A Math Summary Booklet

QUADRATIC FUNCTIONS 01 Completing The Square Quadratic Inequalities Reverse Quadratic Inequality Application Step 1: Flush everything to the left and rearrange according to 𝑎𝑥2 + 𝑏𝑥 + 𝑐 Step 2: Simplify and rearrange according to 𝑎𝑥2 + 𝑏𝑥 + 𝑐 Step 3: Solve your quadratic inequalities Note: Always ensure 𝒙𝟐 is Positive, If it’s negative, divide and FLIP Your INEQUALITY SIGN. Step 1: Given that 𝑥 < −3 or 𝑥 > 2 Step 2: 𝑥 + 3 𝑥 − 2 > 0 (Reverse and Form Back Original) Step 3: 𝑥2 + 𝑥 − 6 > 0 (Expand) Maximum Point Starting Point Sub 𝑡 Sub 𝑦 = 0 −2𝑥2 + 4𝑥 + 8 = −2(𝑥2 − 2𝑥 − 4) = −2[(𝑥2 − 2𝑥 + (−2 2 )2 − (−2 2 )2 − 4] = −2[(𝑥 − 1)2 − 5] = −2(𝑥 − 1)2 + 10 When 𝒙𝟐 𝒊𝒔 𝒏𝒐𝒕 𝟏 Reverse Inequalities

Simultaneous Equations CHAPTER 1: QUADRATIC FUNCTIONS 02 Algebra Word Problems Validation Important Concepts𝑥2 + 𝑦2 = 34 … (1) 𝑦 + 3𝑥 = 14 … (2) Using (2) 𝑦 = 14 − 3𝑥 Substitute into (1) 𝑥2 + 14 − 3𝑥 2 = 34 𝑥2 + 196 − 84𝑥 + 9𝑥2 = 34 10𝑥2 − 84𝑥 + 162 = 0 𝑥 − 3 5𝑥 − 27 = 0 𝑥 = 3 or 𝑥 = 27 5 Substitute into (2) 𝑦 + 3 3 = 14 𝑦 + 3 27 4 = 14 𝑦 = 14 − 9 = 5 𝑦 = 14 − 3 27 4 = − 11 5 Answer: 𝑥 = 3, 𝑦 = 5 𝑥 = 5 2 5 , 𝑦 = −2 1 5 The line 2𝑥 + 3𝑦 = 8 meets the curve 2𝑥2 + 3𝑦2 = 110 at the point A and B. Find the coordinates of 𝐴 and B. 2𝑥 = 8 − 3𝑦 𝑥 = 8 − 3𝑦 2 Substitute into (2) 2 8 − 3𝑦 2 2 + 3𝑦2 = 110 2 64 − 48𝑦 + 9𝑦2 4 + 3𝑦2 = 110 128 − 96𝑦 + 18𝑦2 + 12𝑦2 = 440 30𝑦2 − 96𝑦 − 312 = 0 10𝑦2 − 32 − 104 = 0 𝑦 + 2 5𝑦 − 26 = 0 𝑦 = −2, 𝑦 = 26 5 Substitute into (1) 𝑥 = 8−3(−2) 2 = 7 𝑥 = 8−3(26 5 ) 2 = − 19 5 Answer: 7, −2 and (−3 4 5 , 5 1 5) For Simultaneous Equations, • Validate by Substituting Your Final Answer back into the Original Question • If the question is related to coordinates, ensure that you leave your answers in (𝑥, 𝑦) Concept: • There are 2 methods to solve for Simultaneous, either Substitution Method or Elimination Method. • I highly recommend to use Substitution Method as I find that it is faster and easier.

Completing The Square CHAPTER 1: QUADRATIC FUNCTIONS 03 Easy Advance Validation Important ConceptsSimplify 𝑥2 + 4𝑥 − 12 = 𝑥2 + 4𝑥 + 4 2 2 − 4 2 2 − 12 = 𝑥 + 2 2 − 16 Hence, Solve 𝑥2 + 4𝑥 − 12 = 0 𝑥 + 2 2 − 16 = 0 𝑥 + 2 2 = 16 𝑥 + 2 = 4 or 𝑥 + 2 = −4 • After you get your final answer, re-expand back to make sure it gives you back the original answer. • If you are solving, you can check your solution by using the Quadratic Equation Function in your calculator... they should be the same. Concept: 1. The condition for Completing The Square is that the coefficient of 𝑥2 MUST BE +1. If it is not +1, we need to FACTORISE the value to make it +1. 2. Be careful of the values you substitute in the bracket. Always include the SIGN. 3. When solving and completing the square, always solve by Square Rooting the values. NEVER expand back and solve by factorisation. That defeats the purpose of Completing The Square. Simplify 𝑥2 − 6𝑥 + 8 = 𝑥2 −6𝑥 + − 6 2 2 − − 6 2 2 + 8 = 𝑥 − 3 2 − 1 Hence, Solve 𝑥 − 3 2 − 1 = 0 𝑥 − 3 2 = 1 𝑥 − 3 = 1 or 𝑥 − 3 =