SOLVING QUADRATIC EQUATION

The introduction of = sign to the expressions above will change them to equation, and hence we will be required to solve the equations for the given values of the unknown. Because quadratic equation is of power 2, we will get two values of the unknown. In order to solve quadratic equation, we will first factorize as done in the previous section, then we later proceed to solve.

Note that a quadratic equation has the general form ax2 + bx + c = 0

          Example

Solve the quadratic equations below:

       1.    x2 + 8x – 20 = 0

       2.    2a2 – 7a = 15

       3.    36p2 – 25 = 0

       4.    7d = 2d2 – 28

       5.    81 = 25z2

       6.    0 = 24e – 36e2

            Solution

 

1.       x2 + 8x – 20 = 0  

          Factorize the left side of the equation.

 

I believe by now you have mastered the factorization steps, so I will insert the factors into the equations directly. The long process of using product and sum to bring out the factors will be skipped.

 

            x2 – 2x + 10x – 20 = 0

            (x2 – 2x) + (10x – 20) = 0

            x(x – 2) + 10(x – 2) = 0

            (x – 2) (x + 10) = 0

 

Now that factorization is complete, we will now equate each expression in the bracket to zero and separating them with or

            (x – 2) =0                   or                   (x + 10) = 0

            x = 0 + 2                     or                    x = 0 – 10

            x = 2                           or                    x = - 10

 

            x = 2 or – 10

 

the two values of x (2 and –10) above are the solutions to the equation and are called the roots of the equation.

 

 

2.       2a2 – 7a = 15

          Re-arrange the equation so that one side is 0

            2a2 – 7a – 15 = 0

            2a2 + 3a – 10a – 15 = 0

            (2a2 + 3a) – (10a + 15) = 0

            a(2a + 3) – 5(2a + 3) = 0

            (2a + 3)(a – 5) = 0

            2a + 3 = 0                  or                    a – 5 = 0

            2a = 0 – 3                  or                    a = 0 + 5

            2a = –3                      or                    a = 5

            a = –3/2

 

            a = - 3/2 or 5

           

 

3.         36p2 – 25 = 0

            62p2 – 52 = 0

            (6p)2 - 52 = 0

            (6p – 5)(6p + 5) = 0

            6p – 5 = 0                  or                    6p + 5 = 0

            6p = 0 + 5                  or                    6p = 0 – 5

            6p = 5                        or                    6p = - 5

            p = 5/6                       or                    p = - 5/6

 

            p = – 5/6 or 5/6

 

 4.        d = 2d2 – 28

            Re-arrange

            0 = 2d2 – d – 28

            0 = 2d2 – d – 28 

            0 = 2d2 + 7d – 8d – 28

            0 = (2d2 + 7d) – (8d + 28)

            0 = d(2d + 7) – 4(2d + 7)

            0 = (2d + 7)(d – 4)

            0 = 2d + 7                  or                    0 = d – 4

            0 – 7 = 2d                  or                    0 + 4 = d

            - 7 = 2d                      or                    4 = d

            -7/2 = d

            d = -7/2                     or                    d = 4

 

            d = 4 or – 7/2

 

 5.         81 = 25z2

            0 = 81 – 25z2

            0 = 92 – 52z2

            0 = (9 – 5z)( 9 + 5z)

            0 = 9 – 5z                  or                    0 = 9 + 5z

            0 + 5z = 9                  or                    0 – 9 = 5z

            5z = 9                         or                    - 9 = 5z

            z = 9/5                       or                    -9/5 = z

                       

z = 9/5 or – 9/5

 

Note: for equations involving difference of two squares, we will always get the same value but the signs will be different.

           

   6.      0 = 24e – 36e2

            0 = 12e (e – 3)

            0 = 12e                      or                    0 = e – 3

            0/12 = e                     or                    0 + 3 = e

            0 = e                           or                    3 = e

 

                        e = 3 or 0

 

                       

                        Exercise J

Find the roots of the equations below.

            1.        3v2 + 19v + 20 = 0

            2.         3r2 = 11r – 6

            3.        14 – 15t = 9t2

            4.        16 = 81z2

            5.         28d2 + 15d = 0

            6.        4e – 21 =  – e2 

            7.        k2 – 3k – 10 = 0

            8.        62 – 2k = 5k2

            9.        49p2 = 225

            10.      32 – 4d = 32

            11.      2x2 – 5x + 3 = 0

            12.      t2 = 9t – 8

            13.      1 + 2c = c2

            14.      y2 = 3y

            15.      n2 + 7n = 30

            16.      48 + 8d = d2

            17.      8y = y2 + 16 

            18.      18 – 5m + 2m2 = 0

            19.      81 = 169r2

            20.      e2 – 25e + 136 = 0

            21.      2a2 = 19a – 9

            22.      0 = 9n2 – 21

            23.      81 = 121k

            24.      9p + 52 = p2

            25.      2y2 + 15 = 31y

            26.      30 – c = c2

            27.      8j2 – 44j = 0

            28.      6d2 = 4 + 5d

            29.      50 = 72p2

            30.      m2 + 18m + 77 = 0

            31.      8 = 5g2 – 6g

 

                              Forming quadratic equations from the given roots

Here, the roots of the quadratic equation will be given, we will then use the roots to get the equation. This is just the reverse of the previous section.

