Class 10 Math Chapter 2 Study Guide: Theory of Quadratic Equations & Cube Roots of Unity | FBISE & Punjab Board Notes PDF

Theoretical Foundations & Core Concepts

The Theory of Quadratic Equations forms the algebraic backbone of Higher Secondary and Matriculation Mathematics in Pakistan. A standard quadratic equation is defined as $ax^2 + bx + c = 0$ where $a \neq 0$ and $a, b, c \in \mathbb{R}$.

1. The Discriminant ($\text{Disc}$) and Nature of Roots

The quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ reveals that the expression inside the square root determines the nature of the roots. This expression is called the Discriminant ($\Delta$):

$$\Delta = b^2 - 4ac$$

• Case 1 ($\Delta > 0$ and a perfect square): The roots are real, rational, and unequal (distinct).
• Case 2 ($\Delta > 0$ and not a perfect square): The roots are real, irrational, and unequal (distinct). Irrational roots always occur in conjugate pairs ($m \pm \sqrt{n}$).
• Case 3 ($\Delta = 0$): The roots are real, rational, and equal ($x = -\frac{b}{2a}$).
• Case 4 ($\Delta < 0$): The roots are imaginary / complex conjugates ($p \pm iq$).

2. Cube Roots of Unity ($\omega$ and $\omega^2$)

Solving the equation $x^3 - 1 = 0 \implies (x - 1)(x^2 + x + 1) = 0$ yields three roots: one real root ($x = 1$) and two complex conjugate roots:

$$\omega = \frac{-1 + i\sqrt{3}}{2} \quad \text{and} \quad \omega^2 = \frac{-1 - i\sqrt{3}}{2}$$

Fundamental Properties of Cube Roots of Unity:

1. Sum of roots is zero: $1 + \omega + \omega^2 = 0 \implies \omega + \omega^2 = -1, \quad 1 + \omega = -\omega^2, \quad 1 + \omega^2 = -\omega$.
2. Product of roots is unity: $1 \cdot \omega \cdot \omega^2 = \omega^3 = 1$. For any integer $k$, $\omega^{3k} = 1$.
3. Each complex cube root of unity is the square of the other.

3. Symmetric Functions of Roots & Synthetic Division

For $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$:

$$\text{Sum of roots } (S) = \alpha + \beta = -\frac{b}{a}$$ $$\text{Product of roots } (P) = \alpha \cdot \beta = \frac{c}{a}$$

A quadratic equation with given roots $\alpha, \beta$ can be constructed as: $x^2 - Sx + P = 0$.

Synthetic Division is a shortcut method for dividing a polynomial $P(x)$ by a linear divisor $(x - a)$. It utilizes only the coefficients of $P(x)$ to determine the Quotient $Q(x)$ and Remainder $R$.

Formula Summary & Quick Reference

Concept / Target	Mathematical Formula / Relation	Key Board Exam Rule / Condition
Discriminant	$\Delta = b^2 - 4ac$	Determines root nature without solving equation.
Sum of Cube Roots	$1 + \omega + \omega^2 = 0$	Substitute $\omega + \omega^2 = -1$ to simplify powers.
Product of Cube Roots	$\omega^3 = 1$	Reduce higher powers: $\omega^n = \omega^{n \bmod 3}$.
Sum of Quadratic Roots	$S = \alpha + \beta = -\frac{b}{a}$	Used in forming symmetric relations.
Product of Quadratic Roots	$P = \alpha \cdot \beta = \frac{c}{a}$	Used alongside sum of roots.
Quadratic Equation Formation	$x^2 - Sx + P = 0$	$S = \text{Sum of given roots}, P = \text{Product}$.

Step-by-Step Solved Board Exam Questions

Question 1 (Short Question - BISE Lahore 2022 Group I)

Find the value of $k$ if the roots of the equation $(k+1)x^2 + 2(k+3)x + (k+8) = 0$ are equal.

Solution:

Step 1: Identify coefficients $a, b, c$.

$$a = k+1, \quad b = 2(k+3), \quad c = k+8$$

Step 2: Write the discriminant formula.

$$\text{Disc} = b^2 - 4ac$$

Step 3: Substitute coefficients into the discriminant expression.

$$\text{Disc} = [2(k+3)]^2 - 4(k+1)(k+8)$$ $$\text{Disc} = 4(k^2 + 6k + 9) - 4(k^2 + 8k + k + 8)$$ $$\text{Disc} = 4(k^2 + 6k + 9) - 4(k^2 + 9k + 8)$$ $$\text{Disc} = 4[(k^2 + 6k + 9) - (k^2 + 9k + 8)]$$ $$\text{Disc} = 4[k^2 - k^2 + 6k - 9k + 9 - 8]$$ $$\text{Disc} = 4[-3k + 1]$$

Step 4: Apply the condition for equal roots ($\text{Disc} = 0$).

$$4[-3k + 1] = 0$$ $$-3k + 1 = 0 \implies 3k = 1 \implies k = \frac{1}{3}$$

Final Answer: $k = \frac{1}{3}$

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Question 2 (Long Question - FBISE 2023)

Evaluate: $\left(1 - 3\omega - 3\omega^2\right)^5$ and prove that $x^3 - y^3 = (x - y)(x - \omega y)(x - \omega^2 y)$.

