Quadratic Equation | NEB Class 11 Mathematics Notes

Chapter Overview

Namaste students! Today we are diving into one of the most fundamental pillars of Algebra: Quadratic Equations. In simpler terms, we are moving from basic arithmetic to a world where we use variables (like $x$) to solve for unknowns. A polynomial is essentially a mathematical 'sentence' made of terms, and when we set that sentence equal to zero, it becomes an equation that we can solve.

Importance of the Chapter: This chapter is the 'bread and butter' of the NEB Class 11 Mathematics syllabus. It's not just a standalone topic; the concepts of roots, factors, and discriminants you learn here will be used in Calculus, Coordinate Geometry, and even complex numbers. In the NEB Board Exams, you can expect a mix of MCQs, short conceptual questions, and long analytical problems from this section.

Real-life Applications: Ever wondered how engineers calculate the path of a rocket? Or how architects design the curve of a bridge? The path followed by a projectile (like a football kicked in the air) is a Parabola, which is the graph of a quadratic equation. It is also used in business to find the 'break-even' point where profit equals zero.

What this chapter covers

Classification of Polynomials by DegreeThe Remainder Theorem and Factor TheoremQuadratic Equations and the Quadratic FormulaThe Discriminant and Nature of RootsSymmetric Functions of RootsRelation between Roots and CoefficientsCommon Roots Conditions between Quadratic Equations

Learning Objectives

1

Define and classify polynomials by their degree and identify leading coefficients.

2

Apply the Remainder Theorem and Factor Theorem to solve algebraic division and factorization problems.

3

Derive and apply the Quadratic Formula using the method of completing the square.

4

Determine the Nature of Roots of quadratic equations using the discriminant.

5

Establish and utilize the relation between roots and coefficients ($\alpha + \beta$ and $\alpha\beta$) to simplify expressions.

6

Solve problems involving symmetric functions of roots and construct new equations with modified roots.

7

Analyze and apply the conditions for common roots between two quadratic equations.

The Essence of Quadratics: Geometry and Roots

Key Insight

At its core, a quadratic equation $ax^2 + bx + c = 0$ is the algebraic representation of a parabola. The roots of the equation correspond to the points where this parabola intersects the $x$-axis. This bridge between algebra (solving equations) and geometry (graphical intercepts) is a fundamental connection in coordinate mathematics.

Think of it this way: Calculating the discriminant $D = b^2 - 4ac$ allows you to predict the exact geometric behavior of the parabola without actually drawing it. If $D > 0$, the parabola cuts the $x$-axis at two distinct points. If $D = 0$, it merely touches the $x$-axis at a single point (vertex). If $D < 0$, the parabola floats entirely above or below the $x$-axis, meaning it has no real intercepts (imaginary roots).

In NEB exams, the nature of roots is a heavily tested concept. Understanding this geometric connection makes remembering the three states of the discriminant intuitive and straightforward.

Key Definitions

Polynomial

An algebraic expression of the form $a_{0}+a_{1}x+a_{2}x^{2}+...+a_{n}x^{n}$ where $n$ is a non-negative integer and $a_n \neq 0$. The powers of the variable must be non-negative integers.

Degree

The highest power of the variable $x$ in a polynomial. It dictates the maximum number of roots the polynomial equation can have.

Leading Coefficient

The coefficient $a_n$ associated with the term of the highest power ($x^n$) in a polynomial.

Zero of a Polynomial

A value $a$ such that substituting $x = a$ into the polynomial yields $p(a) = 0$. Geometrically, it is the $x$-intercept of the polynomial's graph.

Quadratic Equation

A polynomial equation of degree 2, generally written in the standard form $ax^2 + bx + c = 0$, where $a, b, c$ are real coefficients and $a \neq 0$.

Identity

An equation that is true for every possible value of the variable within its domain. Unlike conditional equations, an identity has infinitely many solutions (e.g., $(x-2)^2 = x^2 - 4x + 4$).

Remainder Theorem vs. Factor Theorem

The Remainder and Factor theorems are closely related tools used to simplify polynomial expressions and find roots without performing full polynomial long division.

Remainder Theorem

$$Remainder = $f(a)$$$

Purpose: Calculates the remainder when a polynomial $f(x)$ is divided by a linear divisor of the form $(x - a)$.

Condition: The remainder $f(a)$ can be any real value, positive, negative, or zero.

Core Rule: If polynomial $f(x)$ is divided by $(x - a)$, the remainder is simply the value $f(a)$.

Example: Dividing $f(x) = x^2 + 3x + 5$ by $(x - 1)$ gives remainder $f(1) = 9$.

VS

Factor Theorem

$$$f(a) = 0 \iff (x-a)$ is a factor$$

Purpose: Determines whether a given linear term $(x - a)$ is an exact divisor (factor) of the polynomial $f(x)$.

Condition: The remainder must be exactly equal to zero.

Core Rule: $(x - a)$ is a factor of $f(x)$ if and only if $f(a) = 0$.

Example: For $f(x) = x^2 - 3x + 2$, we find $f(1) = 0$, proving $(x-1)$ is an exact factor.

Relationship: The Factor Theorem is a special case of the Remainder Theorem. When the remainder evaluated by the Remainder Theorem is exactly zero, it confirms factorization.

