IP AMaths Notes (Upper Sec, Year 3-4): 05) Quadratic Equations and Functions

Quadratics appear everywhere: simultaneous equations, optimisation, and curve sketching. Keep vertex form and discriminant analysis at your fingertips.

Forms to remember

• General: \( y = ax^{2} + bx + c \), \( a \neq 0 \).
• Vertex (completed square): \( y = a\bigl(x - h\bigr)^{2} + k \) where vertex is \( (h, k) \).
• Factored: \( y = a(x - \alpha)(x - \beta) \) with roots \( \alpha, \beta \).
• Discriminant: \( \Delta = b^{2} - 4ac \) controls the number of real roots.

Worked example 1 — Completing the square

Write \( y = 2x^{2} - 8x + 5 \) in vertex form and state the minimum value.

1. Factor leading coefficient: \( y = 2\bigl(x^{2} - 4x\bigr) + 5 \).
2. Complete square inside: \( x^{2} - 4x = (x - 2)^{2} - 4 \).
3. Substitute: \( y = 2\bigl((x - 2)^{2} - 4\bigr) + 5 = 2(x - 2)^{2} - 8 + 5 \).
4. Simplify: \( y = 2(x - 2)^{2} - 3 \).

Minimum occurs at \( x = 2 \) with \( y_{\min} = -3 \).

Worked example 2 — Discriminant constraint

Find the range of \( k \) such that \( x^{2} - (2k + 3)x + k = 0 \) has real and distinct roots.

1. Compute discriminant: \( \Delta = \bigl(-(2k + 3)\bigr)^{2} - 4(1)(k) = (2k + 3)^{2} - 4k \).
2. Expand: \( \Delta = 4k^{2} + 12k + 9 - 4k = 4k^{2} + 8k + 9 \).
3. Require \( \Delta > 0 \). The quadratic \( 4k^{2} + 8k + 9 \) is always positive because its discriminant is negative: \( 8^{2} - 4(4)(9) = 64 - 144 = -80 \).
4. Therefore \( \Delta > 0 \) for all real \( k \).

Equation has two distinct real roots for every real \( k \).

Try this

Sketch \( y = -x^{2} + 6x - 5 \) and determine the interval of \( x \) where \( y > 0 \) without using a calculator.