SSC-Standard Mathematics for Forest Posts · Chapter 3

Algebra II: Quadratics, Simultaneous Equations, Inequalities and Sets

What to remember

A quadratic equation is ax² + bx + c = 0 with a ≠ 0. Its roots are x = (−b ± √D) / 2a, where D = b² − 4ac is the discriminant.
For the roots α and β: sum = −b/a and product = c/a. The sign of D tells the nature of the roots without solving.
Two linear equations in two unknowns have one solution, no solution, or infinitely many, depending on the ratios of their coefficients. A set with n elements has 2ⁿ subsets, and n(A ∪ B) = n(A) + n(B) − n(A ∩ B).

Quadratic equations and their roots

A quadratic equation has degree 2. If a = 0 it becomes a linear equation, so a ≠ 0 is always required. There are three standard ways to solve it.

1. Factorisation. Split the middle term. For x² − 5x + 6 = 0, find two numbers with product 6 and sum −5. These are −2 and −3, so (x − 2)(x − 3) = 0 and x = 2 or x = 3.
2. Quadratic formula. x = (−b ± √(b² − 4ac)) / 2a. It works for every quadratic.
3. Completing the square. Write x² + bx as (x + b/2)² − b²/4. The quadratic formula comes from this method.

Sum and product of roots. If α and β are the roots of ax² + bx + c = 0:

α + β = −b/a
αβ = c/a
A quadratic with given roots is x² − (sum)x + (product) = 0. Roots 3 and −2 give x² − x − 6 = 0.

Useful results built from the sum S and product P:

α² + β² = S² − 2P
1/α + 1/β = S/P
(α − β)² = S² − 4P

Worked example 1. For x² − 5x + 6 = 0, S = 5 and P = 6. Then α² + β² = 25 − 12 = 13.

Worked example 2. Two numbers have sum 25 and product 144. They are the roots of x² − 25x + 144 = 0. Here D = 625 − 576 = 49, so x = (25 ± 7)/2 = 16 or 9.

Nature of roots

The discriminant D = b² − 4ac decides the nature of the roots (a, b, c real).

Value of D	Nature of roots	Graph (parabola)
D > 0	Two real and distinct roots	Cuts the x-axis at two points
D = 0	Two real and equal roots, x = −b/2a	Touches the x-axis at one point
D < 0	No real roots (two complex roots)	Does not meet the x-axis

If D is a perfect square and a, b, c are rational, the roots are rational. If D > 0 but not a perfect square, the roots are irrational.

Worked example. For x² + kx + 16 = 0 to have equal roots, D = k² − 64 = 0, so k = ±8.

Word problems. Form the equation, solve it, and reject any root that makes no sense (a negative length or age). A rectangle with perimeter 34 and area 60 has sides that are roots of x² − 17x + 60 = 0, which gives 12 and 5. If two positive numbers differ by 3 and their squares add to 117, then x² + (x + 3)² = 117, so x² + 3x − 54 = 0 and x = 6. The numbers are 6 and 9.

Simultaneous linear equations

Two equations a₁x + b₁y = c₁ and a₂x + b₂y = c₂ can be solved by substitution, elimination or cross-multiplication.

Condition	Lines	Number of solutions
a₁/a₂ ≠ b₁/b₂	Intersecting	Exactly one (unique)
a₁/a₂ = b₁/b₂ ≠ c₁/c₂	Parallel	None (inconsistent)
a₁/a₂ = b₁/b₂ = c₁/c₂	Coincident	Infinitely many

Worked example (elimination). x + y = 10 and x − y = 4. Adding gives 2x = 14, so x = 7 and y = 3.

Worked example (substitution). 2x + 3y = 12 and 3x − y = 7. From the second, y = 3x − 7. Then 2x + 9x − 21 = 12, so 11x = 33, x = 3 and y = 2.

Worked example (cost). 5 pencils and 7 pens cost 50; 7 pencils and 5 pens cost 46 (in rupees). Adding gives 12(p + q) = 96, so p + q = 8. Subtracting gives p − q = −2. So pencil = 3 and pen = 5.

Digit problem. A two-digit number is 10x + y. If the digits sum to 9 and reversing the digits adds 27, then y − x = 3. So x = 3, y = 6 and the number is 36.

Inequalities

Solve a linear inequality like an equation, with one important rule: multiplying or dividing by a negative number reverses the inequality sign.

3x − 7 > 8 gives 3x > 15, so x > 5.
−2x > 6 gives x < −3.

Modulus (absolute value) inequalities (a > 0):

|x| < a means −a < x < a.
|x| > a means x < −a or x > a.
|x − c| ≤ a means c − a ≤ x ≤ c + a. For example, |x − 1| ≤ 3 gives −2 ≤ x ≤ 4.
|2x − 1| = 5 gives 2x − 1 = 5 or 2x − 1 = −5, so x = 3 or x = −2.