Formula Sheet & Key Properties

General Quadratic Form

ax^2 + bx + c = 0, \quad a \neq 0

The standard representation of a quadratic equation with real coefficients.

1Any equation containing $x^2$ as its highest degree term can be simplified to this standard format by grouping terms.

Quadratic Formula

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Used to calculate the roots of any quadratic equation.

1Start with standard form: $ax^2 + bx + c = 0$.

2Divide by $a$: $x^2 + \frac{b}{a}x + \frac{c}{a} = 0$.

3Isolate the variable terms: $x^2 + \frac{b}{a}x = -\frac{c}{a}$.

4Complete the square: Add $\left(\frac{b}{2a}\right)^2$ to both sides.

5$x^2 + 2(x)\left(\frac{b}{2a}\right) + \left(\frac{b}{2a}\right)^2 = \left(\frac{b}{2a}\right)^2 - \frac{c}{a}$.

6Factor the left side and find common denominator on the right: $\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$.

7Take square roots: $x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a}$.

8Solve for $x$: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. (Proved)

Discriminant and Nature of Roots

$$D = b^2 - 4ac$$

Indicates whether the roots are real, equal, or imaginary.

1If $D > 0$: The expression under the radical is positive, yielding two distinct real roots.

2If $D = 0$: The radical term disappears, leaving a single repeating real root: $x = -\frac{b}{2a}$.

3If $D < 0$: The negative value under the radical introduces the imaginary unit $i$, resulting in conjugate complex roots.

Relation between Roots and Coefficients

\alpha + \beta = -\frac{b}{a}, \quad \alpha\beta = \frac{c}{a}

Formulates the sum and product of roots in terms of the coefficients.

1Let $\alpha = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$ and $\beta = \frac{-b - \sqrt{b^2 - 4ac}}{2a}$.

2Add the roots: $\alpha + \beta = \frac{-b + \sqrt{D} - b - \sqrt{D}}{2a} = \frac{-2b}{2a} = -\frac{b}{a}$.

3Multiply the roots: $\alpha\beta = \frac{(-b + \sqrt{D})(-b - \sqrt{D})}{4a^2} = \frac{b^2 - D}{4a^2} = \frac{b^2 - (b^2 - 4ac)}{4a^2} = \frac{4ac}{4a^2} = \frac{c}{a}$.

Quadratic Equation Formation

x^2 - (\alpha + \beta)x + \alpha\beta = 0

Builds a quadratic equation if its roots $\alpha$ and $\beta$ are known.

1Starting from factors: $(x - \alpha)(x - \beta) = 0$.

2Expand the binomials: $x^2 - \beta x - \alpha x + \alpha\beta = 0$.

3Group the linear terms: $x^2 - (\alpha + \beta)x + \alpha\beta = 0$. (Proved)

Symmetric Functions of Roots

\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta, \quad \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)

Symmetric identities used to evaluate algebraic combinations of roots.

1For squares: Expand $(\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2$, then isolate $\alpha^2 + \beta^2$.

2For cubes: Expand $(\alpha + \beta)^3 = \alpha^3 + \beta^3 + 3\alpha^2\beta + 3\alpha\beta^2$, factor out $3\alpha\beta$, and isolate $\alpha^3 + \beta^3$.

Common Root Condition

$$(bc' - b'c)(ab' - a'b) = (ca' - c'a)^2$$

The strict algebraic condition for two equations $ax^2 + bx + c = 0$ and $a'x^2 + b'x + c' = 0$ to share a common root.

1Let $\alpha$ be the common root: $a\alpha^2 + b\alpha + c = 0$ and $a'\alpha^2 + b'\alpha + c' = 0$.

2Using Cramer's Rule for cross-multiplication on $\alpha^2$, $\alpha$, and $1$:

3$\frac{\alpha^2}{bc' - b'c} = \frac{\alpha}{ca' - c'a} = \frac{1}{ab' - a'b}$.

4From these ratios: $\alpha^2 = \frac{bc' - b'c}{ab' - a'b}$ and $\alpha = \frac{ca' - c'a}{ab' - a'b}$.

5Equating $\alpha^2 = (\alpha)^2$: $\frac{bc' - b'c}{ab' - a'b} = \left(\frac{ca' - c'a}{ab' - a'b}\right)^2$.

6Simplify: $(bc' - b'c)(ab' - a'b) = (ca' - c'a)^2$. (Proved)

Step-by-Step: Solving Polynomial and Quadratic Problems

Use this structured decision sequence to determine the appropriate mathematical tool for any polynomial or quadratic question.

Is the problem asking to find a remainder or check divisibility?

Yes

Apply the Remainder Theorem ($R = f(a)$) or Factor Theorem ($f(a) = 0$).

No

Continue to next step ↓

Does the question ask about the nature of roots (real, equal, complex)?

Yes

Compute the discriminant $D = b^2 - 4ac$ and analyze its sign relative to zero.

No

Continue to next step ↓

Does the problem require relating roots to algebraic coefficients?

Yes

Write down $\alpha + \beta = -b/a$ and $\alpha\beta = c/a$, and use symmetric identities to solve.

No

Continue to next step ↓

Does the question involve common roots between two equations?

Yes

Substitute a shared root variable $\alpha$ into both equations, or directly apply the common root condition formula.

