2. Quadratic Equations (equations with x 2 ) This unit will teach you how to solve equations like 2x 2 – 7x – 3 = 0 a 2 – 9 = 0 2b 2 + 5b = 0 x 2 – 5x = 6 etc…

3. But first….we need two bits of BACKGROUND KNOWLEDGE!!

4. Background #1 2  0 = 0 0  5 = 0 0 If we multiply any two numbers together and the result is zero, then….. One of the numbers must be zero! (Maybe both are) If a  b = 0 then we can conclude…. Either a = 0 OR b = 0 Now, using letters rather than numbers, If (a – 2 )  (a – 1) = 0 then we can conclude…. Either a – 2 = 0 OR a – 1 = 0 a  0 = 0 a = 2a = 1 Making a the subject of each equation.

5. Background #2 You need to revise how to factorise quadratics. Study THE FORMATS in this table: TYPE Distinguishing Features ExampleAnswer 1 2 terms each with x common factor x 2 – 5xx(x – 5) 2 2 terms, only one x both terms are squares always a minus sign x 2 – 4(x – 2)(x + 2) 3 3 terms: x 2, x and a constant (number) x 2 – 3x – 28(x – 7)(x + 4) 4 3 terms like Type 3 Number in front of x 2 2x 2 – 7x – 4(2x + 1)(x – 4) SHOW ME

6. Example #1 Solve x 2 – 16 = 0 STEP 1Use Background 2 information to classify x 2 – 16 as a TYPE 2 & factorise to (x – 4)(x + 4) STEP 2Rewrite the original equation as (x – 4)(x + 4) = 0 STEP 3Noting that this really means (x – 4)  (x + 4) = 0 we can now use Background 1 knowledge to conclude that x – 4 = 0ORx + 4 = 0 STEP 4 x = 4 OR x = – 4 Make x the subject of each

7. Example #2 Solve x 2 + 7x = 0 STEP 1Use Background 2 information to classify x 2 + 7x as a TYPE 1 & factorise to x ( x + 7) STEP 2Rewrite the original equation as x(x + 7) = 0 STEP 3Noting that this really means x  (x + 7) = 0 we can now use Background 1 knowledge to conclude that x = 0ORx + 7 = 0 STEP 4 OR x = – 7 Make x the subject of the 2 nd one x = 0

8. Example #3 Solve x 2 + 4x + 3 = 0 STEP 1Use Background 2 information to classify x 2 + 4x + 3 as a TYPE 3 & factorise to (x + 1) ( x + 3) STEP 2Rewrite the original equation as (x + 1)(x + 3) = 0 STEP 3As this means (x + 1)  (x + 3) = 0 we can conclude that x + 1 = 0ORx + 3 = 0 STEP 4 OR x = – 3 Make x the subject x = –1

9. Example #4 Solve 2x 2 –7x + 3 = 0 STEP 1Use Background 2 information to classify 2x 2 – 7x + 3 as a TYPE 4 & factorise to (2x – 1 ) ( x – 3) STEP 2 Rewrite the original equation as (2x – 1 )(x – 3 ) = 0 STEP 32x – 1 = 0ORx – 3 = 0 STEP 4 OR x = 3 Make x the subject x = ½ SHOW ME

10. Example #5 Solve 2x 2 – 50 = 0 STEP 1 Becomes x 2 – 25 = 0 STEP 2Rewrite as (x – 5 )(x + 5 ) = 0 STEP 3x – 5 = 0ORx + 5 = 0 STEP 4 OR x = – 5 Make x the subject x = 5 Divide through by 2. See note Now this is a TYPE 2 & factorises to (x – 5)(x + 5)

11. Example #6 Solve 3x 2 – 9x – 30 = 0 STEP 1 x 2 – 3x – 10 is now a TYPE 3 & factorises to (x – 5)(x + 2) STEP 2Rewrite as (x – 5 )(x + 2 ) = 0 STEP 3x – 5 = 0ORx + 2 = 0 STEP 4 OR x = – 2 Make x the subject x = 5 Divide through by 3. SEE NOTE SHOW ME x 2 – 3x – 10 = 0

12. Example #7 Solve 2x 2 – 7x = 4 STEP 1Regroup to get 0 on right hand side. Becomes 2x 2 – 7x – 4 = 0. Do as Type 4 STEP 2Rewrite the original equation as (2x + 1 )(x – 4 ) = 0 STEP 3Noting that this really means (2x + 1)  (x – 4 ) = 0 we can now conclude that 2x + 1 = 0ORx – 4 = 0 STEP 4 OR x = 4 Make x the subject x = -½ SHOW ME See note

13. Example #8 Solve STEP 1First, KILL THE FRACTION by multiplying throughout by x x 2 = 21 – 4x STEP 2 Second, get 0 on the right hand side: x 2 + 4x – 21 = 0 STEP 3Do as a Type 3: (x + 7)(x – 3 ) = 0 STEP 4x + 7 = 0 OR x – 3 = 0 x = – 7ORx = 3 PLEASE EXPLAIN

14. PROBLEM SOLVING !! Example #9 Two numbers have a sum of 12 and a product of 35. Find the numbers. With these questions, always begin by saying “Let one number be equal to x”. Since they both add to 12, that means the other number must be 12 – x. So we can also say “Let the other number be equal to 12 – x”.

