2.4 Quadratic Equations, Relations Between Roots & Factor/Remainder Theorems

Quadratic Equations, Relations Between Roots & Factor/Remainder Theorems

Quadratic equations and polynomial algebra form the bedrock of the intermediate algebra syllabus (FSc Part I). On the UET ECAT, questions test candidate speed in evaluating the nature of roots via the discriminant, manipulating symmetric functions of roots, applying synthetic division, and identifying extraneous roots in reducible equations. Applying algebraic identities rather than manually computing square roots is vital to maintaining the 1-minute-per-question pace demanded by the exam.

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1. Standard Form and the Quadratic Formula

A quadratic equation in one variable $x$ is a second-degree polynomial equation written in the standard form:

$ax^2 + bx + c = 0, \quad a, b, c \in \mathbb{R}, \quad a \neq 0$

Completing the square yields the universal quadratic formula:

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

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2. The Discriminant and Nature of Roots

The term inside the radical, $\Delta = b^2 - 4ac$, is the discriminant of the quadratic equation. Assuming rational coefficients $a, b, c \in \mathbb{Q}$ with $a \neq 0$, the value of $\Delta$ completely dictates the nature of the roots $\alpha$ and $\beta$.

Comprehensive Discriminant Classification

ECAT Key Principle: If the coefficients $a, b, c$ are real and one root is $p + iq$, the second root is guaranteed to be its complex conjugate $p - iq$. Similarly, if coefficients are rational and one root is an irrational surd $p + \sqrt{q}$ ($q$ non-square), the other root must be $p - \sqrt{q}$.

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3. Relations Between Roots and Coefficients (Vieta's Formulas)

Let $\alpha$ and $\beta$ be the roots of $ax^2 + bx + c = 0$. Using the quadratic formula:

$\alpha = \frac{-b + \sqrt{\Delta}}{2a}, \quad \beta = \frac{-b - \sqrt{\Delta}}{2a}$

Sum and Product Formulas

1. Sum of Roots ($S$):

$S = \alpha + \beta = \frac{-b + \sqrt{\Delta} - b - \sqrt{\Delta}}{2a} = \frac{-2b}{2a} = -\frac{b}{a}$

1. Product of Roots ($P$):

$P = \alpha \beta = \left(\frac{-b + \sqrt{\Delta}}{2a}\right)\left(\frac{-b - \sqrt{\Delta}}{2a}\right) = \frac{(-b)^2 - (b^2 - 4ac)}{4a^2} = \frac{4ac}{4a^2} = \frac{c}{a}$

Forming a Quadratic Equation from Given Roots

Any quadratic equation with roots $\alpha$ and $\beta$ can be constructed directly from their sum $S$ and product $P$:

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

$x^2 - Sx + P = 0$

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4. Symmetric Functions of Roots

A function $f(\alpha, \beta)$ is symmetric with respect to $\alpha$ and $\beta$ if interchanging $\alpha$ and $\beta$ leaves the expression unchanged ($f(\alpha, \beta) = f(\beta, \alpha)$). In ECAT examinations, candidates are frequently asked to compute symmetric expressions or form new equations whose roots are related to $\alpha$ and $\beta$.

Fundamental Symmetric Conversions

1. Sum of Squares:

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = S^2 - 2P$

1. Sum of Cubes:

$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = S^3 - 3SP$

1. Difference of Roots:

$|\alpha - \beta| = \sqrt{(\alpha + \beta)^2 - 4\alpha\beta} = \sqrt{S^2 - 4P} = \frac{\sqrt{b^2 - 4ac}}{|a|} = \frac{\sqrt{\Delta}}{|a|}$

1. Sum of Reciprocals:

$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{S}{P} = -\frac{b}{c}$

1. Sum of Reciprocal Squares:

$\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2} = \frac{S^2 - 2P}{P^2} = \frac{b^2 - 2ac}{c^2}$

1. Sum of Fourth Powers:

$\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha\beta)^2 = (S^2 - 2P)^2 - 2P^2$

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5. Synthetic Division Algorithm

Synthetic division is an abbreviated procedure for dividing a polynomial $P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$ by a linear divisor of the form $x - c$.

Algorithmic Steps

1. Write the root of the divisor, $c$, to the left.
2. Write the coefficients $a_n, a_{n-1}, \dots, a_0$ in a horizontal row, inserting $0$ for any missing powers.
3. Bring down the leading coefficient $a_n$.
4. Multiply $a_n$ by $c$ and place the product beneath $a_{n-1}$; add the column to get the next coefficient $b_{n-1}$.
5. Repeat the multiply-and-add process until the final column is reached.
6. The final number is the remainder $R$, and the preceding row represents the coefficients of the quotient polynomial $Q(x)$ of degree $n - 1$.

$P(x) = (x - c)Q(x) + R$

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6. Remainder and Factor Theorems

The Remainder Theorem

If a polynomial $P(x)$ of degree $n \ge 1$ is divided by a linear divisor $x - c$, the remainder is independent of $x$ and is equal to $P(c)$:

$R = P(c)$

If the divisor is $ax - b$ with $a \neq 0$, the remainder is:

$R = P\left(\frac{b}{a}\right)$

The Factor Theorem

The linear binomial $x - c$ is a factor of the polynomial $P(x)$ if and only if the remainder is zero:

$x - c \text{ is a factor of } P(x) \iff P(c) = 0$

Similarly, $ax - b$ is a factor of $P(x) \iff P\left(\frac{b}{a}\right) = 0$.

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7. Equations Reducible to Quadratic Form

Many higher-degree or non-linear equations can be transformed into quadratic equations via an appropriate substitution $y = g(x)$.

Type 1: Biquadratic Equations ($ax^4 + bx^2 + c = 0$)

Substitute $y = x^2$ to obtain $ay^2 + by + c = 0$. Solve for $y$, then compute $x = \pm \sqrt{y}$.

Type 2: Reciprocal Equations

An equation where coefficients of terms equidistant from the beginning and end are equal:

$ax^4 + bx^3 + cx^2 + bx + a = 0$

Divide throughout by $x^2$:

$a\left(x^2 + \frac{1}{x^2}\right) + b\left(x + \frac{1}{x}\right) + c = 0$

Substitute $y = x + \frac{1}{x}$. Then $\left(x + \frac{1}{x}\right)^2 = y^2 \implies x^2 + \frac{1}{x^2} = y^2 - 2$. The equation transforms to:

$a(y^2 - 2) + by + c = 0 \implies ay^2 + by + (c - 2a) = 0$

Solve for $y$, then solve $x + \frac{1}{x} = y \implies x^2 - yx + 1 = 0$ for $x$.

Type 3: Exponential Equations

Equations in which the variable appears in the exponent, such as $4^{x+1} - 9 \cdot 2^x + 2 = 0$:

Rewrite as $4 \cdot (2^x)^2 - 9(2^x) + 2 = 0$. Let $y = 2^x$:

$4y^2 - 9y + 2 = 0 \implies (4y - 1)(y - 2) = 0 \implies y = \frac{1}{4} \quad \text{or} \quad y = 2$

• $2^x = \frac{1}{4} = 2^{-2} \implies x = -2$
• $2^x = 2^1 \implies x = 1$

Type 4: Radical Equations and Extraneous Roots

Equations involving square roots, such as $\sqrt{x + 3} + \sqrt{x + 8} = 5$, require squaring both sides.

Caution: Squaring can introduce extraneous roots (solutions that satisfy the squared equation but not the original equation). On the ECAT, you must mentally check candidate roots in the original radical expression.