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# 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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