Solution:

Part A: Evaluation of $\left(1 - 3\omega - 3\omega^2\right)^5$

Factor out $-3$ from the last two terms inside the parentheses:

$$= \left(1 - 3(\omega + \omega^2)\right)^5$$

Since $1 + \omega + \omega^2 = 0 \implies \omega + \omega^2 = -1$, substitute this identity:

$$= \left(1 - 3(-1)\right)^5 = \left(1 + 3\right)^5 = 4^5 = 1024$$

Part B: Proof of Identity $x^3 - y^3 = (x - y)(x - \omega y)(x - \omega^2 y)$

Take the Right Hand Side ($\text{R.H.S.}$):

$$\text{R.H.S.} = (x - y)(x - \omega y)(x - \omega^2 y)$$

Multiply the second and third factors first:

$$(x - \omega y)(x - \omega^2 y) = x^2 - \omega^2 xy - \omega xy + \omega^3 y^2$$ $$= x^2 - xy(\omega^2 + \omega) + (1)y^2 \quad (\because \omega^3 = 1)$$ $$= x^2 - xy(-1) + y^2 \quad (\because \omega + \omega^2 = -1)$$ $$= x^2 + xy + y^2$$

Now substitute this back into the expression:

$$\text{R.H.S.} = (x - y)(x^2 + xy + y^2) = x^3 - y^3 = \text{L.H.S.}$$

Hence Proved.

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Question 3 (Long Question - BISE Rawalpindi 2021)

If $\alpha, \beta$ are the roots of the equation $px^2 + qx + r = 0$ ($p \neq 0$), form a quadratic equation whose roots are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$.

Solution:

Step 1: Find sum and product of original roots.

$$\alpha + \beta = -\frac{q}{p}, \quad \alpha\beta = \frac{r}{p}$$

Step 2: Find the sum ($S$) of the new roots.

$$S = \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta}$$

Using the algebraic identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$:

$$S = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta} = \frac{\left(-\frac{q}{p}\right)^2 - 2\left(\frac{r}{p}\right)}{\frac{r}{p}}$$ $$S = \frac{\frac{q^2}{p^2} - \frac{2r}{p}}{\frac{r}{p}} = \frac{\frac{q^2 - 2pr}{p^2}}{\frac{r}{p}} = \left(\frac{q^2 - 2pr}{p^2}\right) \cdot \left(\frac{p}{r}\right) = \frac{q^2 - 2pr}{pr}$$

Step 3: Find the product ($P$) of the new roots.

$$P = \left(\frac{\alpha}{\beta}\right) \cdot \left(\frac{\beta}{\alpha}\right) = 1$$

Step 4: Form the required equation using $x^2 - Sx + P = 0$.

$$x^2 - \left(\frac{q^2 - 2pr}{pr}\right)x + 1 = 0$$

Multiplying the entire equation by $pr$:

$$pr x^2 - (q^2 - 2pr)x + pr = 0$$

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Question 4 (Synthetic Division - BISE Gujranwala 2022)

Use synthetic division to find the value of $k$ if $x + 2$ is a factor of $P(x) = 3x^3 + kx^2 - 7x + 6$.

Solution:

Step 1: Determine the zero of the divisor.

$$x + 2 = 0 \implies x = -2$$

Step 2: Set up synthetic division scheme with coefficients of $P(x)$.

Coefficients of $P(x)$ are $3, k, -7, 6$. Divisor root $a = -2$.

$$\begin{array}{c|rrrr} -2 & 3 & k & -7 & 6 \\ & \downarrow & -6 & -2k + 12 & 4k - 10 \\ \hline & 3 & k - 6 & -2k + 5 & 4k - 4 \end{array}$$

Step 3: Extract remainder $R$.

$$R = 4k - 4$$

Step 4: Apply the Factor Theorem.

Since $(x + 2)$ is a factor of $P(x)$, the remainder must equal zero ($R = 0$):

$$4k - 4 = 0 \implies 4k = 4 \implies k = 1$$

Final Answer: $k = 1$

Interactive Practice Quiz (Clickable MCQs)

Frequently Asked Questions (FAQs)

Q1: What is the main difference between rational and irrational roots in discriminant analysis?
Ans: Both rational and irrational roots occur when the discriminant is positive ($\Delta > 0$). However, if $\Delta$ is a perfect square (e.g., $1, 4, 9, 16, 25$), taking its square root yields a rational number, resulting in rational roots. If $\Delta$ is not a perfect square (e.g., $2, 3, 5, 7$), the square root remains radical ($\sqrt{\Delta}$), leading to irrational conjugate roots of the form $m \pm \sqrt{n}$.

Q2: Why must missing terms be represented by a zero coefficient in Synthetic Division?
Ans: Synthetic division relies strictly on positional notation corresponding to powers of $x$ in descending order ($x^n, x^{n-1}, \dots, x^0$). If a power term (like $x^2$ in $x^3 - 7x + 6$) is missing, placing a $0$ as its coefficient acts as a essential placeholder. Omitting it shifts all degree terms, resulting in an incorrect quotient and remainder.

Q3: How are marks allocated for algebraic identities involving $\omega$ in Matric Board Examinations?
Ans: Pakistani Board examiners follow step-wise marking schemes: 1. Stating the fundamental identity used (e.g., $1 + \omega + \omega^2 = 0$ or $\omega^3 = 1$) carries $1$ mark. 2. Correct substitution into the expression carries $1$ mark. 3. Step-by-step simplification leading to the final result carries $1$ to $2$ marks depending on whether it is a short or long question.