
Factorizing a quadratic expression and finding the roots of a quadraticequation are closely related. Example 1 Find the solution to the equations (a) x + 6cx + 8 = 0, (b) x  4x + 4 = 0. Solution (a) The quadratic expression will factorize as follows. x + 6x + 8 = (x + 2)(x + 4). The solution to the equation may now be obtained; If x + 6x + 8 = 0 then (x + 2)(x + 4) = 0 . Thus either (x + 2) = 0, or (x + 4) = 0. The solution to the equation is thus x = 2 or x = 4. (b) In this example the expression is x  4x + 4 = (x  2)(x  2) = (x  2) . The solution to the equation x  4x + 4 = 0 is thus x = 2. In this case, the equation is said to have equal roots. Exercise 1 Find the solution to each of the following equations. (a) 2x + 5x + 3 = 0 (b) 3x + 7x + 2 = 0 (c) 3y  5y  2 = 0 (d) 4z  23z + 15 = 0 (e) 64z + 4z  3 = 0 (f) 4w  25 = 0 Solution (a) We can easily see that 2x + 5x + 3 = (2x + 3)(x + 1) . Thus if 2x + 5x + 3 = 0 then we have either (2x + 3) = 0 , or (x + 1) = 0. For the first of these 2x + 3 = 0 2x = 3 (adding 3 to both sides) (dividing both sides by 2) . The solution to the second is obviously x = 1. (b) We can easily see that 3x + 7x + 2 = (3x + 1)(x + 2) , so that if 3x + 7x + 2 = 0 , then (3x + 1)(x + 2) = 0 . Thus either 3x + 1 = 0 or x + 2 = 0 . For the first of these 3x + 1 = 0 3x = 1 (adding 1 to both sides) (dividing both sides by 3) . The solution to the second part is obviously x = 2. The solution to the original equation is thus x = 2 or x = 1/3. (c) We can easily see that 3y  5y  2 = (3y + 1)(y  2) = 0 . Thus either 3y + 1 = 0, or y  2 = 0. For the first part, 3y + 1 = 0 , 3y = 1 (adding 1 to both sides) , (dividing both sides by 3) . The solution to the second part is obviously y = 2. The quadratic equation 3y  5y  2 = 0 thus has the solution y = 1/3 or y = 2. (d) We can easily see that 4z  23z + 15 = (4z  3)(z  5) = 0 . Thus either 4z  3 = 0, or z  5 = 0. Proceeding as in the previous examples, the solution to the first partis z = 3/4 and to the second part is z = 3. The solution to 4z  23z + 15 = 0 is therefore z = 3/4 or z = 3. (e) We can easily see that 64z + 4z  3 = (16z  3)(4z + 1) = 0 . Thus either 16z  3 = 0 or 4z + 1 = 0. For the first part 16z  3 = 0 , 16z = 3 (adding 3 to both sides) , (dividing both sides by 16) . For the second part 4z + 1 = 0 4z = 1 (adding 1 to both sides) , z = 14 (dividing both sides by 4) . The solution to the equation 64z + 4z  3 = 0 is thus z = 3/16 or z = 1/4. (f) We can easily see that 4w  25 = (2w  5)(2w + 5) = 0. The solution to this is w = 5/2 or w = 5/2 , i.e. w = ±5/2 . Quiz Which of the following is the solution to the quadratic equation 12x + 17x  14 = 0 ? (a) 2, 7/12 (b) 2, 7/12 (c) 2, 7/12 (d) 2, 7/12 Solution This quadratic is the one that occurs in the first quiz. There it was seen that 12x + 17x  14 = (x + 2)(12x  7), so either x + 2 = 0 or 12x  7 = 0 . The solution to the first is x = 2 and to the second is x = 7/12. 