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Gr 11 equations

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Examples and notes on solving quadratic equations

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Gr 11 equations

  1. 1. Quadratic equations<br />Grade 11 – Paper 1<br />
  2. 2. Quadratic equations<br />Quadraticequations in one variable are equations of degree 2, whichmeans the highest power of the variable is 2.<br />The standard formis<br />ax2 + bx + c = 0 where a ≠ 0<br />Theseequations have 2 solutions, <br />theycanbe:<br />different<br />the same<br />non-real<br />The solutions are known as the roots of the equation.<br />
  3. 3. Method:<br />The solution of quadraticequations by factorisingisbased on the factthat if<br />p.q = 0 <br />theneither p or q must bezero.<br />Thereforetherewillbe 2 solutions.<br />*This onlyworks if the RHS is 0 <br />
  4. 4. Solve for x:<br />x – 3 = 0<br />(x – 3)(x + 2) = 0<br />x(x + 7) = 0<br />x2 - 4x = 0<br />x2 + x – 6 = 0<br />(x – 3)(x +5)(x – 1) = 0<br />If ONE factor is equal to zero, the whole expression will be zero.<br />
  5. 5. Method 1: Factorising<br />x2 – 2x – 35 = 0<br /> (x – 7)(x + 5) = 0<br />x – 7 = 0 or<br /> x = 7<br />x(x – 1) = 20<br /> x2 – x – 20 = 0<br /> (x – 5)(x + 4) = 0<br />∴ x – 5 = 0<br /> x = 5<br />x + 5 = 0<br /> x = -5<br />x + 4 = 0<br /> x = -4<br />
  6. 6. Example 3<br />(x + 1)(x – 2) +3(x – 1)(x + 1) = 3(x – 2)<br />x2 – 2x + x – 2 + 3x2 – 3 = 3x – 6 <br /> 4x2 – 4x + 1 = 0<br /> (2x – 1)(2x – 1) = 0<br />2x – 1 = 0<br />2x – 1 = 0<br />2x = 1<br />2x = 1<br />OR<br /> x = ½<br /> x = ½<br />Both roots are equal<br />
  7. 7. Classwork<br />Pg 68 Ex 4.29, 11, 17, 21, 24<br />
  8. 8. Ex 4.2 (9 & 11)<br />9) 3x2 – 12 = 0<br /> 3(x2 – 4) = 0<br /> 3(x – 2)(x + 2) = 0<br />x – 2 = 0 or<br /> x = 2<br />11) 2x2 = 18 (÷2)<br /> x2 = 9<br /> x = ±√9<br /> ∴ x = ±3<br />x + 2 = 0<br /> x = -2<br />Remember a square root have a positive and a negative answer.<br />
  9. 9. Ex 4.2 (17 & 21)<br /> 7x - x2 – 6 = 0<br /> - x2 + 7x – 6 = 0<br /> -(x2 – 7x + 6) = 0<br /> -(x – 6)(x – 1) = 0<br />x – 6 = 0 or<br /> x = 6<br />21) x(x – 1) = 4(3x – 10)<br /> x2 – x = 12x - 40<br /> x2 – 13x + 40 = 0<br /> (x – 5)(x – 8) = 0<br />∴ x – 5 = 0 or x – 8 = 0<br /> x = 5 x = 8 <br />x – 1= 0<br /> x = 1<br />
  10. 10. Ex 4.2 (24)<br />24) 2(m – 1)(m +1)= 0<br /> 2m2 – 2 = 7m +14 - 1<br /> 2m2 – 7m – 15 = 0<br /> (2m + 3)(m – 5) = 0<br />2m + 3= 0 or m – 5 = 0<br /> 2m = -3 m = 5<br /> m = -3/2<br />Steps:<br />Multiply out.<br />Make the RHS = 0<br />Factorise LHS<br />Set each factor = 0<br />Solve each factor.<br />
  11. 11. Home work<br />Pg 68 Ex 4.22, 4, 7, 8,12, 13, 16, 18, 23<br />
