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2 b b 4 ac x 2 a Chapter 5: Quadratic Equations in One - PDF document

Nov 15, 2023 •199 likes •240 views

SET 3 Chapter 5 Quadratic Equations in One Variable 5.1 Quadratic Equations An equation that contains one variable is said to be a quadratic

  1. SET 3 Chapter 5 Quadratic Equations in One Variable لادحاو ريغتمب ةيعيبرتلا تلبداعم 5.1 Quadratic Equations ةيعيبرتلا تلبداعملا  An equation that contains one variable is said to be a quadratic equation if the highest power of the variable is 2.  This type of equations can be solved: (i) By factorisation )ةبرجتلاب( ليلحتلاب (ii) By using the quadratic formula ماعلا نوناقلاب (iii) By completing the square عبرملا لامكإب (iv) Graphically ينايبلا نسرلاب The general form of quadratic equation is: ax 2 + bx + c = 0 Where a , b and c are constants. Chapter 5: Quadratic Equations in One Variable 1

  2. 5.2 Solving Quadratic Equations by Factorisation ليلحتلاب ةيعيبرتلا تلبداعملا لح  x   and (b)    2 2 Example 1. Solve the equations: (a) x 2 8 0 3 x 11 x 4 0 by factorisation. Solution:  x   2 (a) x 2 8 0      2 x 2 x 8 ( x 4 )( x 2 )  x   2 The quadratic equation x 2 8 0 becomes:    ( x 4 )( x 2 ) 0      Either ( 4 ) 0 x x 4     or ( x 2 ) 0 x 2  x     2 So the roots of x 2 8 0 are x and 2 4    2 (b) 3 x 11 x 4 0      2 3 11 4 ( 3 1 )( 4 ) x x x x    2 The quadratic equation becomes: 3 x 11 x 4 0    ( 3 x 1 )( x 4 ) 0 1      Either ( 3 x 1 ) 0 x 3     or ( x 4 ) 0 x 4 1      2 So the roots of 3 x 11 x 4 0 are x and 4 3 5.3 Solving Quadratic Equations by Using the Quadratic Formula ماعلا نوناقلاب ةيعيبرتلا تلبداعملا لح  Using quadratic formula is a straightforward method for solving any quadratic equation.  The quadratic formula that is used to solve a quadratic equation having the form    2 of is: ax bx c 0 2    b b 4 ac  x 2 a Chapter 5: Quadratic Equations in One Variable 2

  3.  x   Example 2: Solve 2 x 2 8 0 by using the quadratic formula. Solution:  x      gives : 2 2 Comparing x 2 8 0 with ax bx c 0 a = 1, b = 2 and c = ‒ 8. 2    b b 4 ac  Substituting these values into the quadratic formula gives: x 2 a     2 2 2 4(1)( 8)  x 2(1)        2 4 32 2 36 2 6    x 2 2 2  2   2  6 6  x or 2 2  4 8  2    Therefore x or 2 4 (as in example 1(a)). 2  x   2 4 x 7 2 0 Example 3: Solve by using the quadratic formula. Solution:     x   2 2 ax bx c 0 Comparing 4 x 7 2 0 with gives: a = 4, b = 7 and c = 2. 2    b b 4 ac  Substituting these values into the quadratic formula x gives: 2 a 2    7 7 4(4)( 2 )  x 2(4)      7 49 32 7 17   x 8 8     7 17 7 17      x 0 . 360 or 1 . 390 8 8  x       Hence the roots of 2 are and 4 x 7 2 0 x 0 . 360 x 1 . 390 Chapter 5: Quadratic Equations in One Variable 3

  4. 2    Example 4: Solve 2 . 68 x 10 . 3 x 4 . 2 0 by using the quadratic formula. Solution:      a 2 . 68 b 10 . 3 c 4 . 2 , and . Substituting the values of a , b and c into the quadratic formula 2    b b 4 ac  gives: x 2 a       2 ( 1 0 . 3 ) ( 1 0 . 3 ) 4(2.68)( 4 .2)  x 2(2.68)     10 . 3 1 06 . 09 45 . 024 10 . 3 151 . 114 10 . 3 12 . 293    x 5.36 5.36 5.36   10 . 3 1 2 . 293 10 . 3 1 2 . 293      x 4 . 215 or x 0 . 372 5.36 5.36    2.68 2    Thus, the roots of are and x 10.3 x 4.2 0 x 4 . 215 x 0 . 372 Chapter 5: Quadratic Equations in One Variable 4

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