Quadratic Equations and Inequalities

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your ECAT exam.

A quadratic equation has the form $ax^2 + bx + c = 0$ where $a \neq 0$.

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

The discriminant $D = b^2 - 4ac$ determines the nature of roots:

• $D > 0$: two distinct real roots
• $D = 0$: two equal (one repeated) real roots
• $D < 0$: two complex conjugate roots

Sum and Product of Roots: If $\alpha$ and $\beta$ are roots: $$\alpha + \beta = -\frac{b}{a} \quad \text{(sum of roots)}$$ $$\alpha \beta = \frac{c}{a} \quad \text{(product of roots)}$$

Nature of Roots — Quick Tests:

• $\alpha + \beta > 0$ and $\alpha \beta > 0$: both roots are positive
• $\alpha + \beta < 0$ and $\alpha \beta > 0$: both roots are negative
• $\alpha \beta < 0$: roots have opposite signs

⚡ ECAT exam tips:

• To form a quadratic from given roots: $x^2 - (\text{sum})x + \text{product} = 0$
• If one root is $\sqrt{2}$, the other is $-\sqrt{2}$ (irrational roots occur in conjugate pairs)
• For equal roots: $b^2 = 4ac$; this is the condition for the quadratic to have a repeated root

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🟡 Standard — Regular Study (2d–2mo)

For ECAT students who want genuine understanding.

Solving Quadratic Inequalities:

Method: First solve the equation $ax^2 + bx + c = 0$ to find critical points. Then test intervals.

Example: Solve $x^2 - 5x + 6 < 0$. Factor: $(x-2)(x-3) < 0$. Critical points: $x = 2, 3$. Test: for $x < 2$: both factors negative → product positive. For $2 < x < 3$: $(x-2) > 0$, $(x-3) < 0$ → product negative. For $x > 3$: both positive → product positive. Solution: $2 < x < 3$.

Key inequality result: For a quadratic $ax^2 + bx + c$:

• If $a > 0$ and $D < 0$: $ax^2 + bx + c > 0$ for all real $x$ (always positive)
• If $a < 0$ and $D < 0$: $ax^2 + bx + c < 0$ for all real $x$ (always negative)

Maximum and Minimum Values: For $y = ax^2 + bx + c$ with $a > 0$: minimum at $x = -b/(2a)$, minimum value = $\frac{4ac - b^2}{4a}$. For $a < 0$: maximum at $x = -b/(2a)$, maximum value = $\frac{4ac - b^2}{4a}$.

Transformation of Equations:

If $\alpha, \beta$ are roots of $ax^2 + bx + c = 0$:

• Sum of roots squared: $\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = \frac{b^2}{a^2} - \frac{2c}{a}$
• Product of roots cubed: $\alpha^3 \cdot \beta^3 = (\alpha\beta)^3 = \frac{c^3}{a^3}$
• Reciprocal roots equation: $cx^2 + bx + a = 0$

⚡ Common student mistakes:

1. Forgetting that $a \neq 0$ in a quadratic — if $a = 0$, it’s linear, not quadratic
2. Not checking whether roots are real before applying certain conditions
3. Solving inequalities by dividing by a variable without considering sign changes
4. Confusing the sign in the quadratic formula: $x = (-b \pm \sqrt{D})/(2a)$

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for ECAT mastery of quadratic equations.

The Discriminant — Geometric Interpretation:

The discriminant $D = b^2 - 4ac$ is actually $(2a \cdot \text{vertex } x\text{-coordinate} + b)^2$ in a shifted form. In vertex form $a(x-h)^2 + k = 0$, the discriminant equals $4a^2$ times the difference between the vertex $y$-value and zero.

Condition for Common Roots:

If two quadratics $a_1x^2 + b_1x + c_1 = 0$ and $a_2x^2 + b_2x + c_2 = 0$ share a common root: $$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$$ Or equivalently, using the resultant: $$a_1b_2 - a_2b_1 + b_1c_2 - b_2c_1 + c_1a_2 - c_2a_1 = 0$$ This is the condition for the two equations to have a common root.

Location of Roots:

Let $f(x) = ax^2 + bx + c$:

• Both roots greater than $\alpha$: $f(\alpha) > 0$ and $-b/(2a) > \alpha$ (vertex to the right of $\alpha$)
• Both roots less than $\alpha$: $f(\alpha) > 0$ and $-b/(2a) < \alpha$
• One root less than $\alpha$ and one greater: $f(\alpha) < 0$
• Both roots between $\alpha$ and $\beta$ (where $\alpha < \beta$): $f(\alpha) > 0$, $f(\beta) > 0$, and vertex between $\alpha$ and $\beta$

Descartes’ Rule of Signs (for positive roots):

The number of positive real roots of $f(x) = 0$ is at most the number of sign changes in $f(x)$, and differs from it by an even number. For $f(x) = x^3 - 6x^2 + 11x - 6 = 0$: sign changes: $+ \to - \to + \to -$: 3 changes → 3 or 1 positive roots. Indeed: roots are 1, 2, 3.

Irrational Roots:

Irrational roots of a quadratic with rational coefficients occur in conjugate pairs. If $\sqrt{p}$ is a root (where $p$ is not a perfect square), then $-\sqrt{p}$ is also a root.

Vieta’s Formulas for Higher Degree:

For a cubic $ax^3 + bx^2 + cx + d = 0$ with roots $\alpha, \beta, \gamma$:

• $\alpha + \beta + \gamma = -b/a$
• $\alpha\beta + \beta\gamma + \gamma\alpha = c/a$
• $\alpha\beta\gamma = -d/a$

Quadratic Inequalities — Interval Testing:

For $ax^2 + bx + c > 0$:

• If $a > 0$ and $D \leq 0$: always true (or never zero)
• If $a > 0$ and $D > 0$: $x < x_1$ or $x > x_2$ (where $x_1 < x_2$ are roots)
• If $a < 0$ and $D > 0$: $x_1 < x < x_2$

For $ax^2 + bx + c < 0$:

• Reverse the inequalities above

Problem Type — Worked Example:

Find the range of $k$ for which $x^2 - (k-2)x + (k^2 - 4) = 0$ has both roots positive.

Conditions:

1. $D \geq 0$: $(k-2)^2 - 4(k^2-4) \geq 0 \Rightarrow k^2 - 4k + 4 - 4k^2 + 16 \geq 0 \Rightarrow -3k^2 - 4k + 20 \geq 0 \Rightarrow 3k^2 + 4k - 20 \leq 0$. Roots of $3k^2 + 4k - 20 = 0$: $k = \frac{-4 \pm \sqrt{16 + 240}}{6} = \frac{-4 \pm 16}{6}$. So $k = 2$ or $k = -10/3$. Between them: $-10/3 \leq k \leq 2$.

2. Sum of roots $> 0$: $(k-2) > 0 \Rightarrow k > 2$.

3. Product of roots $> 0$: $k^2 - 4 > 0 \Rightarrow k > 2$ or $k < -2$.

Combining: $k > 2$ (from sum) AND $k > 2$ OR $k < -2$ (from product) → $k > 2$. Combined with $k \leq 2$: no solution! So there is no value of $k$ for which both roots are strictly positive. Check endpoint $k=2$: $x^2 - 0x + 0 = 0 \Rightarrow x = 0$ (repeated root, not strictly positive). So answer: no real $k$ satisfies all three conditions.

ECAT Previous Year Patterns:

• Quadratic formula: very common
• Discriminant and nature of roots: very common
• Sum and product of roots: common
• Forming equations from roots: common
• Quadratic inequalities: common
• Maximum/minimum values: periodic

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