Reviewing Algebra II- Part Three: Solving & Simplifying Radical Equations & Expressions

 In Part One the focus was on relatively benign algebraic fractions and in Part Two the equally benign exponential equations (solved without use of logarithms). Now we will move on to the more challenging solution of radical equations, which may also involve surds. So a starting point is distinguishing radicals from surds.

Example: Solve the radical equation:

 Ö(5x - 11)   - Ö(x - 3)   = 4  

The equation contains a root (in Ö5  as a coefficient of x) which is both a radical and surd.

Solving the equation is straightforward.  We begin by isolating the radical on the left side, making it the subject, viz.

Ö(5x - 11)   =  Ö(x - 3)   + 4  

Next equate the squares of the members:

5x - 11   =  x - 3   + 8Ö (x - 3)   + 16

Isolating 8Ö (x - 3), making it the subject:

 8Ö (x - 3)   = 5x - 11  -  x + 3  -16

Simplify the right side and combine like terms:

 8Ö (x - 3)   =  4x -  24

Divide each side by 4: 

  2 Ö (x - 3) = x - 6 

 Squaring each side: 

4 (x - 3) = x ²  - 12x  + 36 

 Apply distributive law:

4x -  12 = x ²  - 12x  + 36 

Add 4x - 12 to each side:

x ²  - 16x  + 48 =  0

Factoring the left hand side:

(x - 12)  (x - 4)  =  0  

Setting x - 12 = 0 and solving:

x = 12

Doing the same for x -4:

x = 4

Substitute values back into original equation, i.e.: 

Ö(5x - 11)   =  Ö(x - 3)   + 4  

We obtain:

Ö(60 - 11)  -  Ö(12 - 3)   =  Ö49   -  Ö9  = 7 - 3  = 4 

Therefore 12 is one root.

But, checking x = 2: 

Ö(20 - 11)  -  Ö(4 - 3)   =  Ö9   -  Ö1 = 3 - 1  = 2

2  ≠ 4, so x= 12 is the only root

The method is longer than one might expect because squaring a radical-surd equation doesn't always result in an equivalent equation.

Practice Problems: 

Solve each for x and verify the root is correct.

1) Ö(2x - 3)   = 5

2)  Ö(x + 3)   + Öx   = 5

3)   3   - Ö(x + 1) =  0

4)   Ö(x + 5)   =   1 +  Öx

5)  Ö(x + 1)   + Ö(2x + 3) - Ö(8x + 1)    = 0