved)
A train travels 480 km at a uniform speed. If the speed had been 8 km/h less, it would have taken 3 hours more. Find the speed of the train.
Solution: Let the speed be x km/h. From Concept 2, x2 − 8x − 1280 = 0.
D = 64 + 5120 = 5184, and √5184 = 72.
x = (8 ± 72)/2, so x = 40 or x = −32. Speed cannot be negative, so x = 40.
Check: 480/40 = 12 hours; 480/32 = 15 hours; the difference is 3 hours. The speed of the train is 40 km/h.
Example 12: Two number problems (NCERT Exercise 4.2)
(i) Find two numbers whose sum is 27 and product is 182. (ii) Find two consecutive positive integers, the sum of whose squares is 365.
Solution:
(i) Let one number be x; the other is 27 − x. Then x(27 − x) = 182 ⇒ 27x − x2 = 182 ⇒ x2 − 27x + 182 = 0. Product 182, sum −27: −13 and −14. So (x − 13)(x − 14) = 0, giving x = 13 or x = 14. Either way the pair is 13 and 14 (check: 13 + 14 = 27, 13 × 14 = 182).
(ii) Let the integers be x and x + 1. Then x2 + (x + 1)2 = 365 ⇒ 2x2 + 2x + 1 = 365 ⇒ 2x2 + 2x − 364 = 0 ⇒ x2 + x − 182 = 0. Product −182, sum 1: 14 and −13. So (x + 14)(x − 13) = 0, giving x = 13 or x = −14. The integers must be positive, so x = 13. The integers are 13 and 14 (check: 169 + 196 = 365).
Competency-Based Questions (with answers)
1. Case-based: The school garden
A school is laying out a rectangular vegetable garden. The length is 5 m more than the breadth and the area is 84 m2.
(a) Form the quadratic equation for the breadth x. x(x + 5) = 84, so x2 + 5x − 84 = 0.
(b) Find the length and breadth. Product −84, sum 5: 12 and −7. (x + 12)(x − 7) = 0, so x = 7 (rejecting −12). Breadth 7 m, length 12 m.
(c) How much fencing is needed to go round the garden once? Perimeter = 2(12 + 7) = 38 m.
2. Case-based: How high does the ball go
A ball is thrown straight up. Its height above the ground after t seconds is h = 20t − 5t2 metres.
(a) At what times is the ball 15 m high? 20t − 5t2 = 15 ⇒ 5t2 − 20t + 15 = 0 ⇒ t2 − 4t + 3 = 0 ⇒ (t − 1)(t − 3) = 0. At t = 1 s on the way up and t = 3 s on the way down.
(b) Can the ball reach 25 m? 5t2 − 20t + 25 = 0 ⇒ t2 − 4t + 5 = 0. D = 16 − 20 = −4 < 0. No real time exists, so the ball never reaches 25 m.
(c) What is the greatest height? For a height of h metres, 5t2 − 20t + h = 0 has D = 400 − 20h, which is ≥ 0 only when h ≤ 20. So the ball cannot rise above 20 m. At h = 20, t2 − 4t + 4 = 0 has D = 0 and the repeated root t = 2. The greatest height is 20 m, reached at t = 2 s.
3. Case-based: The motorboat and the stream
A motorboat whose speed in still water is 18 km/h takes 1 hour more to go 24 km upstream than to return downstream to the same spot.
(a) Write the upstream and downstream speeds if the stream flows at x km/h. Upstream (18 − x) km/h; downstream (18 + x) km/h.
(b) Form the equation. 24/(18 − x) − 24/(18 + x) = 1. Taking the LCM: 24[(18 + x) − (18 − x)] = (18 − x)(18 + x), so 48x = 324 − x2, that is x2 + 48x − 324 = 0.
(c) Find the speed of the stream. D = 2304 + 1296 = 3600, √D = 60. x = (−48 ± 60)/2 = 6 or −54. Speed is positive, so the stream flows at 6 km/h. Check: 24/12 = 2 h upstream, 24/24 = 1 h downstream, a difference of 1 hour.
