How to Solve Radical Equations

√ ax + b   ±  √ cx + d   =  √ ex + f 

Radical equations are algebraic equations of the form

√ ax + b   ±  √ cx + d   =  √ ex + f ,

where a, b, c, d, e, and f are real numbers. These equations can be solved by squaring both sides of the equation, rearranging terms, and then squaring both sides again. This process will simplify the problem to a quadratic equation in most cases, or a linear equation in some special cases.

You can apply the steps below to solve these equations, or use the calculator on the left, which returns the possible solutions for x when you input values for a, b, c, d, e, and f.

Example Problem #1

Solve the equation √ 2x + 15   ±  √ x - 1   =  √ x + 44  for x. The first step is the square both sides of the equation to obtain

(√ 2x + 15   ±  √ x - 1 )²  =  x + 44
2x + 15 ± 2√ 2x² + 13x - 15  + x - 1 = x + 44
± 2√ 2x² + 13x - 15  = -2x + 30

Now square both sides again:

4(2x² + 13x - 15) = (-2x + 30)²
8x² + 52x - 60 = 4x² - 120x + 900
4x² + 172x - 960 = 0
x² + 43x - 240 = 0

This quadratic equation has two solutions, x = 5 and x = -48. For this particular problem, the first solution satisfies the "+" equation and the second satisfies the "-" equation, though the "-" equation involves imaginary numbers. Here is what you get when you plug x = 5 and x = -48 into the equations:

√ 2(5) + 15   +  √ 5 - 1   =  √ 5 + 44 
√ 25   +  √ 4   =  √ 49 
5 + 2 = 7

√ 2(-48) + 15   -  √ -48 - 1   =  √ -48 + 44 
√ -81   -  √ -49   =  √ -4 
9i - 7i = 2i

In practical contexts, if one solution yields imaginary numbers it is often discarded.

Example Problem #2

Solve √ 5x + 11   ±  √ 69 - 4x   =  13. If you are using the calculator above, you plug in the parameters a = 5, b = 11, c = -4, d = 69, e = 0, and f = 169 (since √169 = 13).

If you want to solve the equation algebraically, first square both sides as in Example 1.

(√ 5x + 11   ±  √ 69 - 4x )²  =  169
5x + 11 ± 2√ -20x² + 301x + 759  + 69 - 4x = 169
± 2√ -20x² + 301x + 759  = 89 - x

Square both sides again:

4(-20x² + 301x + 759) = 7921 - 178x + x²
-80x² + 1204x + 3036 = 7921 - 178x + x²
-81x² + 1382x - 4885 = 0

This has two solutions: x = 5 and x = 977/81. In this problem, both solutions satisfy the "+" equation and there are no imaginary numbers involved in either solution. Observe:

√ 5(5) + 11   +  √ 69 - 4(5) 
= √ 36   +  √ 49 
= 6 + 7
= 13

√ 5(977/81) + 11   +  √ 69 - 4(977/81) 
= √ 5776/81   +  √ 1681/81 
= 76/9 + 41/9
= 117/9
= 13

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