7.3 Real Life Examples Example 3. The length of a - - PDF document

Chapter 7: Quadratic Equations

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SET 1

Chapter 7

Quadratic Equations

لاةـيعيـبرتلا تلبداـعم

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Chapter 7: Quadratic Equations

7.1 Introduction ةـمدـقم  A quadratic equation is one in which the highest power of the unknown quantity is 2.  For example, x2 – 3x + 1 = 0 is a quadratic equation.  The general form of quadratic equations is:

a x 2 + b x + c = 0

 There are five methods for solving quadratic equations: (i) by square root. رذـجلا ذـخأبيـعيبزتلا (ii) by factorisation (where possible). لـيلحتلاب (iii) by using the quadratic formula. ماـعلا نوناقلاب (iv) by completing the square. عـبزملا لاـمكإب (v) graphically. يـناـيبلا نـسزلاب 7.2 Solution by the Quadratic Formula ماـعلا نوناـقلا مادـختسإب لـحلا  The Quadratic formula is written as follows: a ac b b x 2 4

2 

  

Example 1. Solve

8 2

2

   x x

by using the quadratic formula. Solution For the equation

8 2

2

   x x

: a = 1 , b = 2 and c = ‒ 8 . Substituting these values into the quadratic formula a ac b b x 2 4

2 

   gives: 2(1) 8) 4(1)( (2) 2) (

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     x 2 6 2 2 36 2 2 32 4 2           x 2 6 2    x

• r

2 6 2   So, 2   2 4 x

• r

4     2 8 x

Chapter 7: Quadratic Equations

3 Example 2. Solve

2 7 4

2

   x x

by using the quadratic formula. Solution For the equation

2 7 4

2

   x x

: a = 4 , b = 7 and c = 2 . Substituting these values into the quadratic formula a ac b b x 2 4

2 

   gives: 2(4) ) 2 4(4)( (7) 7) (

2 

   x 8 17 7 8 32 49 7        x 8 17 7    x

• r

8 17 7   Therefore, 8 17 7    x

• r

8 17 7    x

7.3 Real Life Examples عـقاوـلا نـم لـئاـسم

Example 3. The length of a rectangular piece of land is 4 m longer than its width, and its area is 45 m2. Find the dimensions

• f this land.

Solution Let the length = x  the width = x – 4 The area = length × width The area = x (x – 4) = 45 x2 – 4x – 45 = 0 a = 1 , b = ‒ 4 and c = ‒ 45 . Substituting these values into the quadratic formula a ac b b x 2 4

2 

   gives: 2(1) ) 45 4(1)( 4) ( ) 4 (

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       x 2 14 4 2 196 4 2 80 1 6 1 4        x 9 2 4 1 4    x

• r

5 2 4 1 4     x (ignore) Therefore, length = 9 m and the width = 9 – 4 = 5 m

x x – 4

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Chapter 7: Quadratic Equations

Example 4 The hypotenuse of a right-angled triangle is 6 cm longer than the shorter leg, and the longer leg is 3 cm more than the shorter leg. Find the length of each side of this triangle. Solution Let the length of the shorter leg = x Then, the length of the longer leg = x + 3 , and the length of the hypotenuse = x + 6 Applying Pythagoras theorem gives: (x + 6)2 = x2 + (x + 3)2 x2 + 12x + 36 = x2 + x2 + 6x + 9 x2 – 6x – 27 = 0 a = 1 , b = ‒ 6 and c = ‒ 27 . Substituting these values into the quadratic formula a ac b b x 2 4

2 

   gives: 2(1) ) 27 4(1)( 6) ( ) 6 (

2

       x 2 12 6 2 144 6 2 8 10 6 3 6        x 9 2 2 1 6    x

• r

3 2 2 1 6     x (ignore) Therefore, the shorter leg = 9 cm , the longer leg = 9 + 3 = 12 cm , and the hypotenuse = 9 + 6 = 15 cm

x + 3 x