No

If you need to find the roots directly, apply the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Top 5 Memory Tricks for Quadratics

Trick: These structural associations and rules-of-thumb help prevent algebraic slips during examinations.

Steps to Remember:

Trick 1 - Vowel Rule for Roots: Sum corresponds to the Second term coefficient (with a negative sign): $-\frac{b}{a}$. Product corresponds to the Past (constant) term: $\frac{c}{a}$.
Trick 2 - Nature of Roots Placement: Think of the discriminant $D$ relative to a horizontal scale: $D = 0$ is the 'Dead center' boundary (equal roots); $D > 0$ means 'Greater is Real' (two distinct roots); $D < 0$ means 'Less is Imaginary' (no real roots).
Trick 3 - Reciprocal Coefficient Flip: If a quadratic equation has coefficients where the first equals the last ($a = c$), its roots are reciprocals of each other (i.e., $\alpha$ and $\frac{1}{\alpha}$). Remember: 'First equals Last $\rightarrow$ Flip the roots.'
Trick 4 - Sum of Coefficients Shortcut: If you notice that the coefficients of $ax^2 + bx + c = 0$ sum up to zero ($a + b + c = 0$), then $x = 1$ is always one of the roots, and the other root is $\frac{c}{a}$.
Trick 5 - Symmetric Square Formation: When finding $\alpha^2 + \beta^2$, visualize a square area: you take the whole sum square $(\alpha + \beta)^2$ and subtract the two rectangular products $-2\alpha\beta$ to clean up the middle.

Mnemonic: Remember: Discriminant checks nature. Sum is negative second over first. Product is last over first. If coefficients sum to zero, 1 is a root.

Common Mistakes in NEB Polynomial Exams

Avoid these typical algebraic pitfalls to preserve maximum credit on your solutions.

Mistake 1 - Sign Errors in Sum of Roots: Forgetting the negative sign in the relation $\alpha + \beta = -b/a$. For example, in $x^2 - 5x + 6 = 0$, the sum is $+5$, not $-5$. Always explicitly write out $b$ with its sign before applying the formula.
Mistake 2 - Non-Polynomial Identification: Classifying expressions with negative or fractional exponents (such as $x + 1/x$ or $\sqrt{x} + 2$) as polynomials. Polynomial powers must exclusively be non-negative integers ($0, 1, 2, \dots$).
Mistake 3 - Discriminant Calculation with Negative 'c': Computing $b^2 - 4ac$ when $c$ is negative and forgetting to convert the operation to addition. For $x^2 + 2x - 8 = 0$, the calculation is $2^2 - 4(1)(-8) = 4 + 32 = 36$, not $4 - 32$.
Mistake 4 - Confusing Identical Roots with Single Root: Stating a quadratic equation with $D = 0$ has only 'one root'. It technically has two real roots that happen to be equal (repeated roots). This distinction is critical in higher-level algebra.
Mistake 5 - Misinterpreting Common Root Conditions: Using the common root formula with arbitrary coefficient orders. Always align both quadratic equations into the standard $ax^2 + bx + c = 0$ format before identifying $a, b, c$ and $a', b', c'$.

Solved Examples (Easy → Medium → Hard)

1

Determine the nature of the roots of the quadratic equation $x^2 - 8x + 16 = 0$.

Easy

Solution

Identify the coefficients from the standard form $ax^2 + bx + c = 0$:
$a = 1, \quad b = -8, \quad c = 16$.

Compute the discriminant $D$:
$D = b^2 - 4ac$
$D = (-8)^2 - 4(1)(16)$
$D = 64 - 64 = 0$.

Since the discriminant is exactly zero, the roots are real and equal.

∴ Roots are real and equal.

2

If the equation $x^2 + (k+3)x + 3k = 0$ has equal roots, find the value of $k$.

Medium

Solution

For the quadratic equation to have equal roots, its discriminant must be zero ($D = 0$).

Identify coefficients:
$a = 1, \quad b = (k+3), \quad c = 3k$.

Set the discriminant formula to zero:
$b^2 - 4ac = 0$
$(k+3)^2 - 4(1)(3k) = 0$

Expand and simplify:
$k^2 + 6k + 9 - 12k = 0$
$k^2 - 6k + 9 = 0$

Factor the quadratic expression:
$(k-3)^2 = 0$
$k - 3 = 0 \implies k = 3$.

∴ $k = 3$

3

Prove that the equation $(x-p)(x-q) + (x-q)(x-r) + (x-r)(x-p) = 0$ has equal roots if and only if $p = q = r$.

Hard

Solution

First, expand each product of binomials:
$(x^2 - (p+q)x + pq) + (x^2 - (q+r)x + qr) + (x^2 - (r+p)x + rp) = 0$.

Group the like terms to express in standard form $Ax^2 + Bx + C = 0$:
$3x^2 - 2(p+q+r)x + (pq + qr + rp) = 0$.

For equal roots, set the discriminant $D = B^2 - 4AC = 0$:
$[-2(p+q+r)]^2 - 4(3)(pq + qr + rp) = 0$
$4(p+q+r)^2 - 12(pq + qr + rp) = 0$.

Divide the entire equation by 4:
$(p+q+r)^2 - 3(pq + qr + rp) = 0$.

