Math 112  10.  Worksheet on "Quadratic Types" of Eq.

 

Basically, simply note that any equation of the form

u2 + bu +c = 0

can be solved using quadratic methods.

The only hard part is "seeing " the quadratic form.

 

Example 1:

 

x - 3 Ãx + 2 = 0,       if we let u = Ãx, we'd have u2 - 3u + 2 = 0, which can easily be solved for u, then just as easily for x.

 

Solving by factoring we have, u2 - 3u + 2 = (u - 2)(u - 1) = 0, which has solutions u = 2 and u = 1

 

substituting u = Ãx, we have

 

Ãx = 2                          and à x  = 1  :  squaring both sides to eliminate the radical, we have

 

x = 4                            and  x = 1

 

Checking both these solutions in the original problem gives

 

4 - 3 (2) + 2 = 0 is true, so x = 4 is a valid solution and

 

1 - 3 (1) + 2 = 0 is also true so x = 1 is also a valid solution to the problem.

 

Example 2: 

y2/3 + 9y1/3 + 8 = 0  is just u2 + 9 u +8 = 0 if we let u represent y1/3 .

 

solving by factoring we  have (u + 1)(u + 8) = 0, which has solutions u = -1 and u = - 8

 

substituting u = y 1/3 we get

 

y 1/3  = -1         and y 1/3  = -8     :   since we want y = y1 , we must cube both sides to make the exponents become 1, and we get

 

y = -1              and y = - 512

 

Because we raised both sides to an odd power, (3), we do not have to check our answers since there will be no extraneous solutions.

 

Example 3: 

15 t-4  - 23 t - 2 + 4 = 0   is just 15u2 - 23 u + 4 = 0 if we let u represent t -2

 

which factors as (3u - 4)( 5 u - 1) = 0  which yields u = 4/3 and u = 1/5

 

Replacing u by t -2 gives us the equations

 

t -2 = 4/3          and t -2 = 1/5  :  Since we want t = t1  , we must raise both sides to the -1/2

 

which gives us t = ± (3/4)1/2    and t = ± (5)1/2 

Since we did not raise both sides to an even power, we do not have to check for extraneous roots.

 

Example 4: 

  (m2 + 2m)2 - 6(m2 + 2m) = 16 is just u2 - 6u = 16 if we let u = (m2 + 2m)

 

which yields u2 - 6u -16 = 0 in standard form, and factors as

(u - 8)( u + 2) = 0  which yields u = 8 and u = -2

Replacing u by (m2 = 2m) gives us the equations

 

m2 + 2m -8 = 0 and m2 +2m +2 = 0

 

m2 + 2m -8 = (m + 4)( m - 2) = 0 gives us m = -4 and m = 2 but m2 + 2m +2 = 0 will not factor.  Using the quadratic formula yields m = -1 ± i

 

Problems:

 

1.  2y2/3 + 5y1/3 -12 = 0   solution:  y = 27/8 and y = -64

 

2.  (m2 - 2m)2 + 2(m2 - 2m) = 15        solution:  m = 3, -1, 1 ± 2i

 

3.  y-2 - 2y-1  -  3 = 0   solution:  y = 1/3 and y = -1