Solving Irrational Equations

Irrational equations

An irrational or radical equation in one unknown is an equation in which the unknown letter occurs in a radicand. Thus

3x + 2\sqrt{x} = 16, \qquad \sqrt{1 - x} + \sqrt{x + 3} = 2, \qquad \sqrt[3]{x^2 - 8} = 0

are irrational equations.

Example 1. Solve \sqrt{2x - 5} - 3 = 0.

Transposing, \sqrt{2x - 5} = 3. Squaring both members, 2x - 5 = 9. Solving, x = 7.

Check: substituting 7 for x in the original equation, \sqrt{14 - 5} - 3 = 3 - 3 = 0.

In irrational equations it is understood that each radical expression, not preceded by the sign \pm, is to have one sign and only one; therefore each radical has one value and only one. That value is the principal root of the radical: \sqrt{9} = 3, never -3. This fact is of importance in checking.

Example 2. Solve 2\sqrt[3]{8x^3 + \dfrac{19x^2}{2}} - 2x - 2 = 2x - 1.

Transposing and collecting, 2\sqrt[3]{8x^3 + \frac{19x^2}{2}} = 4x + 1. Cubing each member, 64x^3 + 76x^2 = 64x^3 + 48x^2 + 12x + 1. Transposing and collecting, 28x^2 - 12x - 1 = 0, or (2x - 1)(14x + 1) = 0. Therefore x = \frac12 or -\frac{1}{14}. Both values check: an odd root such as a cube root can be negative, so cubing introduces no false roots here.

"Impossible equations"

It is easily possible to write a statement involving radicals which has the form of an equation but is not one. Thus \sqrt{x + 1} + \sqrt{x + 3} + 1 = 0 looks like an equation, but no value of x can satisfy it: it says that the sum of three positive numbers is zero. In the attempt to solve an apparent equation one may use the usual methods and obtain a result which does not satisfy the original statement. Not until one tries to verify the result is the falsity discovered.

Example 3. Solve 1 + \sqrt{x + 2} = \sqrt{x}.

Transposing, 1 - \sqrt{x} = -\sqrt{x + 2}. Squaring, 1 - 2\sqrt{x} + x = x + 2, so -2\sqrt{x} = 1. Squaring again, 4x = 1 and x = \frac14.

Check: 1 + \sqrt{\tfrac14 + 2} = 1 + \tfrac32 = \tfrac52, but \sqrt{\tfrac14} = \tfrac12. The statement is false: the equation has no root. (Already -2\sqrt{x} = 1 was impossible, since a principal square root is never negative.)

Extraneous roots

Example 4. Solve \sqrt{x - 1} + \sqrt{3x + 1} - 2 = 0. \quad(1)

Therefore x = 1 or 5.

Check: for x = 1: \sqrt0 + \sqrt4 - 2 = 0. So 1 is a root. For x = 5: \sqrt4 + \sqrt{16} - 2 = 4, and 4 \ne 0. So 5 is not a root. It was introduced by squaring equation (5): 5 satisfies x - 1 = +2\sqrt{x - 1}, not x - 1 = -2\sqrt{x - 1}.

Values for the unknown introduced during the solution of an equation are called extraneous roots. Equation (1) is typical of many radical equations which, when solved by rationalizing, give the roots not only of the original equation but also of the equations obtained from it by giving each radical the sign \pm.

Example 5. Solve \sqrt{x + 2} + \sqrt{3 - x} = 3.

Transposing, \sqrt{3 - x} = 3 - \sqrt{x + 2}. Squaring, 3 - x = 9 - 6\sqrt{x + 2} + x + 2; collecting and dividing by -2, x + 4 = 3\sqrt{x + 2}. Squaring, x^2 + 8x + 16 = 9x + 18, so x^2 - x - 2 = 0 and x = 2 or -1.

Check: \sqrt4 + \sqrt1 = 3 and \sqrt1 + \sqrt4 = 3. Both are roots: rationalizing does not necessarily introduce extraneous roots.

We cannot tell how many roots a radical equation has without solving it, nor whether a statement involving radicals is an equation at all. Results obtained are roots if they satisfy the original statement, and not otherwise.

Rule

Transpose the terms so that one radical expression (the least simple one) is the only term in the first member of the equation.

Next raise both members of this equation to the same power as the index of the radical in the first member.

If radical expressions still remain, repeat the two preceding operations until an equation is obtained which is free from radicals. Then solve this equation.

Check. Substitute in the original equation and reduce the resulting radicals to their simplest form. Never simplify by raising both members of the equation to a power, for extraneous roots introduced by that process would not then be detected. Reject all extraneous roots.

In this course the radicals stand for real numbers: a value of x that makes the radicand of a square root (or a fourth root) negative is rejected, while an odd root of a negative number is real (\sqrt[3]{-125} = -5).

Typing answers. List the roots that check, separated by commas: 2, -1. If every value you found is extraneous, type none. The checker tells you when a value you typed is an extraneous root.