GCSE Solving Quadratic Equations by Factorisation www drfrostmaths

GCSE: Solving Quadratic Equations by Factorisation www. drfrostmaths. com Dr J Frost (jamie@drfrostmaths. com) @Dr. Frost. Maths Last modified: 13 th January 2020

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Recap of Factorisation Factorise the following (intentionally similar looking!) expressions: ? ? ? Recall that we find two numbers which add to give -4 and multiply to give +4.

Test Your Understanding ? 1 ? 2 3 ?

Mini Exercise : : Quadratic Factorisation Recap 1 2 ? ? 3 4 5 ? ? 6 7 ? ? 8 9 10 11 12 N ? ? ?

Solving Quadratic Equations But now we’re moving on. . . Quadratic Expression Equation Click to Move On

Starter = 0 If two things multiply to give 0, what do you know? At least one of those ? things must be 0.

Solving Equations Therefore, how could we make this equation true? ? or ? ?

Quickfire Questions Solving the following. ? ? ?

Exercise 1 Solving the following equations. 1 2 3 4 5 6 7 8 ? ? ? ?

Solving non-factorised equations We’ve seen that solving equations is not too difficult when we have it in the form: [factorised expression] = 0 ? Step 1 ? Step 2 ? Step 3 Step 1: Ensure 0 is on one side of the equation. Step 2: Factorise. Step 3: Consider each factorised term being 0 (as before)

Further example Solve the following. Method 1 Method 2 ? ?

Extension In pairs, discuss what solutions there are to the following equation. ?

Test Your Understanding 1 ? 2 ? 3 ?

Exercise 2 Solve the following equations. 1 2 3 4 5 6 7 8 9 10 11 12 13 x 2 + 7 x + 12 = 0 x 2 + x – 6 = 0 x 2 + 10 x + 21 = 0 x 2 + 2 x + 1 = 0 x 2 – 3 x = 0 x 2 + 7 x = 0 2 x 2 – 2 x = 0 x 2 – 49 = 0 4 x = x 2 10 x 2 – x – 3 = 0 12 y 2 – 16 y + 5 = 0 64 – z 2 = 0 2 x 2 = 8 ? x = -3 or? x = 2 x = -7 or? x = -3 x = -1 ? x = 0 or ? x=3 x = 0 or ? x = -7 x = 0 or ? x=1 x = -7 or? x = 7 x = 0 or ? x=4 x = -1/2 ? or x = 3/5 y = 1/2 or ? y = 5/6 z = 8 ? x = 2 ? x = -3 or x = -4 14 15 16 17 18 19 20 21 N N N 16 x 2 – 1 = 0 x 2 + 5 x = 14 2 x 2 + 7 x = 15 2 x 2 = 8 x + 10 4 x 2 + 7 = 29 x y 2 + 56 = 15 y 63 – 2 y = y 2 8 = 3 x 2 + 10 x x 6 = 9 x 3 – 8 x 4 = 5 x 2 – 4 x 3 = x 2 + x – 1 x 3 + 1 = – x 2 x 4 + 2 x 3 = 8 x + 16 ? x = -7 or? x = 2 x = -5 or? x = 3/2 x = -1 or? x = 5 x = 1/4 ? or x = 7 y = 7 or? y=8 x = -9 or? x = 7 x = -4 or? x = 2/3 x = 1 or? x=2 ? 2 x = 1 or x = 1 ? x = -1 ? x = 2 ? x = 1/4

Harder Equations This is still a quadratic equation! What should we do here? ? Expand everything out first, and simplify. ? ? As before, factorise.

Further Examples ?

Further Examples ? Whenever you have fractions in an equation, your instinct should be to multiply everything through by the denominator of that fraction.

Test Your Understanding 1 ? 2 ?

Exercise 3 Solve the following equations. 1 2 3 4 5 6 7 8 9 10 11 12 N 1 ? ? ? N 2 ? ?

Forming Quadratic Equations from Context A rectangle’s length is 3 cm more than its width. Its area is 28 cm 2. Form and solve a quadratic equation to find the length. We don’t know the length of width, but we’re told the length is 3 more than the width. How could we represent this algebraically? ? ? We’re told the area is 28. So you should form an equation using this information. ? Lengths have to be positive. I tend to write the other solution but put a strike through it to reject it. Ensure you re-read what the question is asking for!

Harder Example Edexcel ? ?

Even Harder Example According to the Edexcel examiner’s report, around 1 in 200 students got (a) to (c) correct. ? ?

Test Your Understanding 1 3 [Edexcel IGCSE Jan 2012 -3 H Q 21 a Edited] ? 2 ? ?

Exercise 4 Determine x 1 3 Determine the length of the hypotenuse. 5 Area = 96 Answer: ? x=3 2 ? ? 4 N Area = 28 ? Answer: x = 5 Given the two triangles have the same area, determine x. ? Answer: x = 2 [Maclaurin] An arithmetic sequence is one in which the difference between successive terms remains constant (for example, 4, 7, 10, 13, …). Suppose that a right-angled triangle has the property that the lengths of its sides form an arithmetic sequence. Prove that the sides of the triangle are in the ratio 3: 4: 5. Solution: Making sides x – a, x and x + a, we obtain x = 4 a by Pythagoras. Thus sides are 3 a, 4 a, 5 a which are in desired ratio. ?

Final Test Your Topic Understanding …of solving by factorising. 1 ? 2 ?