METHOD OF COMPLETING THE SQUARE

If we try to solve the equation 2k2 – 6k + 3 = 0 using the factorization method explained in the previous pages, we will realize that there are no factors.

The factorization method, though simple, cannot solve all quadratic equation questions, hence the need for some other methods that can solve such equations.

The general form of a quadratic equation ax2 ± bx ± c = 0 still remains the same.

            a is the coefficient of x2

            b is the coefficient of x

            c is the constant

the above terms a, b and c will be used in the next examples so master them.

 

 

Example

Solve the equation 2k2 – 6k + 3 = 0 using completing the square.

 

Solution

2k2 – 6k + 3 = 0

 

Step1: re-arrange the equation such that c is away from others.

 

            2k2 – 6k = –3

 

Step2: make sure a is 1. If a has value other than 1, divide through by such value.

In the case of the equation above, divide through by 2 

          

          \dpi{100} \frac{2k^2}{2} - \frac{6k}{2}=\frac{-3}{2} 

          k2 – 3k = –3/2 

 

Step3: take the square of half of b and add to both sides of the equation.

            b = -3

            half of b = -3/2

            square of half of b = (-3/2)2

                                          = 9/4

            Add the 9/4 to both sides of the equation.

 

                        k2 – 3k + 9/4 = -3/2 + 9/4

 

Step4: at the left side of the equation, put k and – 3/2 (half of b) in a bracket and square the bracket. At the right side, add -3/2 and 9/4

                 

                        (k – 3/2)2 =  \dpi{100} \frac{-6+9}{4}

                        (k – 3/2)2 =  3/4

Step5: cancel the square at the left side by taking the square root of both sides

                      \dpi{100} \sqrt{\left ( k-\frac{3}{2} \right )^2} = \sqrt{\frac{3}{4}} \\ \\ \\ \left ( k-\frac{3}{2} \right ) = \pm \frac{\sqrt{3}}{2}

Note: the square root of a number will always take ±. i.e.\dpi{100} \sqrt{9} = \pm 3, \ \sqrt{2} = \pm 1.414 \ \ etc.   

 

Step6: make k the subject of the equation. Place the 3/2 immediately after the = sign.

   

              \dpi{100} k = \frac{3}{2} \pm \frac{\sqrt{3}}{2}\ \ \ \ \ \ \ \ \ lcm \ is \ 2 \\ \\ k = \frac{3\pm \sqrt{3}}{2}

 

Step7: at this point split the equation in to two. The first with + and the second with – 

          \dpi{100} k = \frac{3+\sqrt{3}}{2} \ \ \ \ \ \ \ or \ \ \ \ \ \ \ k = \frac{3-\sqrt{3}}{2}  

 

**you can stop here, or proceed to find the square root of 3 so that your answer will be in decimal. It depends on where the question wants you to stop. if the question does not specify where to stop, then proceed.

 

                  \dpi{100} \fn_cm k = \frac{3+1.732}{2} \ \ \ \ \ \ \ or \ \ \ \ \ \ \ k = \frac{3-1.732}{2} \\ \\ \ \ k = \frac{4.732}{2} \ \ \ \ \ \ \ or \ \ \ \ \ \ \ k = \frac{1.268}{2} \\ \\

               k = 2.366           or         k = 0.634

 

Example 2 

Solve the equation d2 + 5d = 2 

 

Solution 

              d2 + 5d = 2

** Step 1:  c is already at the right side of the equation, so we proceed to step 2 

                    d2 + 5d = 2

** Step 2: There is no need to divide both sides by anything since a is already 1. So we move to step 3

                  d2 + 5d = 2

** Step 3:  take the square of half of b and add to both sides of the equation. 

                 b = 5 

                 half of b = 5/2

                square of half of b = 25/4

       add the 25/4 above to both sides of the equation 

                     d2 + 5d = 2

                      d^2 + 5d + \frac{25}{4} = 2 + \frac{25}{4}  

Step4: at the left side of the equation, put d and + 5/2 (half of d) in a bracket and square the bracket. At the right side, add 2 and 25/4

                  \left ( d + \frac{5}{2} \right )^2 = 2 + \frac{25}{4} \\ \\ \left ( d + \frac{5}{2} \right )^2 = \frac{8 + 25}{4}

