2.2 Linear & Quadratic Equations and Systems

2.2 Linear & Quadratic Equations and Systems

Equations and algebraic systems are central to the Bocconi Admission Test. Competitive test-takers do not merely solve equations mechanically; they exploit structural relationships between coefficients, roots, and geometry to deduce solutions rapidly. This section provides rigorous coverage of first-degree equations, quadratic theory, discriminant analysis, Vieta's formulas, and simultaneous 2x2 systems.

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Linear Equations in One Variable

A first-degree equation in $x$ reduces to the canonical form $ax = b$ ($a, b \in \mathbb{R}$). Its solvability falls into three mutually exclusive categories:

Fractional Equations and Extraneous Roots

In rational equations such as $\frac{2x}{x - 3} = 4 + \frac{6}{x - 3}$, candidates must state the domain restriction $x \neq 3$ first. Multiplying by $(x - 3)$ gives $2x = 4(x - 3) + 6 \implies 2x = 4x - 6 \implies 2x = 6 \implies x = 3$. Because $x = 3$ is forbidden by the denominator, it is extraneous, leaving $S = \emptyset$.

Literal Equations: Expressing One Quantity as a Function of the Others

Bocconi's published algebra syllabus contains a bullet that candidates routinely overlook: "given a relation among several quantities, know how to find one as a function of the others." These are literal equations — equations whose coefficients are letters rather than numbers — and the exam tests them both directly and inside numerical-reasoning items, where a financial identity has to be rearranged before any arithmetic is possible.

The method is the ordinary balancing method, executed symbolically:

1. Clear denominators and brackets so the target letter appears only in products and sums.
2. Collect every term containing the target letter on one side and everything else on the other.
3. Factor the target letter out of its side.
4. Divide by the bracket, and state the condition under which that division is legal.

Worked Example A (linear in the target). Solve $S = \frac{a(1 - r^n)}{1 - r}$ for $a$: $a = \frac{S(1 - r)}{1 - r^n} \quad (r \neq 1, \; r^n \neq 1)$

Worked Example B (target appears twice). The lens-style relation $\frac{1}{p} + \frac{1}{q} = \frac{1}{f}$ appears in disguised commercial forms. Solve for $p$: multiply by $pqf$ to get $qf + pf = pq$, collect the $p$ terms as $pf - pq = -qf$, factor to $p(f - q) = -qf$, and divide: $p = \frac{qf}{q - f} \quad (q \neq f)$

Worked Example C (the commercial case). Operating margin satisfies $m = \frac{R - C}{R}$. Solving for $R$: $mR = R - C \implies C = R(1 - m) \implies R = \frac{C}{1 - m}$ for $m \neq 1$. This single rearrangement is what makes the missing-cell table problems in Chapter 7 solvable in seconds.

The Condition Trap: Every division by a bracket carries a hidden assumption that the bracket is non-zero. On a parameter question, the excluded value is very often the answer the item is really testing — in Example C, $m = 1$ would require zero cost, which is why no finite revenue satisfies it.

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Quadratic Equations: Derivation and Formula

A second-degree equation has canonical form $ax^2 + bx + c = 0$ ($a, b, c \in \mathbb{R}, a \neq 0$).

Completing the Square

Dividing by $a$ gives $x^2 + \frac{b}{a}x = -\frac{c}{a}$. Adding the square of half the linear coefficient, $\left(\frac{b}{2a}\right)^2 = \frac{b^2}{4a^2}$, to both sides yields: $\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$ Taking the square root produces the Quadratic Formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Reduced Quadratic Formula (Even $b = 2k$)

When the linear coefficient is even, define $\Delta/4 = k^2 - ac$. The formula simplifies to $x = \frac{-k \pm \sqrt{k^2 - ac}}{a}$, eliminating large square calculations under exam pressure.

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Discriminant Analysis ($\Delta = b^2 - 4ac$)

The discriminant $\Delta = b^2 - 4ac$ dictates the root structure and parabola geometry ($y = ax^2 + bx + c$):

If $a, b, c \in \mathbb{Q}$, roots are rational if and only if $\Delta$ is a rational perfect square. Otherwise, roots form irrational conjugate pairs $p \pm q\sqrt{d}$.