Quadratic inequalities. Factorise, mark the roots on a number line and test the intervals. For (x − a)(x − b) < 0 with a < b, the answer lies between the roots. For (x − a)(x − b) > 0, the answer lies outside them.

x² − 5x + 6 < 0 gives 2 < x < 3.
x² − 9 ≥ 0 gives x ≤ −3 or x ≥ 3.
x² < 0 has no real solution, because a square is never negative.

Sets

A set is a well-defined collection of objects. Standard terms:

Empty set (∅): no elements. It is a subset of every set.
Subset (A ⊆ B): every element of A is in B. A proper subset is a subset that is not equal to the set itself.
Universal set (U): the set that contains all objects under discussion.
Union (A ∪ B): elements in A or B or both. Intersection (A ∩ B): elements in both.
Difference (A − B): elements in A but not in B. Complement (A′): U − A.

Result	Meaning
n(A ∪ B) = n(A) + n(B) − n(A ∩ B)	Counting formula for two sets
Number of subsets = 2ⁿ	n = number of elements
Number of proper subsets = 2ⁿ − 1	Excludes the set itself
(A ∪ B)′ = A′ ∩ B′	De Morgan's first law
(A ∩ B)′ = A′ ∪ B′	De Morgan's second law
A ∪ A′ = U and A ∩ A′ = ∅	Complement laws

Worked example. In a class of 50, 30 play cricket and 25 play football, and everyone plays at least one game. Then 50 = 30 + 25 − n(both), so n(both) = 5.

Sets A = {1, 2, 3, 4} and B = {3, 4, 5, 6} give A ∪ B = {1, 2, 3, 4, 5, 6}, A ∩ B = {3, 4} and A − B = {1, 2}.

Exam traps

Forgetting a ≠ 0 when deciding whether an equation is quadratic.
Mixing up the sum of roots (−b/a) with the product (c/a), or dropping the minus sign in −b/a.
Saying roots are "not real" when D = 0. For D = 0 the roots are real and equal.
Not reversing the inequality sign when dividing by a negative number.
Writing |x| > a as −a < x < a. That is the answer for |x| < a.
Counting the set itself as a proper subset. The number of proper subsets is 2ⁿ − 1.
Treating the empty set as having no subsets. It has exactly one subset, itself.
Forgetting to subtract n(A ∩ B) in the union formula, which counts the common elements twice.

One-liners

1. In ax² + bx + c = 0, the condition for a quadratic is a ≠ 0.
2. Discriminant D = b² − 4ac.
3. Sum of roots = −b/a; product of roots = c/a.
4. D > 0: real and distinct roots; D = 0: real and equal; D < 0: no real roots.
5. α² + β² = (α + β)² − 2αβ.
6. A quadratic with roots α, β is x² − (α + β)x + αβ = 0.
7. Unique solution if a₁/a₂ ≠ b₁/b₂; no solution if a₁/a₂ = b₁/b₂ ≠ c₁/c₂.
8. Infinitely many solutions if a₁/a₂ = b₁/b₂ = c₁/c₂.
9. Multiplying an inequality by a negative number reverses the sign.
10. |x| < a means −a < x < a.
11. A set with n elements has 2ⁿ subsets.
12. n(A ∪ B) = n(A) + n(B) − n(A ∩ B).