4. Source-based: The Babylonian problem
The NCERT chapter notes that the Babylonians knew how to find two positive numbers with a given positive sum and a given positive product, and that this is the same as solving x2 − px + q = 0, where p is the sum and q is the product.
(a) Find two numbers with sum 13 and product 40. x2 − 13x + 40 = 0 ⇒ (x − 5)(x − 8) = 0. The numbers are 5 and 8.
(b) For a sum of 10, what is the largest possible product? x2 − 10x + q = 0 has real roots only if D = 100 − 4q ≥ 0, that is q ≤ 25. The largest product is 25, when both numbers are 5.
5. Assertion-Reason
Assertion (A): The equation x2 + 4x + 5 = 0 has no real roots.
Reason (R): A quadratic equation ax2 + bx + c = 0 has no real roots when b2 − 4ac < 0.
Choose: (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.
Answer: (a) Both A and R are true, and R is the correct explanation of A. D = 16 − 20 = −4 < 0.
6. Assertion-Reason
Assertion (A): (x − 1)2 = x2 − 5 is not a quadratic equation.
Reason (R): An equation is quadratic whenever an x2 term appears in it as written.
Answer: (c) A is true but R is false. Expanding gives x2 − 2x + 1 = x2 − 5, so −2x + 6 = 0, which is linear. The test is applied after simplifying, so R is false.
7. Error analysis: The lost root
A student solves 3x2 = 12x like this: “Divide both sides by 3x, so x = 4.” Identify the error and give the complete answer.
Answer: Dividing by 3x assumes x ≠ 0, which throws away the root x = 0. Correct method: 3x2 − 12x = 0 ⇒ 3x(x − 4) = 0 ⇒ x = 0 or x = 4. A quadratic can have two roots, so both must be found.
8. Competency MCQ: Which equation has two distinct real roots
(a) x2 + x + 1 = 0 (b) 2x2 − 4x + 2 = 0 (c) x2 − 3x − 4 = 0 (d) 4x2 + 4x + 5 = 0
Answer: (c). D values: (a) 1 − 4 = −3; (b) 16 − 16 = 0; (c) 9 + 16 = 25; (d) 16 − 80 = −64. Only (c) has D > 0; its roots are 4 and −1. Shortcut: when a and c have opposite signs, −4ac is positive, so D is positive and the roots are always real and distinct.
Important Questions for Board Exams
1-Mark Questions
- Find the discriminant of 2x2 − 4x + 3 = 0. D = (−4)2 − 4(2)(3) = 16 − 24 = −8.
- If x = 2 is a root of 3x2 + kx − 2 = 0, find k. 12 + 2k − 2 = 0, so k = −5.
- Find the values of k for which x2 + kx + 9 = 0 has equal roots. k2 − 36 = 0, so k = ±6.
- State the nature of the roots of x2 − 4x + 4 = 0. D = 16 − 16 = 0, so two equal real roots, each 2.
2-Mark Questions
- Find the roots of 6x2 − x − 2 = 0 by factorisation. a × c = −12, sum −1: 3 and −4. 6x2 + 3x − 4x − 2 = 3x(2x + 1) − 2(2x + 1) = (3x − 2)(2x + 1). Roots 2/3 and −1/2.
- Solve √3 x2 + 10x + 7√3 = 0. a × c = 21, sum 10: 7 and 3, with 3x = √3 × √3 x. √3 x2 + 3x + 7x + 7√3 = √3 x(x + √3) + 7(x + √3) = (x + √3)(√3 x + 7). Roots −√3 and −7/√3 = −7√3/3.
- Solve x2 − 2ax + (a2 − b2) = 0. The split needs two numbers with product a2 − b2 = (a + b)(a − b) and sum −2a: these are −(a + b) and −(a − b). So [x − (a + b)][x − (a − b)] = 0, and the roots are a + b and a − b.
3-Mark Questions
- Solve 2x2 + x − 4 = 0 using the quadratic formula. a = 2, b = 1, c = −4. D = 1 + 32 = 33 > 0. x = (−1 ± √33)/4. The roots are (−1 + √33)/4 and (−1 − √33)/4.