Expand the squared trinomial:
$p^2 + q^2 + r^2 + 2pq + 2qr + 2rp - 3pq - 3qr - 3rp = 0$
$p^2 + q^2 + r^2 - pq - qr - rp = 0$.

Multiply the entire expression by 2 to construct perfect squares:
$2p^2 + 2q^2 + 2r^2 - 2pq - 2qr - 2rp = 0$
$(p^2 - 2pq + q^2) + (q^2 - 2qr + r^2) + (r^2 - 2rp + p^2) = 0$
$(p-q)^2 + (q-r)^2 + (r-p)^2 = 0$.

Since the terms are squared real numbers, they cannot be negative. The only way their sum can equal zero is if each individual squared term is zero:
$p-q = 0 \implies p = q$
$q-r = 0 \implies q = r$
$r-p = 0 \implies r = p$.

Thus, the roots are equal if and only if $p = q = r$.

∴ Proved by showing $D = 0 \iff (p-q)^2 + (q-r)^2 + (r-p)^2 = 0$, which requires $p = q = r$.

MCQ Practice

Mixed Level — MCQ

1

Which of the following algebraic expressions is NOT a polynomial?

Correct!

Incorrect. Correct: $x + \frac{1}{x}$

Solution

By definition, a polynomial cannot contain variables with negative or fractional powers. The term $\frac{1}{x}$ has exponent $-1$, disqualifying the expression.

∴ $x + \frac{1}{x}$

2

What is the degree of the polynomial $f(x) = (x-1)(x-2)(x-3)$?

Correct!

Incorrect. Correct: 3

Solution

Expanding $(x-1)(x-2)(x-3)$ gives $x^3 - 6x^2 + 11x - 6$. The highest exponent of the variable $x$ is 3.

∴ 3

3

If the roots of the quadratic equation $2x^2 - kx + 8 = 0$ are equal, what is the value of $k$?

Correct!

Incorrect. Correct: $\pm 8$

Solution

For equal roots, set $b^2 - 4ac = 0$. Substituting $a=2$, $b=-k$, and $c=8$ gives: $k^2 - 4(2)(8) = 0 \implies k^2 - 64 = 0 \implies k^2 = 64 \implies k = \pm 8$.

∴ $\pm 8$

4

If one root of the quadratic equation $ax^2 + bx + c = 0$ is zero, which of the following must be true?

Correct!

Incorrect. Correct: $c = 0$

Solution

Let the roots be $\alpha$ and $\beta$. If $\alpha = 0$, then the product of roots $\alpha\beta = 0 \cdot \beta = 0$. Since $\alpha\beta = \frac{c}{a}$, we must have $\frac{c}{a} = 0 \implies c = 0$.

∴ $c = 0$

5

The sum of the roots of the quadratic equation $3x^2 + 9x - 5 = 0$ is:

Correct!

Incorrect. Correct: -3

Solution

Using the relation between roots and coefficients: $\text{Sum} = -\frac{b}{a}$. For the equation $3x^2 + 9x - 5 = 0$, $a = 3$ and $b = 9$. Thus, $\text{Sum} = -\frac{9}{3} = -3$.

∴ -3

6

If the roots of a quadratic equation are equal in magnitude but opposite in sign, then:

Correct!

Incorrect. Correct: $b = 0$

Solution

Let the roots be $\alpha$ and $-\alpha$. Their sum is $\alpha + (-\alpha) = 0$. Since the sum of roots is given by $-\frac{b}{a}$, we set $-\frac{b}{a} = 0 \implies b = 0$.

∴ $b = 0$

7

The roots of a quadratic equation are imaginary if its discriminant is:

Correct!

Incorrect. Correct: Negative

Solution

In the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$, the term under the radical is $D = b^2 - 4ac$. If $D < 0$, we get the square root of a negative number, which yields imaginary roots.

∴ Negative

8

Which quadratic equation has roots of 2 and -5?

Correct!

Incorrect. Correct: $x^2 + 3x - 10 = 0$

Solution

Calculate the sum and product of the given roots:
$\text{Sum} = 2 + (-5) = -3$
$\text{Product} = 2 \times (-5) = -10$.
The quadratic equation formula is $x^2 - (\text{Sum})x + \text{Product} = 0$. Substituting these values:
$x^2 - (-3)x + (-10) = 0 \implies x^2 + 3x - 10 = 0$.

∴ $x^2 + 3x - 10 = 0$

9

If one root of the quadratic equation $ax^2 + bx + c = 0$ is $2 + i$, what is the other root?

Correct!

Incorrect. Correct: $2 - i$

Solution

By the Complex Conjugate Root Theorem, if a polynomial has real coefficients, any complex roots must occur in conjugate pairs. Since $2 + i$ is a root, its conjugate $2 - i$ must also be a root.

∴ $2 - i$

10

What is the algebraic condition for two quadratic equations to share both roots in common?

Correct!

Incorrect. Correct: $\frac{a}{a'} = \frac{b}{b'} = \frac{c}{c'}$

Solution

When two quadratic equations share both roots, they are identical equations scaled by a constant factor. Therefore, the ratio of their corresponding coefficients must be equal: $\frac{a}{a'} = \frac{b}{b'} = \frac{c}{c'}$.

