Crazy “Quadratic” equations

 

Introduction

 

          Some equations are not really quadratic. However, they can be transformed to quadratic equations and can then be solved easily.  Here are some examples.

 

 

(1)     Solve:     9x4 – 145x2 + 16 = 0

 

          Solution :               9(x2)2 – 145(x2) + 16 = 0             (Quadratic with variable x2)

                                          (9x2 – 1) (x2 – 16) = 0

                                          9x2 = 1         or          x2 = 16

                                  \    x = ± 1/3       or           x = ± 4

(2)     Solve:     4x – 2x – 2 = 0

          Solution:                (2x)2 – (2x) – 2 = 0                        (Quadratic with variable 2x)

                                          (2x + 1) ( 2x – 2) = 0

                                          2x = - 1          or            2x = 2

                                          2x = - 1 has no real solution

                                                          (the complex root is beyond the discussion here),

                                  \    x = 1 is the only real root.   

                                         

 

Challenge

 

               I think you have got the craziness on changing equations to quadratics. Can you solve the following harder equations employing similar technique?

 

Solve the equations for real roots:

1.   

2.    2(4x) + 6x = 9x

3.   

4.    (x + 1)(x + 2)(x + 3)(x + 4) = 360

5.   

 

Try first, then look for short solutions below.

 

 

Short Solutions

 

1.   

       Put  t = x2 + 6x + 1, the equation becomes:

              

       For t ¹ 0, -2, -7, we get:     4t(t + 1) + 5t(t + 7) = 3(t + 7) (t + 1)

       Simplify, we have:                         2t2 + 5t – 7 = 0              (Gee, quadratic!)

       Solving for t:                                  t = -7/2 , 1

       Putting back x:                       x2 + 6x + 1 = -7/2 ,      x2 + 6x + 1 =1

       Quadratic again:                     2x2 + 12x + 9 = 0,         x2 + 6x = 0

       Solving for x,                         x = (-6 ±Ö18)/2,  -6, 0.

 

2.    2(4x) + 6x = 9x

       Rearrange, you get:               (3x)2 – (3x)(2x) – 2(2x)2 = 0         (Quadratic in 3x and 2x )

       Factorize the quadratic:        [3x – 2(2x)] [3x + 2x] = 0

                                                       3x = 2(2x)                                  (3x + 2x = 0 has no solution)

       Taking logarithm,                   x log 3 = log 2 + x log 2  

                                                       x = log 2 / (log3 – log 2) » 1.71          

      

3.   

      

      

      

      

       Squaring, we have  

       \                        (Note that one of solutions has been rejected)

 

 

4.    (x + 1)(x + 2)(x + 3)(x + 4) = 360

       [(x+1)(x+4)][(x+2)(x+3)] = 360

       [x2 + 5x + 4][x2 + 5x + 6] = 360

       Put  t = x2 + 5x + 4, we get    t (t + 2) = 360

                                                               t2 + 2t – 360 = 0

                                                               (t + 20)(t – 18) = 0

                                                               t = -20 or 18

\                 x2 + 5x + 4 = -20         or           x2 + 5x + 4 = 18

                x2 + 5x + 24 = 0            or            x2 + 5x – 14 = 0

The first equation has no real roots and the second equation give the solution:

                x = -7, 2

 

5.   

      

      

      

      

      

       For  x ¹ 1, 2, 3,      x[(x-2)(x-3) + (x-1)(x-3) + (x-1)(x-2)] = 0

                                               x[3x2 –12x +11] = 0

                                       \    x = 0  or      (6±Ö3)/3