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Quadratic Equations (equations with x 2 )

Learn to solve quadratic equations involving x^2 with examples and background knowledge explained. Factorization methods and problem-solving techniques included.

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Quadratic Equations (equations with x 2 )

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  1. Quadratic Equations (equations with x2) This unit will teach you how to solve equations like 2x2 – 7x – 3 = 0 a2 – 9 = 0 2b2 + 5b = 0 x2– 5x = 6 etc…

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

  3. Background #1 2  0 = 0 2  0 = 0 0  5 = 0 a 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) Now, using letters rather than numbers, If a  b = 0 then we can conclude…. Either a = 0 OR b = 0 If (a – 2 )  (a – 1) = 0 then we can conclude…. Either a – 2 = 0 OR a – 1 = 0 Making a the subject of each equation. a = 2 a = 1

  4. Background #2 You need to revise how to factorise quadratics. Study THE FORMATS in this table: SHOW ME SHOW ME

  5. Example #1 Solve x2 – 16 = 0 STEP 1 Use Background 2 information to classify x2 – 16 as a TYPE 2 & factorise to (x – 4)(x + 4) STEP 2 Rewrite the original equation as (x – 4)(x + 4) = 0 STEP 3 Noting that this really means (x – 4) (x + 4) = 0 we can now use Background 1 knowledge to conclude that x – 4 = 0 OR x + 4 = 0 STEP 4 Make x the subject of each x = 4 x = – 4 OR

  6. Example #2 Solve x2 + 7x = 0 STEP 1 Use Background 2 information to classify x2 + 7x as a TYPE 1 & factorise to x ( x + 7) STEP 2 Rewrite the original equation as x(x + 7) = 0 STEP 3 Noting that this really means x(x + 7) = 0 we can now use Background 1 knowledge to conclude that x = 0 OR x + 7 = 0 STEP 4 Make x the subject of the 2nd one x = 0 x = – 7 OR

  7. Example #3 Solve x2 + 4x + 3 = 0 STEP 1 Use Background 2 information to classify x2 + 4x + 3 as a TYPE 3 & factorise to (x + 1) ( x + 3) STEP 2 Rewrite the original equation as (x + 1)(x + 3) = 0 STEP 3 As this means (x + 1)(x + 3) = 0 we can conclude that x + 1 = 0 OR x + 3 = 0 STEP 4 Make x the subject x = –1 x = – 3 OR

  8. Example #4 Solve 2x2 –7x + 3 = 0 STEP 1 Use Background 2 information to classify 2x2 – 7x + 3 as a TYPE 4 & factorise to (2x – 1 ) ( x – 3) SHOW ME STEP 2 Rewrite the original equation as (2x – 1 )(x– 3 ) = 0 STEP 3 2x – 1 = 0 OR x – 3 = 0 STEP 4 Make x the subject x = ½ x = 3 OR

  9. Divide through by 2. See note Example #5 Solve 2x2 – 50 = 0 STEP 1 Becomes x2 – 25 = 0 Now this is a TYPE 2 & factorises to (x – 5)(x + 5) STEP 2 Rewrite as (x – 5 )(x + 5 ) = 0 STEP 3 x – 5 = 0 OR x + 5 = 0 STEP 4 Make x the subject x = 5 x = – 5 OR

  10. Divide through by 3. SEE NOTE Example #6 Solve 3x2 – 9x – 30 = 0 STEP 1 x2 – 3x – 10 = 0 x2 – 3x – 10 is now a TYPE 3 & factorises to (x – 5)(x + 2) SHOW ME STEP 2 Rewrite as (x – 5 )(x + 2 ) = 0 STEP 3 x – 5 = 0 OR x + 2 = 0 STEP 4 Make x the subject x = 5 x = – 2 OR

  11. Example #7 Solve 2x2 – 7x = 4 STEP 1 Regroup to get 0 on right hand side. Becomes 2x2 – 7x – 4 = 0. Do as Type 4 See note SHOW ME STEP 2 Rewrite the original equation as (2x + 1 )(x– 4 ) = 0 STEP 3 Noting that this really means (2x + 1) (x – 4 ) = 0 we can now conclude that 2x + 1 = 0 OR x – 4 = 0 STEP 4 Make x the subject x = -½ x = 4 OR

  12. Example #8 Solve STEP 1 First, KILL THE FRACTION by multiplying throughout by x PLEASE EXPLAIN x2 = 21 – 4x STEP 2 Second, get 0 on the right hand side: x2 + 4x – 21 = 0 STEP 3 Do as a Type 3: (x + 7)(x – 3 ) = 0 STEP 4 x + 7 = 0 OR x – 3 = 0 x = – 7 OR x = 3

  13. 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”.