- Find p for which (p + 4)x2 + (p + 1)x + 1 = 0 has equal roots, and find the roots. D = (p + 1)2 − 4(p + 4) = p2 − 2p − 15 = (p − 5)(p + 3) = 0, so p = 5 or p = −3 (both keep p + 4 ≠ 0). For p = 5: 9x2 + 6x + 1 = (3x + 1)2, root −1/3 (twice). For p = −3: x2 − 2x + 1 = (x − 1)2, root 1 (twice).
- The product of Sunita’s age (in years) two years ago and her age four years from now is one more than twice her present age. Find her present age. Let her age be x. (x − 2)(x + 4) = 2x + 1 ⇒ x2 + 2x − 8 = 2x + 1 ⇒ x2 = 9 ⇒ x = 3 (rejecting −3). Sunita is 3 years old. Check: 1 × 7 = 7 = 2(3) + 1.
5-Mark Questions
- An express train takes 1 hour less than a passenger train to travel 132 km between Mysuru and Bengaluru, ignoring the time they stop at intermediate stations. The average speed of the express train is 11 km/h more than that of the passenger train. Find the average speed of each train. Let the passenger train’s speed be x km/h; the express train’s speed is (x + 11) km/h. Time difference: 132/x − 132/(x + 11) = 1. LCM: 132 × 11 = x(x + 11), so x2 + 11x − 1452 = 0. D = 121 + 5808 = 5929 = 772. x = (−11 ± 77)/2 = 33 or −44. Speed is positive, so the passenger train runs at 33 km/h and the express at 44 km/h. Check: 132/33 = 4 h, 132/44 = 3 h.
- Two water taps together can fill a tank in 75/8 hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank alone. Find the time each tap takes on its own. Let the smaller tap take x hours; the larger takes (x − 10) hours. In one hour they fill 1/x + 1/(x − 10) = 8/75 of the tank. So 75(2x − 10) = 8x(x − 10) ⇒ 150x − 750 = 8x2 − 80x ⇒ 8x2 − 230x + 750 = 0 ⇒ 4x2 − 115x + 375 = 0. D = 13225 − 6000 = 7225 = 852. x = (115 ± 85)/8 = 25 or 3.75. If x = 3.75, the larger tap would take −6.25 hours, which is impossible. So the smaller tap takes 25 hours and the larger 15 hours. Check: 1/25 + 1/15 = 3/75 + 5/75 = 8/75.
Common Mistakes and Examiner Tips
- Deciding “quadratic or not” before simplifying. x(x + 1) + 8 = (x + 2)(x − 2) looks quadratic and is linear; (x + 2)3 = x3 − 4 looks cubic and is quadratic. Expand, collect everything on one side, then look at the highest power.
- Forgetting a ≠ 0 in “find k” questions. In kx(x − 2) + 6 = 0, setting D = 0 gives k = 0 or 6. k = 0 kills the x2 term, so the answer is k = 6 only. Always test each value of k against the leading coefficient.
- Dropping the sign of b or c. In 2x2 − 6x + 3 = 0, b = −6 and −b = 6. In x2 + 7x − 60 = 0, c = −60 and −4ac = +240. Write “a = , b = , c = ” with signs before substituting.
- Dividing only part of the numerator by 2a. x = −b ± √D / 2a is wrong; the whole of (−b ± √D) is divided by 2a. Use brackets.
- Cancelling x from both sides. Dividing x2 = 5x by x loses the root x = 0. Bring everything to one side and factorise.
- Setting each factor equal to the right-hand number. From (x − 3)(x + 2) = 6 you cannot write x − 3 = 6. The zero product rule works only when the right side is 0. Expand first: x2 − x − 12 = 0 = (x − 4)(x + 3), roots 4 and −3.
- Keeping a root that the situation rejects. Negative lengths, ages, speeds, times, and fractional counts must be struck out, with a reason written: “x = −12 is rejected as distance cannot be negative.” Writing the reason shows the examiner why the root was dropped.