Practice questions

Which condition is necessary for ax² + bx + c = 0 to be a quadratic equation?
1. b ≠ 0
2. c ≠ 0
3. a = b
4. a ≠ 0
Answer
D. a ≠ 0
If a = 0 the x² term vanishes and the equation becomes linear.
The discriminant of the quadratic ax² + bx + c = 0 is
1. a² − 4bc
2. b² − 4ac
3. b² + 4ac
4. 4ac − b
Answer
B. b² − 4ac
D = b² − 4ac decides the nature of the roots.
The roots of x² − 5x + 6 = 0 are
1. 2 and 3
2. −2 and −3
3. 1 and 6
4. −1 and 6
Answer
A. 2 and 3
x² − 5x + 6 = (x − 2)(x − 3).
The sum of the roots of 2x² − 7x + 3 = 0 is
1. 3/2
2. 7/2
3. −3/2
4. −7/2
Answer
B. 7/2
Sum of roots = −b/a = 7/2.
The product of the roots of 3x² + 5x − 12 = 0 is
1. 4
2. −5/3
3. −4
4. 12
Answer
C. −4
Product = c/a = −12/3 = −4.
The nature of the roots of x² − 6x + 9 = 0 is
1. real and distinct
2. not real
3. irrational and unequal
4. real and equal
Answer
D. real and equal
D = 36 − 36 = 0, so the roots are real and equal (x = 3).
The roots of 2x² − 3x + 5 = 0 are
1. real and distinct
2. rational
3. real and equal
4. not real
Answer
D. not real
D = 9 − 40 = −31 < 0, so there are no real roots.
For what value of k does x² + kx + 16 = 0 have equal roots (k > 0)?
1. 4
2. 8
3. 16
4. 32
Answer
B. 8
D = k² − 64 = 0 gives k = 8 for k > 0.
The quadratic equation whose roots are 3 and −2 is
1. x² − x − 6 = 0
2. x² − x + 6 = 0
3. x² + x − 6 = 0
4. x² − 5x − 6 = 0
Answer
A. x² − x − 6 = 0
Sum = 1 and product = −6, so x² − x − 6 = 0.
One root of x² − 7x + k = 0 is 3. The value of k is
1. 10
2. 21
3. 4
4. 12
Answer
D. 12
The other root is 7 − 3 = 4, so k = 3 × 4 = 12.
If α and β are the roots of x² − 5x + 6 = 0, then α² + β² equals
1. 25
2. 37
3. 13
4. 11
Answer
C. 13
α² + β² = S² − 2P = 25 − 12 = 13.
If x + y = 10 and x − y = 4, the value of xy is
1. 21
2. 16
3. 40
4. 24
Answer
A. 21
Adding gives x = 7, so y = 3 and xy = 21.
The solution of 2x + 3y = 12 and 3x − y = 7 gives x equal to
1. 2
2. 4
3. 3
4. 5
Answer
C. 3
y = 3x − 7, so 2x + 9x − 21 = 12 and x = 3 (y = 2).
For what value of k do 2x + 3y = 5 and 4x + 6y = k have infinitely many solutions?
1. 5
2. 10
3. 6
4. 12
Answer
B. 10
Need 2/4 = 3/6 = 5/k, so k = 10.
For what value of k do 3x + ky = 7 and 6x + 4y = 11 have no solution?
1. 4
2. 6
3. 3
4. 2
Answer
D. 2
Need 3/6 = k/4, so k = 2. Then 7/11 ≠ 1/2, so the lines are parallel.
The solution of 3x − 7 > 8 is
1. x > 15
2. x > 5
3. x < 5
4. x < 3
Answer
B. x > 5
3x > 15 gives x > 5.
The solution of −2x > 6 is
1. x > 3
2. x < 3
3. x < −3
4. x > −3
Answer
C. x < −3
Dividing by −2 reverses the sign: x < −3.
The solution of |x| < 3 is
1. x < −3 or x > 3
2. x < 3 only
3. x > −3 only
4. −3 < x < 3
Answer
D. −3 < x < 3
|x| < a means −a < x < a.
The sum of the solutions of |2x − 1| = 5 is
1. 1
2. 5
3. −1
4. 3
Answer
A. 1
2x − 1 = 5 or −5 gives x = 3 or −2; the sum is 1.
The solution of x² − 5x + 6 < 0 is
1. 2 < x < 3
2. x > 3
3. x < 2
4. x < 2 or x > 3
Answer
A. 2 < x < 3
The product (x − 2)(x − 3) is negative only between the roots.
The number of integers satisfying −2 < x ≤ 3 is
1. 4
2. 6
3. 5
4. 3
Answer
C. 5
The integers are −1, 0, 1, 2, 3.
If n(A) = 20, n(B) = 15 and n(A ∩ B) = 5, then n(A ∪ B) is
1. 40
2. 30
3. 35
4. 25
Answer
B. 30
20 + 15 − 5 = 30.
The number of subsets of a set with 4 elements is
1. 8
2. 4
3. 15
4. 16
Answer
D. 16
2⁴ = 16.
The number of proper subsets of a set with 3 elements is
1. 8
2. 7
3. 6
4. 3
Answer
B. 7
2³ − 1 = 7.
If A = {1, 2, 3, 4} and B = {3, 4, 5, 6}, then A − B is
1. {1, 2}
2. {5, 6}
3. {3, 4}
4. {1, 2, 5, 6}
Answer
A. {1, 2}
A − B has the elements of A that are not in B.
In a class of 50 students, 30 play cricket and 25 play football. If every student plays at least one game, how many play both?
1. 20
2. 10
3. 5
4. 15
Answer
C. 5
50 = 30 + 25 − n(both), so n(both) = 5.
According to De Morgan's law, (A ∪ B)′ equals
1. A′ ∩ B′
2. A ∪ B′
3. A′ ∪ B′
4. A ∩ B
Answer
A. A′ ∩ B′
The complement of a union is the intersection of the complements.
Two positive numbers have sum 25 and product 144. The larger number is
1. 12
2. 18
3. 9
4. 16
Answer
D. 16
They are roots of x² − 25x + 144 = 0, which gives 16 and 9.
A rectangle has perimeter 34 and area 60. The length of its longer side is
1. 15
2. 12
3. 6
4. 10
Answer
B. 12
l + b = 17 and lb = 60 give 12 and 5.
Two positive numbers differ by 3 and the sum of their squares is 117. Their product is
1. 60
2. 40
3. 54
4. 36
Answer
C. 54
x² + (x + 3)² = 117 gives x = 6; the numbers are 6 and 9, and the product is 54.
The digits of a two-digit number add up to 9, and reversing the digits increases the number by 27. The number is
1. 18
2. 27