∴ $\frac{a}{a'} = \frac{b}{b'} = \frac{c}{c'}$

Short Questions (2-4 Marks Each)

1

Find the remainder when $2x^3 - 3x^2 + x - 5$ is divided by $(x-2)$.

[2 marks]

Solution

Let $f(x) = 2x^3 - 3x^2 + x - 5$. By the Remainder Theorem, dividing $f(x)$ by $(x-2)$ yields a remainder equal to $f(2)$.

Evaluate $f(2)$:
$f(2) = 2(2)^3 - 3(2)^2 + 2 - 5$
$f(2) = 2(8) - 3(4) + 2 - 5$
$f(2) = 16 - 12 + 2 - 5$
$f(2) = 1$.

The remainder is 1.

∴ Remainder = 1

2

If $(x+1)$ is a factor of $x^3 + kx^2 - x + 2$, find the value of $k$.

[2 marks]

Solution

Let $f(x) = x^3 + kx^2 - x + 2$. By the Factor Theorem, since $(x+1)$ is a factor of $f(x)$, we must have $f(-1) = 0$.

Substitute $x = -1$ into the polynomial:
$f(-1) = (-1)^3 + k(-1)^2 - (-1) + 2 = 0$
$-1 + k(1) + 1 + 2 = 0$
$k + 2 = 0 \implies k = -2$.

∴ $k = -2$

3

Form the quadratic equation whose roots are $3 \pm \sqrt{2}$.

[2 marks]

Solution

Let the roots be $\alpha = 3 + \sqrt{2}$ and \beta = 3 - \sqrt{2}$.

Calculate the sum of the roots:
$\text{Sum} = \alpha + \beta = (3 + \sqrt{2}) + (3 - \sqrt{2}) = 6$.

Calculate the product of the roots:
$\text{Product} = \alpha\beta = (3 + \sqrt{2})(3 - \sqrt{2}) = 3^2 - (\sqrt{2})^2 = 9 - 2 = 7$.

Construct the equation:
$x^2 - (\text{Sum})x + \text{Product} = 0$
$x^2 - 6x + 7 = 0$.

∴ $x^2 - 6x + 7 = 0$

4

Find the value of $m$ if the product of the roots of the equation $mx^2 - 5x + 12 = 0$ is 3.

[2 marks]

Solution

For the equation $mx^2 - 5x + 12 = 0$, we have coefficients $a = m, b = -5, c = 12$.

The product of the roots is given by the formula:
$\text{Product} = \frac{c}{a} = \frac{12}{m}$.

We are given that this product is 3:
$\frac{12}{m} = 3 \implies 3m = 12 \implies m = 4$.

∴ $m = 4$

5

Discuss the nature of the roots of the quadratic equation $3x^2 - 5x + 4 = 0$.

[2 marks]

Solution

Identify coefficients: $a = 3, b = -5, c = 4$.

Calculate the discriminant $D$:
$D = b^2 - 4ac$
$D = (-5)^2 - 4(3)(4)$
$D = 25 - 48 = -23$.

Since the discriminant is negative ($D < 0$), the roots are imaginary (complex conjugates) and unequal.

∴ Roots are imaginary and unequal.

6

Find the sum of the squares of the roots of the quadratic equation $x^2 - 5x + 6 = 0$.

[3 marks]

Solution

Identify coefficients: $a = 1, b = -5, c = 6$.

Find the sum and product of the roots $\alpha$ and $\beta$:
$\alpha + \beta = -\frac{b}{a} = -\frac{-5}{1} = 5$
$\alpha\beta = \frac{c}{a} = \frac{6}{1} = 6$.

Use the algebraic identity for the sum of squares:
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$
$\alpha^2 + \beta^2 = (5)^2 - 2(6)$
$\alpha^2 + \beta^2 = 25 - 12 = 13$.

∴ Sum of squares = 13

7

If the roots of the quadratic equation $4x^2 + px + 9 = 0$ are equal, find the value of $p$.

[2 marks]

Solution

For equal roots, set the discriminant $D = b^2 - 4ac = 0$.

Identify coefficients: $a = 4, b = p, c = 9$.

Substitute into the discriminant condition:
$p^2 - 4(4)(9) = 0$
$p^2 - 144 = 0$
$p^2 = 144 \implies p = \pm 12$.

∴ $p = \pm 12$

8

If one root of the quadratic equation $ax^2 + bx + c = 0$ is the reciprocal of the other, what is the relation between $a, b, c$?

[2 marks]

Solution

Let one root be $\alpha$. The other root is its reciprocal, which is $\frac{1}{\alpha}$.

Calculate the product of these roots:
$\text{Product} = \alpha \cdot \left(\frac{1}{\alpha}\right) = 1$.

From the relation between roots and coefficients, the product of roots is $\frac{c}{a}$:
$\frac{c}{a} = 1 \implies c = a$.

Thus, the constant term must equal the leading coefficient.

∴ Relation: $c = a$

9

Determine if $x = 3$ is a zero of the polynomial $p(x) = x^2 - 2x - 3$.

[2 marks]

Solution

To check if $x = 3$ is a zero, substitute $x = 3$ into the polynomial and evaluate the result:
$p(3) = (3)^2 - 2(3) - 3$
$p(3) = 9 - 6 - 3$
$p(3) = 9 - 9 = 0$.