 

            Example

Find the quadratic equation with the following roots:

1.        2 and 3

2.        ½ and ¾

3.         – 1 and – 7

4.        4/5 and – 5 

5.        \dpi{100} \small - 2\frac{1}{3} \ \ and \ \ 1\frac{5}{6} 

6.         1.25 and 1.6 

7.        0 and –3

8.        –7 and 7 

 

            Solution       

1.        2 and 3

** Since 2 and 3 are the roots, that means x = 2 or x =3

            x = 2               or                    x = 3

** Move 2 and 3 to the left side to meet x and leave the right side with 0

Note: the signs will change on crossing the equality sign   

x – 2 = 0        or                    x – 3 = 0

** since the two equations are equal to zero, we can put them together without or.

            (x – 2)(x – 3) = 0

Expand

x2 – 3x – 2x + 6 = 0

x2 – 5x + 6 = 0

 

If we solve the above equation, we will get our initial roots.

 

2.         ½ and ¾

            p = ½                          or                                p = ¾

            p – ½ = 0                   or                                p – ¾ = 0

            (p – ½)(p – ¾) = 0  

              \dpi{100} \small 8 ×p^2 - 8 \times \frac{3p}{4}- 8 \times \frac{p}{2}+ 8 \times \frac{3}{8}= 8 \times 0         

**4 divides 8 to give 2, 2 divides 8 to give 4, 8 divides 8 to give 1

            8p2 – 2 x 3p – 4 x p + 1 x 3 = 0

            8p2 – 6p – 4p + 3 = 0

            8p2 – 10p + 3 = 0

 

 

3.         – 1 and – 7

            e = -1                          or                    e = -7

** move -1 and – 7 to join e at the left side, - will change to +

            e + 1 = 0                    or                    e + 7 = 0

            (e + 1)(e + 7) = 0

            e2 + 7e + e + 7 = 0

            e2 + 8e + 7 = 0  

 

4.        \dpi{100} \small \frac{4}{5} and –5 

         

            \dpi{100} \small k = \frac{4}{5}                        or                    k = –5

           \dpi{100} \small k - \frac{4}{5} = 0                   or                    k + 5 = 0

           (k – \dpi{100} \small \frac{4}{5})( k + 5) = 0  

            k2 + 5k – \dpi{100} \small \frac{4k}{5} – 4 = 0 

           

** multiply through with 5

            5 x k2 + 5 x 5k – \dpi{100} \small 5 \times \frac{4k}{5} – 5 x 4 = 0

            5k2 + 25k – 4k – 20 = 0

            5k2 + 21k – 20 = 0

5.    

      \dpi{100} \small -2\frac{1}{3} \ and \ 1\frac{5}{6} \\ x = -2\frac{1}{3} \ \ \ \ \ or \ \ \ \ \ \ \ x = 1\frac{5}{6} \\ x + 2\frac{1}{3} = 0 \ \ \ or \ \ \ \ x - 1\frac{5}{6} = 0 \\ \left ( x + 2\frac{1}{3}\right )\left ( x - 1\frac{5}{6} \right ) = 0 \\ x^2 - \frac{11x}{6} + \frac{7x}{3} - \frac{77}{18} = 0 \\ ** clear fraction \\ \\ 18 \times x^2 - 18 \times \frac{11x}{6} + 18 \times \frac{7x}{3} - 18 \times \frac{77}{18} = 18 \times 0 \\ \\ 18x^2 - 3 \times 11x + 6 \times 7x - 1 \times 77 = 0 \\ 18x^2 - 33x + 42x - 77 = 0 \\ \\ 18x^2 - 9x - 77 = 0 \\

      

6.        

6,        1.25 and 1.6

** change decimals to fraction  

            1.25                                        1.6

            125/100                               16/10

            5/4                                         8/5

            t = 5/4                        or                    t = 8/5

            t – 5/4 = 0                 or                    t – 8/5 = 0   

            \dpi{100} \left ( t - \frac{5}{4} \right )\left ( t-\frac{8}{5} \right ) = 0 \\ \\ t^2 - \frac{8t}{5} - \frac{5t}{4} + 2 = 0 \\ \\ ** clear \ fraction \ with \ 20 \\ \\ 20 \times t^2 - 20 \times \frac{8t}{5} - 20\times \frac{5t}{4} + 20 \times 2 = 20 \times 0 

             20t2 – 4 x 8t – 5 x 5t + 40 = 0

             20t2 – 32t – 25t + 40 = 0 

           20t2 – 57t + 40 = 0  

 

7,        0 and –3

          n = 0               or                    n = –3

            n = 0               or                    n + 3 = 0

            n(n+3) = 0

            n2 + 3n = 0

 

8,        –7 and 7

            g = –7             or                    g = 7

            g + 7 = 0        or                    g – 7 = 0

            (g + 7)(g – 7) = 0

            g2 – 7g + 7g – 49 = 0

                        note that –7g + 7g = 0

            g2 – 49 = 0

 

Exercise

Find the quadratic equations with the following roots

1.    3 and – 4

2.    –2/3 and 5/6

3.    2 ½  and –4⅓

4.    - 8 and –4

5.    5 and 1

6.    0 and 8

7.    –4/9 and 5/6

8.    –2⅖ and 0

9.     +5 and –5

10.   4 and –1

11.   0.3 and 0.1

12.    1.25 and –0.5

13.    –2.2 and –1.3 

14.    1.1 and 0.6

15.    +12 and –12

16.   –4/5 and 4/5   

 

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NEXT: completing the square method 

 

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