- Writing the wrong condition for the words used. “Equal roots”: D = 0. “Real roots”: D ≥ 0. “Distinct real roots”: D > 0. “No real roots”: D < 0. Mixing ≥ and > changes the answer to “find the range of k” questions.
- Using the formula when the question says “by factorisation”. The method is part of what is marked, so follow the method the question names.
- Stopping at “x = 40”. Finish with units and a sentence: “The speed of the train is 40 km/h.”
- Subtracting times the wrong way round. The slower journey takes longer: write (longer time) − (shorter time) = difference.
Quick Revision Points
- Quadratic equation: ax2 + bx + c = 0, with a, b, c real and a ≠ 0
- b or c can be 0; a can never be 0
- Decide whether an equation is quadratic only after expanding and simplifying
- α is a root if aα2 + bα + c = 0
- Roots of the equation are the zeroes of the polynomial; at most two roots
- Given a root, substitute it to find an unknown coefficient
- Factorisation: split b into two numbers with product ac and sum b
- Zero product rule needs 0 on the right side first
- A repeated factor gives a repeated root: (10x − 1)2 = 0 gives 1/10, 1/10
- Quadratic formula: x = (−b ± √(b2 − 4ac)) / 2a, valid when b2 − 4ac ≥ 0
- The formula comes from completing the square (Sridharacharya)
- Discriminant D = b2 − 4ac
- D > 0: two distinct real roots; D = 0: two equal roots, each −b/2a; D < 0: no real roots
- If a and c have opposite signs, D > 0 automatically
- Equal roots: set D = 0, then reject any k that makes a = 0
- “Is it possible”: form the equation and check D ≥ 0
- Word problems: define x with units, solve, reject impossible roots with a reason, check in the story
Weightage in Board Exams
Quadratic Equations belongs to the Algebra unit of Class 10 Maths, with Polynomials, Pair of Linear Equations in Two Variables, and Arithmetic Progressions. Check the current CBSE course structure for the marks given to the unit; within it, this chapter is assessed in a predictable way.
| Question type | What is usually asked |
|---|---|
| MCQ and Assertion-Reason | Nature of roots; k for equal roots; finding k from a given root; equations that look quadratic but are not |
| Short answers | Factorisation, including surd coefficients; the quadratic formula with a surd answer; age or number problems |
| Long answer | Speed and time, boat and stream, or taps and work: form, solve and reject a root |
| Case-based | A real situation split into parts: form the equation, solve, and use the discriminant to test a value |
In a word problem, the key steps are forming the equation correctly and giving a reason when you reject a root. Practise every NCERT exercise question first, then the speed, boat and taps types above.
Class 10 Maths · Chapter 4 – swipe through all 10 cards to understand the whole chapter.
Standard Form
A quadratic equation is a degree-2 polynomial set equal to zero.
b or c may be 0; a never can, or the equation becomes linear.
- 4x − 3x2 + 2 = 0 in standard form is −3x2 + 4x + 2 = 0.
- Simplify before deciding: x(x + 1) + 8 = (x + 2)(x − 2) is linear.
- (x + 2)3 = x3 − 4 is quadratic because the x3 terms cancel.
Forming the Equation
Pick one unknown, write everything else in terms of it, use the last fact.
Stop at standard form when the question says only represent.
- Consecutive integers with product 306: x2 + x − 306 = 0.
- Rohan and mother, product of ages in 3 years 360: x2 + 32x − 273 = 0.
- Train 480 km, 8 km/h slower takes 3 h more: x2 − 8x − 1280 = 0.
What a Root Is
A real number α is a root if it makes the equation true.
Roots of the equation are the zeroes of the polynomial.
- A quadratic equation has at most two roots.
- 1 is a root of 2x2 − 3x + 1 = 0 since 2 − 3 + 1 = 0.
- Given root 2 of x2 + kx − 6 = 0: 4 + 2k − 6 = 0, so k = 1.
Factorisation
Split the middle term using two numbers with product ac and sum b.
Set each factor to zero only when the right side is 0.
- 6x2 − x − 2 = (3x − 2)(2x + 1), roots 2/3 and −1/2.