Since $p(3) = 0$, $x = 3$ is indeed a zero of the polynomial.

∴ Yes, $x = 3$ is a zero of $p(x)$.

10

Find the value of $k$ if the quadratic equation $x^2 - 5x + k = 0$ has one root equal to zero.

[2 marks]

Solution

If $x = 0$ is a root of the equation, substituting $x = 0$ must satisfy the equation:
$(0)^2 - 5(0) + k = 0$
$0 - 0 + k = 0 \implies k = 0$.

Alternatively, since one root is 0, the product of roots must be 0, which means $\frac{c}{a} = \frac{k}{1} = 0 \implies k = 0$.

∴ $k = 0$

Long Questions (5-8 Marks Each)

1

If $\alpha$ and $\beta$ are the roots of the quadratic equation $ax^2 + bx + c = 0$, find the algebraic value of $\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha}$.

[5 marks]

Solution

We are given that $\alpha$ and $\beta$ are roots of $ax^2 + bx + c = 0$. Therefore:
$\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.

We want to find the value of:
$\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha}$.

Combine the terms over a common denominator:
$\frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} = \frac{\alpha^3 + \beta^3}{\alpha\beta}$.

Using the algebraic identity for sum of cubes:
$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$.

Substitute this into the expression:
$\frac{\alpha^3 + \beta^3}{\alpha\beta} = \frac{(\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)}{\alpha\beta}$.

Now substitute the root-coefficient values:
$\frac{\left(-\frac{b}{a}\right)^3 - 3\left(\frac{c}{a}\right)\left(-\frac{b}{a} ight)}{\frac{c}{a}}$

$= \frac{-\frac{b^3}{a^3} + \frac{3bc}{a^2}}{\frac{c}{a}}$.

Find a common denominator for the numerator:
$-\frac{b^3}{a^3} + \frac{3abc}{a^3} = \frac{3abc - b^3}{a^3}$.

Perform the final division:
$\frac{\frac{3abc - b^3}{a^3}}{\frac{c}{a}} = \frac{3abc - b^3}{a^3} \cdot \frac{a}{c} = \frac{3abc - b^3}{a^2c}$.

∴ $\frac{3abc - b^3}{a^2c}$

2

Prove that the roots of $(a-b)x^2 + 2(a+b)x - (a-b) = 0$ are always real and unequal for real $a \neq b$.

[6 marks]

Solution

Identify coefficients of the quadratic equation:
$A = (a-b), \quad B = 2(a+b), \quad C = -(a-b)$.

Compute the discriminant $D = B^2 - 4AC$:
$D = [2(a+b)]^2 - 4(a-b)[-(a-b)]$
$D = 4(a+b)^2 + 4(a-b)^2$.

Factor out the constant 4:
$D = 4[(a^2 + 2ab + b^2) + (a^2 - 2ab + b^2)]$
$D = 4[2a^2 + 2b^2]$
$D = 8(a^2 + b^2)$.

Analyze the result:
1. Since $a$ and $b$ are real numbers, their squares $a^2$ and $b^2$ are always non-negative ($a^2 \geq 0, b^2 \geq 0$).
2. Since $a \neq b$, both $a$ and $b$ cannot be zero simultaneously. Thus, $a^2 + b^2 > 0$.
3. Multiplying by 8 yields $D = 8(a^2 + b^2) > 0$.

Since the discriminant is strictly positive ($D > 0$), the roots are always real and unequal.

∴ Proved: Discriminant $D = 8(a^2 + b^2) > 0$, ensuring real and unequal roots.

3

If the roots of the quadratic equation $x^2 + px + q = 0$ differ by 1, prove that $p^2 = 4q + 1$.

[5 marks]

Solution

Let the roots be $\alpha$ and $\beta$. We are given that the difference between the roots is 1:
$|\alpha - \beta| = 1$.

Square both sides of this condition:
$(\alpha - \beta)^2 = 1$.

Using the algebraic identity relating difference of squares to sum and product:
$(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$.

Substitute this identity back into the squared condition:
$(\alpha + \beta)^2 - 4\alpha\beta = 1$.

From the equation $x^2 + px + q = 0$, write the sum and product of roots:
$\alpha + \beta = -p$
$\alpha\beta = q$.

Substitute these values into the squared difference equation:
$(-p)^2 - 4q = 1$
$p^2 - 4q = 1$
$p^2 = 4q + 1$. (Proved)

∴ Proved using $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$.

4

Find the values of $k$ such that the equations $x^2 + kx + 1 = 0$ and $x^2 + x + k = 0$ have a common root.

[6 marks]

Solution

Let $\alpha$ be the common root of both equations. Therefore, it must satisfy both equations simultaneously:
1) $\alpha^2 + k\alpha + 1 = 0$
2) $\alpha^2 + \alpha + k = 0$.

Subtract equation (2) from equation (1):
$(\alpha^2 + k\alpha + 1) - (\alpha^2 + \alpha + k) = 0$
$(k - 1)\alpha + (1 - k) = 0$
$(k - 1)\alpha = k - 1$.

Case 1: If $k - 1 \neq 0$ (i.e., $k \neq 1$), we can divide both sides by $(k-1)$:
$\alpha = 1$.

