Quadratics can be solved by:

• factorisation
• using the formula
• completing the square
• graphically (using x intercept/s - only when line touches x axis)

Solving (x-5)(x-5) = 0

• One of the brackets has to = 0 in order to multiply to get 0
• so (x-5) = 0
• so x = 5
• X is inverse of what is in the bracket
• Always show that brackets = 0 as it is an equation not an expression.

• a/co-efficient of x² must be 1
• x² + 5x + 4 = 0

1. Divide b by 2
2. Put x² + b/2x into bracket
3. Expand to find (b/2)²
4. Repeat step 2, but -(b/2)² and +c

• e.g. (x - 2)² - 5 = 0

1. (x - 2)² = 5
2. (x - 2) = √5
3. x = √5 + 2 or x = - √5 + 2

• Positive = b² - 4ac > o  (2 solutions)
• Neutral = b² - 4ac = o (1 solution)
• Negative = b² - 4ac < o (no solution)
• Math error = no solution a.k.a on graph the parabola does not cut the x axis.

When it comes to IGCSE/GCSE Maths, do you remember how to work out the turning point?

• Vertex of the parabola - halfway between the roots.
• If the turning point is the highest point = maximum point.
• If the turning point is the lowest point = minimum point.
   
• By completing the square of a quadratic,
• y = a(x+p)² + q
• a > 0 — minimum point (below x axis)
• a < 0 — maximum point (above x axis)
• Turning point = (p,q)
• a(x+p)² + q
• p = inverse of p (-p)
• q = q

• y = 2x² + 12x + 1

1. Divide everything by co-efficient of x to put it outside the bracket.
2. Complete the square within the bracket.
3. Expand (by co-efficient) to eliminate outside bracket.
4. Then locate p and x.

• Solve the inequality as an equation.
• Check if graph is min/max point (co-efficient of x > or < 0).

x² - b > 0 — x > highest, x < lowest
x² - b < 0 — lowest < x < highest

That's the end of the topic!

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Drafted by Bonnie (Mathematics)