- x2 − 3x − 10 = (x − 5)(x + 2), roots 5 and −2.
- Never divide by x: x2 = 5x gives x = 0 or 5.
Perfect Squares
When both factors are the same, the root appears twice.
Clear fractions first: 2x2 − x + 1/8 = 0 becomes 16x2 − 8x + 1 = 0.
- 16x2 − 8x + 1 = (4x − 1)2, roots 1/4, 1/4.
- 3x2 − 2√6x + 2 = (√3x − √2)2, root √(2/3) twice.
- A repeated root always means the discriminant is 0.
Quadratic Formula
Works for every quadratic with real roots.
Derived by completing the square, as Sridharacharya did.
- x2 + 7x − 60 = 0: D = 289, x = (−7 ± 17)/2 = 5 or −12.
- 2x2 − 6x + 3 = 0: D = 12, x = (3 ± √3)/2.
- Read a, b, c with signs and divide the whole numerator by 2a.
The Discriminant
b2 − 4ac tells you what kind of roots to expect before solving.
Positive D: distinct real roots. D = 0: equal roots, each −b/2a. Negative D: no real roots.
- 2x2 − 4x + 3 = 0: D = −8, no real roots.
- 3x2 − 2x + 1/3 = 0: D = 0, roots 1/3 and 1/3.
- If a and c have opposite signs, D is always positive.
Equal Roots Condition
Set the discriminant equal to zero and solve for the unknown.
Reject any k that makes the x2 coefficient zero.
- 2x2 + kx + 3 = 0: k2 − 24 = 0, so k = ±2√6.
- kx(x − 2) + 6 = 0: 4k(k − 6) = 0, so k = 6 only.
- Real roots means D ≥ 0; distinct real roots means D > 0.
Real-Life Checks
Form the equation, then let the discriminant decide.
A negative D means no real value fits the situation.
- Friends’ ages sum 20, product 4 years ago 48: D = −48, impossible.
- Park of perimeter 80 m and area 400 m2: D = 0, a 20 m square.
- Pole in a 13 m park: D = 289, 5 m from B and 12 m from A.
Rejecting Roots
Keep only the roots that make sense in the story, and say why.
Test every quantity defined from x, not only x itself.
- Train: x2 − 8x − 1280 = 0 gives 40 or −32, speed 40 km/h.
- Taps: x = 25 or 3.75; 3.75 makes x − 10 negative, so 25 h.
- Right triangle: 12 or −5, so sides 12 cm and 5 cm.
📝 Practice Quadratic Equations - 10 board questions
CBSE previous-year and competency-based · with answers & explanations
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Why B: Step 1 (discriminant D = b² – 4ac): D = p² – 24. Step 2 (test p = -5): D = 25 – 24 = 1 = 1², a perfect square, so the roots are rational (the equation is x² + 5x + 6 = 0 with roots -2 and -3).
Why not A: p = 1 gives D = 1 – 24 = -23 < 0, so there are no real roots at all.
Why not C: p = 25 gives D = 625 – 24 = 601, which is not a perfect square, so the roots are irrational; 25 is p² for p = ±5, not p.
Why not D: p = √5 gives D = 5 – 24 = -19 < 0, no real roots.
Remember: rational coefficients plus a perfect-square discriminant give rational roots.
Why C: Step 1 (discriminant D = b² – 4ac): a = k, b = -6, c = -4, so D = 36 – 4(k)(-4) = 36 + 16k. Step 2 (equal roots, D = 0): 36 + 16k = 0, so k = -36/16 = -9/4.
Why not A: 9/4 drops the sign of c = -4, which makes -4ac = -16k instead of +16k.
Why not B: -4 is just the constant term copied as the answer; k = -4 gives D = 36 – 64 = -28.
Why not D: k = -2 gives D = 36 – 32 = 4, not 0.
Remember: put every sign into b² – 4ac before you simplify; two negatives in 4ac make a plus.
Why A: Step 1 (clear denominators, x ≠ 0): multiply by 12x to get x² – 36 = 0. Step 2 (square root of both sides): x² = 36, so x = 6 or x = -6.