Substitute $\alpha = 1$ back into equation (1) to solve for $k$:
$(1)^2 + k(1) + 1 = 0$
$1 + k + 1 = 0 \implies k = -2$.

Case 2: If $k - 1 = 0 \implies k = 1$.
If $k = 1$, both original quadratic equations become identical: $x^2 + x + 1 = 0$. This means they share both roots in common, which naturally satisfies the requirement of having a common root.

Therefore, the possible values of $k$ are $1$ and $-2$.

∴ $k = 1$ or $k = -2$

5

If the ratio of the roots of the equation $ax^2 + bx + c = 0$ is $3:4$, prove that $12b^2 = 49ac$.

[5 marks]

Solution

Let the roots of the equation be $3k$ and $4k$.

From the relation between roots and coefficients:
1) Sum of roots:
$3k + 4k = -\frac{b}{a} \implies 7k = -\frac{b}{a} \implies k = -\frac{b}{7a}$.

2) Product of roots:
$(3k)(4k) = \frac{c}{a} \implies 12k^2 = \frac{c}{a}$.

Substitute the value of $k$ from the sum equation into the product equation:
$12\left(-\frac{b}{7a}\right)^2 = \frac{c}{a}$
$12\left(\frac{b^2}{49a^2} ight) = \frac{c}{a}$
$\frac{12b^2}{49a^2} = \frac{c}{a}$.

Multiply both sides by $a^2$ to clear the denominator:
$\frac{12b^2}{49} = ac \implies 12b^2 = 49ac$. (Proved)

∴ Proved by substituting $k = -\frac{b}{7a}$ into $12k^2 = \frac{c}{a}$.

6

If one root of the quadratic equation $ax^2 + bx + c = 0$ is the square of the other, prove that $b^3 + a^2c + ac^2 = 3abc$.

[8 marks]

Solution

Let the roots of the equation be $\alpha$ and $\alpha^2$.

Write down the relationships between the roots and coefficients:
1) $\alpha + \alpha^2 = -\frac{b}{a}$
2) $\alpha \cdot \alpha^2 = \alpha^3 = \frac{c}{a}$.

Take the cube of both sides of the sum equation (1):
$(\alpha + \alpha^2)^3 = \left(-\frac{b}{a}\right)^3$.

Expand the left side using $(x+y)^3 = x^3 + y^3 + 3xy(x+y)$:
$\alpha^3 + (\alpha^2)^3 + 3\alpha(\alpha^2)(\alpha + \alpha^2) = -\frac{b^3}{a^3}$
$\alpha^3 + (\alpha^3)^2 + 3\alpha^3(\alpha + \alpha^2) = -\frac{b^3}{a^3}$.

Substitute $\alpha^3 = \frac{c}{a}$ and $\alpha + \alpha^2 = -\frac{b}{a}$ into this expanded equation:
$\frac{c}{a} + \left(\frac{c}{a}\right)^2 + 3\left(\frac{c}{a}\right)\left(-\frac{b}{a}\right) = -\frac{b^3}{a^3}$
$\frac{c}{a} + \frac{c^2}{a^2} - \frac{3bc}{a^2} = -\frac{b^3}{a^3}$.

Multiply the entire equation by $a^3$ to clear the denominators:
$a^2c + ac^2 - 3abc = -b^3$.

Rearrange the terms to get the target proof expression:
$b^3 + a^2c + ac^2 = 3abc$. (Proved)

∴ Proved by cubing $\alpha + \alpha^2 = -\frac{b}{a}$ and substituting $\alpha^3 = \frac{c}{a}$.

7

Form the quadratic equation whose roots are the cubes of the roots of the equation $x^2 - px + q = 0$.

[5 marks]

Solution

Let the roots of the original equation $x^2 - px + q = 0$ be $\alpha$ and $\beta$.

From the coefficients, we find:
$\alpha + \beta = p$
$\alpha\beta = q$.

We need to form a new quadratic equation whose roots are $\alpha^3$ and $\beta^3$.

Calculate the sum of the new roots ($S$):
$S = \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$
$S = p^3 - 3qp$.

Calculate the product of the new roots ($P$):
$P = \alpha^3\beta^3 = (\alpha\beta)^3 = q^3$.

The new quadratic equation is represented by:
$x^2 - Sx + P = 0$
$x^2 - (p^3 - 3pq)x + q^3 = 0$.

∴ $x^2 - (p^3 - 3pq)x + q^3 = 0$

8

If the equations $x^2 + ax + bc = 0$ and $x^2 + bx + ca = 0$ have a common root, show that their other roots satisfy the equation $x^2 + cx + ab = 0$ (assume $a \neq b$).

[8 marks]

Solution

Let the common root be $\alpha$. It must satisfy both equations:
1) $\alpha^2 + a\alpha + bc = 0$
2) $\alpha^2 + b\alpha + ca = 0$.

Subtract equation (2) from equation (1):
$(a - b)\alpha + bc - ca = 0$
$(a - b)\alpha - c(a - b) = 0$
$(a - b)(\alpha - c) = 0$.

Since $a \neq b$, we divide by $(a-b)$ to get:
$\alpha = c$.
So, the common root of both equations is $c$.

Let the second root of the first equation ($x^2 + ax + bc = 0$) be $\beta_1$. Since the product of its roots is $bc$:
$\alpha \cdot \beta_1 = bc \implies c \cdot \beta_1 = bc \implies \beta_1 = a$.