  14. 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 – x2 = 35 Remember to make sure you answer the question! Rearranging & changing all signs, x2 – 12x + 35 = 0 Factorising as a Type 3, (x– 5)(x – 7) = 0 FINAL ANSWER: The numbers are 5 and 7. x = 5 or x = 7

  15. LEFT HAND SIDE = RIGHT HAND SIDE 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: On your graphics, by hitting Y= and then entering Y1 = LEFT HAND SIDE Y2 = RIGHT HAND SIDE graphing and using 2nd TRACE INTERSECT, you can find where they meet, and this is the solution/s to the equation!!

  16. Y1 Y2 Solving Quadratics using the Graphics..... Example 1. Solve x2 – 3x – 4 = 0 STEP 1 Hit Y= and enter Y1 = x2 – 3x – 4 and Y2 = 0 (the x-axis) STEP 2 Hit WINDOW and make XMIN – 5, XMAX 5 STEP 3 Hit ZOOM. Choose Option 0 (ZOOMFIT) Make sure you can see where the graphs meet! STEP 4 Hit 2nd TRACE. Choose Option 5 (INTERSECT) & hit ENTER 3 times Doing this twice, with the cursor positioned near each intersection point, will yield the results x = – 1 and x = 4

  17. Solving Quadratics using the Graphics..... Example 2. Solve 2x2 – 5x = 3 STEP 1 Hit Y= and enter Y1 = 2x2 – 5x and Y2 = 3 STEP 2 Hit WINDOW and make XMIN – 5, XMAX 5 STEP 3 Hit ZOOM. Choose Option 0 (ZOOMFIT) Make sure you can see where the graphs meet! STEP 4 Hit 2nd TRACE. Choose Option 5 (INTERSECT) & hit ENTER 3 times Doing this twice, with the cursor positioned near each intersection point, will yield the results x = – 0.5 and x = 3

  18. QUICK QUIZ!!

  19. (1) The solutions to x2 – 16 = 0 are: x = 4 or 0 x = – 4 or 0 x = 4 or – 4 x = 16 or –16

  20. (2) The solutions to x2 – 4x = 0 are: 0 and – 4 4 and – 4 2 and – 2 0 and 4

  21. (3) The solutions to x2 – x = 12 are: 4 and – 3 3 and – 4 2 and – 6 0 and 3

  22. (4) The solutions to 25 – x2 = 0 are 5 and – 5 5 and 0 – 5 and 0 5 and 5

  23. (5) The solutions to 3x2 + x = 10 are - 5/3 and 2 5/3 and – 2 5/3 and 2 - 5/3 and – 2

  24. (6) To solve 2x2 – 5x = 3 on the graphics you could: (there are TWO correct answers) Draw Y1 = 2X2 – 5X – 3 and see where it cuts the y-axis Draw Y1 = 2X2 – 5X – 3 and see where it cuts the x-axis Draw Y1 = 2X2 – 5X and Y2 = 3, and see where they intersect Draw Y1 = 2X2 – 5X and Y2 = 3, and see where they cut the x-axis

  25. Factorising trinomials (Type 4) – example Mult. 2 by 3 to get 6 Factorise 2a2 – 5a + 3 STEP 1 Set up brackets STEP 2 Find two numbers that MULTIPLY to make + 6 ADD to make – 5 Numbers are – 3 and – 2 STEP 3 Insert numbers into brackets in Step 1 STEP 4 Divide denominator (2) fully into 2nd bracket Ans (2a – 3)(a – 1)

  26. 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 x2). 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 3x2 – 9x – 30 can also be done as Type 4s, but it takes more work!! BACK

  27. Factorise 2x2 – 7x – 4 (Example 7) STEP 1 Set up brackets STEP 2 Find two numbers that MULTIPLY to make – 8 ADD to make – 7 Numbers are – 8 and + 1 STEP 3 Insert numbers into brackets in Step 1 STEP 4 Divide denominator (2) fully into 1st bracket Ans (x – 4)(2x + 1)

  28. When solving quadratics, ALWAYS make sure you get 0 on the right hand side, and that your equation has the format ax2 + bx + c = 0 where a, b, c are constants (numbers). BACK

  29. Brilliant!! BACK

  30. Stiff cheddar! Have another go BACK

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