                 (d + 5/2)2 = 33/4 

 

Step5: cancel the square at the left side by taking the square root of both sides  

                        \sqrt{\left ( d + \frac{5}{2} \right )^2} = \sqrt{\frac{33}{4}} \ \\ \\ \\ d + \frac{5}{2} = \pm \frac{\sqrt{33}}{2}

 

Step 6: make k the subject of the equation. Place the 3/2 immediately after the = sign.

            d = - \frac{5}{2} \pm \frac{\sqrt{33}}{2} \\ \\ lcm \ is \ 2 \\ \\ d= \frac{- 5 \pm \sqrt{33}}{2}

 

Step7: at this point split the equation in to two. The first with + and the second with –  

                d= \frac{- 5 + \sqrt{33}}{2} \ \ \ \ or \ \ \ \ d= \frac{- 5 - \sqrt{33}} {2} \\ \\ d= \frac{- 5 + 5.745}{2} \ \ \ \ or \ \ \ \ d= \frac{- 5 - 5.745} {2} \\ \\ d= \frac{0.745}{2} \ \ \ \ or \ \ \ \ d= \frac{- 10.745} {2}

                d= 0.372           or     d = -5.572  

 

                       Example 3

     Solve the equation 70v = 81v2 - 37 

                 

                 Solution 

** Re-arrange 

                    37 =  81v2 - 70v 

                   81v2 - 70v = 37 

 

** divide through by a  

                 v^{2} - \frac{70v}{81} = \frac{37}{81}  

** Add the square of half of b 

              b = 70/81

               half of b = 35/81

               square of half of b = 1225/6561

         v^{2} - \frac{70v}{81} + \frac{1225}{6561} \ \ \ = \ \ \ \frac{37}{81} + \frac{1225}{6561} \\ \\ \left ( v- \frac{35}{81} \right )^2 \ \ \ = \ \ \ \frac{2997+1225}{6561} \\ \\ \left ( v- \frac{35}{81} \right )^2 \ \ \ \ = \ \ \ \ \frac{4222}{6561} \\ \\ Take \ square \ root \ of \ both \ sides \\ \\ \sqrt{ \left ( v- \frac{35}{81} \right )^2} \ \ \ \ = \ \ \ \sqrt { \frac{4222}{6561}} \\ \\ v- \frac{35}{81} \ \ \ \ = \ \ \ \ \pm \frac{64.977}{81} \\ \\ v \ \ = \ \ \ \frac{35}{81} \ \ \ \ \pm \ \ \ \ \frac{64.977}{81} \\ \\ v \ \ = \ \ \ \frac{35 \pm 64.977}{81} \\ \\ Split \ the \ equation \\      

 

         v \ \ = \ \ \ \frac{35 + 64.977}{81} \ \ \ \ or \ \ \ \ v \ \ = \ \ \ \frac{35 - 64.977}{81} \\ \\

        v = 99.977/81    or     v = - 29.977/81  

         v = 1.234          or       v = - 0.3701

 

EXERCISE 1

 

Solve the equations below using completing the square method
1.        x2 + 8x + 3 = 0
2.        p2 - 3p = 12
3.        a2  - 2a - 9 = 0
4.        0 = e2  + 3e + 1
5.        c2  = 6c - 4
6.        2m2  + 5m + 1 = 0
7.        3b2  - 12b - 8 = 0
8.        7 = 4y + 4y2 
9.        - 2n2  = 7n + 2
10.       2 + 5d2  = 9d
11.       6 - z = 3z2 
12.       x2 = 3 - x
13.       23j = 16j2 + 8
14.       9q2  + 13q + 3 = 0
15.      12 - 11r - 2r2 
16.      6 - 3u - 5u2 = 0
17.      1 + 2f = 5f2 
18.      2 = 5m + 6m2 
19.      11g2 - 12g + 3 = 0
20.       2h2  = 3h + 7
21.       0 = 2a2  - 5a - 9
22.       25e2 - 42e + 17 = 0 

 

 

NEXT: QUADRATIC FORMULA METHOD 

PRE: FACTORIZATION METHOD  

 

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