15. Now we’ll put this into symbols As our numbers are x and 12 – x, and we know their product is 35, we can say x (12 – x) = 35 Expanding, 12x – x 2 = 35 Rearranging & changing all signs, x 2 – 12x + 35 = 0 Factorising as a Type 3, (x – 5)(x – 7) = 0 x = 5 or x = 7 FINAL ANSWER: The numbers are 5 and 7. Remember to make sure you answer the question!

16. Solving Quadratics using the Graphics..... The background theory goes like this..... Suppose you have any equation with one unknown (x). The format will always be like this: LEFT HAND SIDERIGHT HAND SIDE= On your graphics, by hitting Y= and then entering Y 1 = LEFT HAND SIDE Y 2 = RIGHT HAND SIDE graphing and using 2 nd TRACE INTERSECT, you can find where they meet, and this is the solution/s to the equation!!

17. Solving Quadratics using the Graphics..... Example 1. Solve x 2 – 3x – 4 = 0 STEP 1Hit Y= and enter Y 1 = x 2 – 3x – 4 and Y 2 = 0 (the x-axis) STEP 2Hit WINDOW and make X MIN – 5, X MAX 5 STEP 3Hit ZOOM. Choose Option 0 (ZOOMFIT) STEP 4Hit 2 nd TRACE. Choose Option 5 (INTERSECT) & hit ENTER 3 times Make sure you can see where the graphs meet! Doing this twice, with the cursor positioned near each intersection point, will yield the results x = – 1 and x = 4 Y1Y1 Y2Y2

18. Solving Quadratics using the Graphics..... Example 2. Solve 2x 2 – 5x = 3 STEP 1Hit Y= and enter Y 1 = 2x 2 – 5x and Y 2 = 3 STEP 2Hit WINDOW and make X MIN – 5, X MAX 5 STEP 3Hit ZOOM. Choose Option 0 (ZOOMFIT) STEP 4Hit 2 nd TRACE. Choose Option 5 (INTERSECT) & hit ENTER 3 times Make sure you can see where the graphs meet! Doing this twice, with the cursor positioned near each intersection point, will yield the results x = – 0.5 and x = 3

20. QUICK QUIZ!!

21. (1) The solutions to x 2 – 16 = 0 are: x = 4 or 0 x = 16 or –16 x = 4 or – 4 x = – 4 or 0

22. (2) The solutions to x 2 – 4x = 0 are: 0 and – 4 0 and 4 2 and – 2 4 and – 4

23. (3) The solutions to x 2 – x = 12 are: 4 and – 3 0 and 3 2 and – 6 3 and – 4

24. (4) The solutions to 25 – x 2 = 0 are 5 and – 5 5 and 5 – 5 and 0 5 and 0

25. (5) The solutions to 3x 2 + x = 10 are - 5/3 and 2 - 5/3 and – 2 5/3 and 2 5/3 and – 2

26. (6) To solve 2x 2 – 5x = 3 on the graphics you could: (there are TWO correct answers) Draw Y 1 = 2X 2 – 5X – 3 and see where it cuts the y-axis Draw Y 1 = 2X 2 – 5X and Y 2 = 3, and see where they cut the x-axis Draw Y 1 = 2X 2 – 5X and Y 2 = 3, and see where they intersect Draw Y 1 = 2X 2 – 5X – 3 and see where it cuts the x-axis

27. Factorising trinomials (Type 4) – example Factorise 2a 2 – 5a + 3 STEP 1Set up brackets STEP 2 STEP 3 Find two numbers that MULTIPLY to make+ 6 ADD to make– 5 Numbers are – 3 and – 2 Insert numbers into brackets in Step 1 Ans (2a – 3)(a – 1) Mult. 2 by 3 to get 6 STEP 4Divide denominator (2) fully into 2 nd bracket

28. NOTE about Type 3 & 4 Sometimes you will get a quadratic (like in Example 5 or 6) which might LOOK LIKE A TYPE 4 (it has a number in front of the x 2 ). But on looking closer, you notice you can divide through by a number, making it into a TYPE 3 or 2. Because Type 3s & Type 2s are easier to factorise than Type 4s, this makes good sense! Examples such as 3x 2 – 9x – 30 can also be done as Type 4s, but it takes more work!! BACK

29. Factorise 2x 2 – 7x – 4 (Example 7) STEP 1Set up brackets STEP 2 STEP 3 Find two numbers that MULTIPLY to make– 8 ADD to make– 7 Numbers are – 8 and + 1 Insert numbers into brackets in Step 1 Ans (x – 4)(2x + 1) STEP 4Divide denominator (2) fully into 1 st bracket

30. When solving quadratics, ALWAYS make sure you get 0 on the right hand side, and that your equation has the format ax 2 + bx + c = 0 where a, b, c are constants (numbers). BACK

31. Brilliant!! BACK

32.  Stiff cheddar! Have another go BACK