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Vieta's Formulas and Symmetric Expressions

For $ax^2 + bx + c = 0$ with roots $x_1, x_2$, Vieta's relations state: $\text{Sum: } S = x_1 + x_2 = -\frac{b}{a}, \quad \text{Product: } P = x_1 x_2 = \frac{c}{a}$ Dividing by $a$ expresses the equation directly as $x^2 - Sx + P = 0$.

Symmetric Polynomial Shortcuts

Never compute individual radical roots when evaluating symmetric combinations of $x_1$ and $x_2$:

1. Sum of Squares: $x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2 = S^2 - 2P = \frac{b^2 - 2ac}{a^2}$
2. Sum of Reciprocals: $\frac{1}{x_1} + \frac{1}{x_2} = \frac{x_1 + x_2}{x_1 x_2} = \frac{S}{P} = -\frac{b}{c}$
3. Sum of Reciprocal Squares: $\frac{1}{x_1^2} + \frac{1}{x_2^2} = \frac{x_1^2 + x_2^2}{(x_1 x_2)^2} = \frac{S^2 - 2P}{P^2} = \frac{b^2 - 2ac}{c^2}$
4. Absolute Difference: $|x_1 - x_2| = \sqrt{(x_1 + x_2)^2 - 4x_1 x_2} = \frac{\sqrt{\Delta}}{|a|}$
5. Sum of Cubes: $x_1^3 + x_2^3 = S(S^2 - 3P) = S^3 - 3SP$

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Systems of Two Equations in Two Unknowns

2x2 Linear Systems

For a system with determinant $D = a_1 b_2 - a_2 b_1$: $\begin{cases} a_1 x + b_1 y = c_1 \\ a_2 x + b_2 y = c_2 \end{cases}$

• $D \neq 0$ (Intersecting Lines): Exactly one unique solution $(x_0, y_0)$.
• $D = 0$ and Parallel: Equal slopes, different intercepts ($a_1/a_2 = b_1/b_2 \neq c_1/c_2$). No solution ($S = \emptyset$).
• $D = 0$ and Coincident: Equal slopes and intercepts ($a_1/a_2 = b_1/b_2 = c_1/c_2$). Infinitely many solutions.

Linear-Quadratic Systems

Intersecting a line $y = mx + q$ and parabola $y = ax^2 + bx + c$ yields the resolving equation $ax^2 + (b - m)x + (c - q) = 0$. Its discriminant $\Delta_s$ classifies geometry:

• $\Delta_s > 0$ (Secant Line): Two distinct intersection points.
• $\Delta_s = 0$ (Tangent Line): One point of tangency.
• $\Delta_s < 0$ (External Line): No real intersection.

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Worked Step-by-Step Examples

Example 1: Parameter Condition for Repeated Roots

Find all values of $k$ for which $(k - 1)x^2 - 2kx + k + 2 = 0$ has two equal real roots:

1. Ensure degree 2: $a = k - 1 \neq 0 \implies k \neq 1$.
2. Set $\Delta/4 = 0$: $(-k)^2 - (k - 1)(k + 2) = k^2 - (k^2 + k - 2) = 2 - k = 0 \implies k = 2$.
3. Check: At $k = 2$, $x^2 - 4x + 4 = 0 \implies (x - 2)^2 = 0$, confirming repeated root $x = 2$.

Example 2: Vieta Evaluation of Symmetric Expressions

For $2x^2 - 8x - 3 = 0$, find $x_1^2 + x_2^2 + 3x_1 x_2$:

1. Extract $S = -(-8)/2 = 4$ and $P = -3/2$.
2. Rewrite: $(x_1 + x_2)^2 + x_1 x_2 = S^2 + P$.
3. Evaluate: $4^2 + (-3/2) = 16 - 1.5 = 29/2$.

Example 3: Line Tangent to Parabola

Find $c$ such that $y = 2x + c$ is tangent to $y = -x^2 + 6x - 1$:

1. Equate: $2x + c = -x^2 + 6x - 1 \implies x^2 - 4x + (c + 1) = 0$.
2. Set $\Delta/4 = 0$: $(-2)^2 - 1(c + 1) = 4 - c - 1 = 3 - c = 0 \implies c = 3$.
3. Intersect point: $x^2 - 4x + 4 = 0 \implies x = 2, y = 2(2) + 3 = 7$.