Why not B: ±4 comes from dividing 12 by 3 instead of multiplying; x = 4 gives 1/3 – 3/4 ≠ 0.
Why not C: ±12 reads the denominator 12 as the answer instead of solving x² = 12 × 3.
Why not D: ±3 reads the numerator 3 as the answer; x = 3 gives 1/4 – 1 ≠ 0.
Remember: x/a = b/x means x² = ab; then take both square roots.
Why A: Step 1 (sum of roots): 7 + 1/7 = 50/7. Step 2 (product of roots): 7 × 1/7 = 1. Step 3 (form and clear the fraction): x² – (50/7)x + 1 = 0; multiply every term by 7 to get 7x² – 50x + 7 = 0.
Why not B: the constant 1 was not multiplied by 7 when the fraction was cleared, so the product of roots becomes 1/7.
Why not C: the sum term enters with a minus sign and the product is +1, not -1.
Why not D: both signs are wrong and the constant was not scaled by 7.
Remember: roots α and β give x² – (α + β)x + αβ = 0; multiply EVERY term by the LCM.
Why C: Step 1 (discriminant D = b² – 4ac): a = 3, b = -2, so D = (-2)² – 4(3)(c) = 4 – 12c. Step 2 (set D = 16): 4 – 12c = 16, so -12c = 12 and c = -1.
Why not A: c = 1 gives D = 4 – 12 = -8, which comes from a sign slip when moving 12c across.
Why not B: c = 0 gives D = 4, not 16.
Why not D: √2 comes from mixing up D with its square root; c = √2 gives D = 4 – 12√2, not 16.
Remember: square b first, including its sign: (-2)² = +4.
Why D: Step 1 (discriminant D = b² – 4ac): a = 1, b = 1, c = 1, so D = 1 – 4 = -3. Step 2 (nature of roots): D < 0, so the equation has no real roots.
Why not A: real and equal roots need D = 0, but here D = -3.
Why not B: irrational roots are real roots, which need D > 0 with D not a perfect square; here D is negative.
Why not C: real and distinct roots need D > 0.
Remember: D > 0 two distinct real roots, D = 0 two equal real roots, D < 0 no real roots.
Why A: Step 1 (split the middle term): find two numbers with product -10 and sum 3, which are 5 and -2, so x² + 5x – 2x – 10 = x(x + 5) – 2(x + 5) = (x – 2)(x + 5). Step 2 (zero product rule): x – 2 = 0 or x + 5 = 0, so x = 2 or x = -5.
Why not B: -2 and 5 are the numbers you get if you set x + 2 = 0 and x – 5 = 0, which are the factors of x² – 3x – 10, the sign of the middle term flipped.
Why not C: 2 and 5 multiply to +10, but the constant term is -10, so the roots must have opposite signs.
Why not D: -2 and -5 also multiply to +10, so they cannot be the roots of an equation with constant -10.
Remember: factor (x – p)(x – q) gives roots p and q; flip the sign inside each bracket.
Why B: Step 1 (sum of roots = -b/a): here a = -1 and b = 4, so the sum = -4/(-1) = 4. Step 2 (check the others): A gives -(-4)/2 = 2, C gives (4/√2)/√2 = 4/2 = 2, D gives -(-4)/4 = 1. Only B gives 4.
Why not A: the sum is -b/a = 4/2 = 2, not 4; reading the sum straight off as -b ignores the division by a.
Why not C: (4/√2) divided by √2 is 4/2 = 2, not 4.
Why not D: -b/a = 4/4 = 1.
Remember: sum of roots = -b/a, product of roots = c/a; always divide by a.
Why A: Step 1 (sum of roots): (2 + √3) + (2 – √3) = 4. Step 2 (product of roots, (a + b)(a – b) = a² – b²): (2 + √3)(2 – √3) = 4 – 3 = 1. Step 3 (form the equation): x² – 4x + 1 = 0.
Why not B: x² + 4x + 1 = 0 has sum of roots -4; the sum must enter with a minus sign.
Why not C: 4x² – 3 = 0 squares the two parts separately and has roots ±√3/2, not 2 ± √3.