Let the second root of the second equation ($x^2 + bx + ca = 0$) be $\beta_2$. Since the product of its roots is $ca$:
$\alpha \cdot \beta_2 = ca \implies c \cdot \beta_2 = ca \implies \beta_2 = b$.

Thus, the non-common (other) roots of the two equations are $a$ and $b$. We need to construct the equation satisfied by $a$ and $b$:
$x^2 - (a + b)x + ab = 0$.

Since the common root $c$ is a solution to the first equation:
$c^2 + ac + bc = 0 \implies c(c + a + b) = 0$.
Since $c \neq 0$ (otherwise the equations would have $c=0$, yielding $a=b$), we have:
$a + b + c = 0 \implies a + b = -c$.

Substitute $a + b = -c$ back into our constructed equation:
$x^2 - (-c)x + ab = 0$
$x^2 + cx + ab = 0$. (Proved)

∴ Proved by finding the other roots are $a$ and $b$, and substituting $a+b = -c$.

9

Prove that the roots of the equation $(x-a)(x-b) = k^2$ are always real for all real values of $a, b, k$.

[5 marks]

Solution

First, expand the given equation and rearrange it into standard quadratic form:
$(x-a)(x-b) = k^2$
$x^2 - (a+b)x + ab = k^2$
$x^2 - (a+b)x + (ab - k^2) = 0$.

Identify coefficients: $A = 1, B = -(a+b), C = ab - k^2$.

Calculate the discriminant $D = B^2 - 4AC$:
$D = [-(a+b)]^2 - 4(1)(ab - k^2)$
$D = (a+b)^2 - 4ab + 4k^2$.

Simplify the algebraic expression:
$D = (a^2 + 2ab + b^2) - 4ab + 4k^2$
$D = a^2 - 2ab + b^2 + 4k^2$
$D = (a - b)^2 + 4k^2$.

Analyze the sign of $D$:
1. Since $a$ and $b$ are real, $(a-b)^2 \geq 0$.
2. Since $k$ is real, $4k^2 \geq 0$.
3. Therefore, $D = (a-b)^2 + 4k^2$ is always the sum of two non-negative terms, meaning $D \geq 0$ for all real values of $a, b, k$.

Since the discriminant is non-negative ($D \geq 0$), the roots of the equation are always real.

∴ Proved: Discriminant $D = (a-b)^2 + 4k^2 \geq 0$, ensuring real roots.

10

Solve the equation $x^2 - 6|x| + 8 = 0$ for real values of $x$.

[6 marks]

Solution

Recall the property of absolute values that $x^2 = |x|^2$. We can rewrite the given equation as:
$|x|^2 - 6|x| + 8 = 0$.

Let $y = |x|$. This substitution transforms the equation into a standard quadratic equation in terms of $y$:
$y^2 - 6y + 8 = 0$.

Factor the quadratic equation:
$(y - 2)(y - 4) = 0$.
This gives two possible values for $y$:
$y = 2 \quad \text{or} \quad y = 4$.

Substitute back $y = |x|$:

Case 1: $|x| = 2$
$x = \pm 2$.

Case 2: $|x| = 4$
$x = \pm 4$.

Thus, the complete set of solutions is $x \in \{-4, -2, 2, 4\}$.

∴ $x = \pm 2$ or $x = \pm 4$

One-Page Revision Cheat Sheet

1

Polynomial Definition: Algebraic expression with non-negative integer powers only. Expressions containing $\sqrt{x}$ or $1/x$ are not polynomials.

2

Remainder Theorem: Dividing a polynomial $f(x)$ by $(x-a)$ yields a remainder equal to $f(a)$ directly without requiring long division.

3

Factor Theorem: A binomial $(x-a)$ is a factor of $f(x)$ if and only if the evaluated value $f(a) = 0$.

4

Standard Quadratic Form: $ax^2 + bx + c = 0$ where $a \neq 0$.

5

The Quadratic Formula: Derived by completing the square, yielding $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

6

The Discriminant ($D = b^2 - 4ac$): Tells the nature of roots. $D > 0 \rightarrow$ Real & Unequal; $D = 0 \rightarrow$ Real & Equal; $D < 0 \rightarrow$ Imaginary conjugate pairs.

7

Root-Coefficient Relations: Sum of roots $\alpha + \beta = -b/a$. Product of roots $\alpha\beta = c/a$. Always pay close attention to signs.

8

Forming New Equations: If sum of roots is $S$ and product is $P$, the corresponding equation is $x^2 - Sx + P = 0$.

9

Symmetric Identities: Expressions like $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$ allow computing root structures without solving for roots.

10

Common Root Condition: Two quadratics share a single root when $(bc' - b'c)(ab' - a'b) = (ca' - c'a)^2$. If both roots are common, then $\frac{a}{a'} = \frac{b}{b'} = \frac{c}{c'}$.

11

Imaginary Conjugate Rule: Complex roots of equations with real coefficients always arrive in pairs. If $p + iq$ is a root, $p - iq$ is guaranteed to be a root.

12

The $a+b+c=0$ Shortcut: If sum of coefficients in $ax^2 + bx + c = 0$ is zero, then $x = 1$ is a root and $x = c/a$ is the other.