3. 36
4. 45
Answer
C. 36
y − x = 3 with x + y = 9 gives x = 3, y = 6.
5 pencils and 7 pens cost ₹50, and 7 pencils and 5 pens cost ₹46. The price of one pencil is
1. ₹3
2. ₹4
3. ₹5
4. ₹6
Answer
A. ₹3
Adding gives p + q = 8; subtracting gives p − q = −2. So p = 3.
Which of the statements is/are correct? 1. The sum of the roots of ax² + bx + c = 0 is −b/a. 2. The product of the roots is c/a.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Both are the standard results for the sum and product of roots.
Which of the statements is/are correct? 1. If D < 0, the roots are real and distinct. 2. If D = 0, the roots are real and equal.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
B. 2 only
D < 0 gives no real roots, so statement 1 is wrong.
Which of the statements is/are correct? 1. Multiplying an inequality by a negative number reverses its sign. 2. The inequality x² < 0 has real solutions.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
A square is never negative, so x² < 0 has no real solution.
Which of the statements is/are correct? 1. The empty set is a subset of every set. 2. The empty set is a proper subset of itself.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
A set is not a proper subset of itself, so statement 2 is false.
Which of the statements is/are correct? 1. A ∩ B is a subset of A. 2. A is a subset of A ∪ B.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Every element of A ∩ B lies in A, and every element of A lies in A ∪ B.
Which of the statements is/are correct? 1. Two lines that are parallel and distinct give a pair of equations with no solution. 2. Two coincident lines give infinitely many solutions.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
Parallel distinct lines never meet; coincident lines share every point.
Which of the statements is/are correct? 1. If D > 0, the roots are always rational. 2. The roots of x² − 2 = 0 are rational.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
D. Neither 1 nor 2
Roots are rational only when D is a perfect square; x² − 2 = 0 has roots ±√2.
Which of the statements is/are correct? 1. |x| > 2 means −2 < x < 2. 2. |x − 1| ≤ 3 means −2 ≤ x ≤ 4.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
B. 2 only
|x| > 2 means x < −2 or x > 2. Statement 2 is correct.
Which of the statements is/are correct? 1. A ∪ A′ = U. 2. A ∩ A′ = ∅.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
C. Both 1 and 2
These are the complement laws.
Which of the statements is/are correct? 1. x² − 5x + 6 < 0 has the solution 2 < x < 3. 2. x² − 5x + 6 > 0 for all x between 2 and 3.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
Between the roots the expression is negative, so statement 2 is false.
Which of the statements is/are correct? 1. A quadratic equation can have at most two real roots. 2. A quadratic equation always has at least one real root.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
With D < 0 there is no real root, so statement 2 is false.
Match the discriminant with the roots: (a) D > 0, (b) D = 0, (c) D < 0 with (i) no real roots, (ii) real and equal, (iii) real and distinct. The correct matching is
1. a-i, b-ii, c-iii
2. a-iii, b-i, c-ii
3. a-ii, b-iii, c-i
4. a-iii, b-ii, c-i
Answer
D. a-iii, b-ii, c-i
D > 0 gives distinct real roots, D = 0 equal roots and D < 0 no real roots.
The sum of the reciprocals of the roots of x² − 5x + 6 = 0 is
1. 1/6
2. −5/6
3. 6/5
4. 5/6
Answer
D. 5/6
1/α + 1/β = S/P = 5/6.

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1.The difference between a number and the square root of the number is 2. The number is:
2.The difference between two numbers is 8. If twice the larger number is 4 more than three times the smaller number, what is the larger number?
3.The expression x4 – 8x2 + m will be a perfect square when the value of m is:
4.The expression √(10 + 2(√6 - √15 - √10)) is equal to:
5.Seven years ago Ramu was five times as old as his son. The present age difference between Ramu and his son is 28 years. Find the sum of their present ages.
6.The ages of two siblings add up to 44 years. If one sibling is 4 years younger than twice the other's age, what is the difference between their ages?
7.The ages of a mother and daughter are 36 years and 12 years, respectively. In how many years will the mother be twice the age of the daughter?
8.Simplify the expression (36p² + 49q²) (6p + 7q) (6p – 7q)
9.Simplify (2pq + 11q)2 - (2pq - 11q)2.
10.The sum of two positive numbers is 26 and their product is 144. The positive difference between them is:

10 left. You can check any time.