  12. 12. Equations with fractions<br />Division by zero is undefined.<br />Start by writing down the restrictions.<br />Multiply with the denominator.<br />
  13. 13. Example 1<br />Limits:<br />x ≠ 0<br />multiply with x<br />(x – 5)(x + 2) = 0<br />(x – 5) = 0<br />(x + 2) = 0<br />x = 5 <br />x = -2<br />
  14. 14. Example 2<br />multiply with (x-2)(x-3)<br />(x – 4)(x – 2) = 0<br />Limits:<br />x ≠ 2<br />(x – 4) = 0<br />(x - 2) = 0<br />x ≠ 3<br />x = 4 <br />x = 2<br />
  15. 15. Example 3<br />multiply with (x+2)(x-2)(x+1)<br />Limits:<br />x ≠ 2<br />x ≠ -2<br />x ≠ -1<br />(x – 3) = 0<br />(x + 1) = 0<br />x = 3 <br />x = -1<br />
  16. 16. Home work<br />Pg 70 Ex 4.32, 4, 6, 10, 17<br />
  17. 17. Method 3: Substitution(another k-method)<br />If an algebraic expression is repeated in the equation we can replace it with “k” to make a simpler equation. <br />Just remember to substitute back once you have calculated k.<br />
  18. 18. 1. (x2 + 2x)2 – 2(x2 + 2x) – 3 = 0<br />Let k = (x2 + 2x) <br />∴ k2 – 2k – 3 = 0<br /> (k – 3)(k + 1) = 0<br />∴k = 3 or k = -1<br />k = (x2 + 2x) <br />k = (x2 + 2x) <br />3 = x2 + 2x <br />-1 = x2 + 2x <br />0= x2 + 2x - 3<br />0= x2 + 2x + 1<br />0= (x + 3)(x – 1)<br />0= (x + 1)(x + 1)<br />x = -3 or x = 1<br />x = -1 or x = -1<br />
  19. 19. 2. (x2 + x)2 – 14(x2 + x) + 24 = 0<br />Let k = (x2 + x) <br />∴ k2 – 14k + 24 = 0<br /> (k – 12)(k – 2) = 0<br />∴k = 12 or k = 2<br />k = (x2 + x) <br />k = (x2 + x) <br />12 = x2 + x <br />2= x2 + x <br />0= x2 + x - 12<br />0= x2 + x – 2<br />0= (x + 4)(x – 3)<br />0= (x - 1)(x + 2)<br />x = -4 or x = 3<br />x = 1 or x = -2<br />
  20. 20. 3. <br />Let k = (x2 - 3x) <br />k2 – 8k - 20 = 0<br /> (k – 10)(k + 2) = 0<br />∴k = 10 or k = -2<br />k = (x2 - 3x) <br />k = (x2 - 3x) <br />Limits:<br />10 = x2 – 3x <br />-2 = x2 – 3x <br />x ≠ 0<br />0= x2 – 3x - 10<br />0= x2 – 3x + 2 <br />0= (x + 2)(x – 5)<br />0= (x – 2)(x – 1)<br />x ≠ 3<br />x = -2 or x = 5<br />x = 2 or x = 1<br />
  21. 21. Home work<br />Pg 72 Ex 4.42, 3, 7, 8, 9<br />
  22. 22. Method 4: Squaring both sides<br />Test:<br />1)<br />x = 3<br />x = -2<br />x = 3 or x = -2 <br />x ≠ -2<br />Squaring both sides might introduce a extra solution which is invalid. <br />
  23. 23. Notes:<br />By definition is non-negative.<br />∴ √9 = 3 and √9 ≠ -3, <br /> but<br /> x2 = 9<br /> √x2 =√9<br /> x = ±3<br />The square root of a negative number is not defined.<br />√x ∈ ℝ ⇒ x ≥ 0<br />(a + b)2 ≠ a2 + b2<br /> (a + b)2 = a2 + 2ab + b2<br />
  24. 24. Limits:<br />x – 3 ≥ 0<br />x ≥ 3<br />
  25. 25. Home work<br />Pg 74 Ex 4.51, 4, 5, 7, 9, 10, 11, 12<br />

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