Why not D: x² – 1 = 0 uses the right product 1 but drops the sum term, giving roots ±1.
Remember: roots α and β give x² – (α + β)x + αβ = 0; surd roots come in pairs that make a rational sum and product.
Why B: (x – 2)³ = x³ – 6x² + 12x – 8. Then x³ – 4x² – x + 1 = x³ – 6x² + 12x – 8 gives 2x² – 13x + 9 = 0 after the x³ terms cancel, which is quadratic.
Why not A: (x + 2)³ = x³ + 6x² + 12x + 8 and 2x(x² – 1) = 2x³ – 2x, so the equation becomes x³ – 6x² – 14x – 8 = 0, which stays cubic.
Why not C: the x² terms cancel, leaving 3x – 1 = 0, which is linear.
Why not D: (x – 2)² = x² – 4x + 4, so x² cancels, leaving 7x – 3 = 0, which is linear.
Remember: an equation that looks cubic can be quadratic, and one that looks quadratic can be linear.
Chapter Navigation
Previous: Pair of Linear Equations in Two Variables Class 10 Notes
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Related Chapters in Class 10 Maths
- Real Numbers Class 10 Notes
- Polynomials Class 10 Notes
- Pair of Linear Equations in Two Variables Class 10 Notes
- Arithmetic Progressions Class 10 Notes
- Triangles Class 10 Notes
- Coordinate Geometry Class 10 Notes
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Frequently Asked Questions
A quadratic equation in x is an equation of the form ax² + bx + c = 0, where a, b and c are real numbers and a ≠ 0. This is its standard form. Any equation p(x) = 0 where p(x) is a polynomial of degree 2 is quadratic. Always simplify first: x(x + 1) + 8 = (x + 2)(x − 2) reduces to x + 12 = 0, so it is linear.
A zero belongs to the polynomial ax² + bx + c and a root belongs to the equation ax² + bx + c = 0, but they are the same numbers. A real number α is a root if aα² + bα + c = 0. For example, 1 is a root of 2x² − 3x + 1 = 0 because 2 − 3 + 1 = 0, and 1 is also a zero of 2x² − 3x + 1.
Find two numbers whose product is a × c and whose sum is b, and use them to split the middle term. Group the four terms in pairs, take out the common factor, and write the quadratic as two linear factors. Set each factor equal to zero. For 2x² − 5x + 3 = 0, the split −2x − 3x gives (2x − 3)(x − 1) = 0, so x = 3/2 or 1.
The roots of ax² + bx + c = 0 are x = (−b ± √(b² − 4ac)) / 2a, provided b² − 4ac ≥ 0. It comes from completing the square, a method Sridharacharya used around C.E. 1025. Read a, b and c with their signs, work out b² − 4ac first, and divide the whole numerator by 2a. For x² + 7x − 60 = 0 it gives (−7 ± 17)/2, that is 5 or −12.
The discriminant of ax² + bx + c = 0 is D = b² − 4ac. If D > 0 there are two distinct real roots. If D = 0 there are two equal real roots, each equal to −b/2a. If D < 0 there are no real roots, because no real number has a negative square. It lets you state the nature of the roots without solving the equation.
Write the discriminant in terms of k, set it equal to zero and solve. For 2x² + kx + 3 = 0, k² − 24 = 0 gives k = ±2√6. Then check that no value of k makes the coefficient of x² zero. In kx(x − 2) + 6 = 0, D = 0 gives k = 0 or 6, and k = 0 is rejected, so k = 6.
The equation gives every number that fits the algebra, but the situation may allow only some of them. Lengths, ages, speeds and times must be positive, and counts must be whole numbers. In the right triangle problem, x² − 7x − 60 = 0 gives 12 and −5, and the side must be 12 cm. Write the reason for rejecting a root so the examiner sees why it was dropped.
Form the quadratic equation from the situation and find its discriminant. If D < 0 there is no real value that fits, so the situation is impossible. Two friends with ages summing to 20 whose product four years ago was 48 give x² − 20x + 112 = 0, where D = 400 − 448 = −48. The situation is not possible.
