Irrational Equations (Square Root Equations)

What is an irrational equation?

An irrational equation is an equation where the unknown x sits under a root sign. Examples: √(x+7) = 3, or √(2x−1) = x−2.

Two small rules first. A square root sign means the non-negative root: √4 = 2 (not −2). And what is inside the root must be 0 or more. That gives the domain: for √(x+2) we need x + 2 ≥ 0, so x ≥ −2.

So a root is never negative. This one fact explains almost every trap in this lesson.

How to solve: isolate, square, solve, check

1. Isolate the root: get it alone on one side.
2. Square both sides. The root sign disappears.
3. Solve the new equation (often a quadratic).
4. Check each answer in the ORIGINAL equation. Keep only those that work.

Example: √(x+7) = 3. Square: x + 7 = 9, so x = 2. Check: √9 = 3. True. Answer: x = 2.

If the equation has a root on both sides, such as √(3x+1) = √(x+5), you can square at once: 3x + 1 = x + 5, so x = 2.

False (extraneous) roots: why we must check

Squaring turns "A = B" into "A² = B²". But A² = B² is also true when A = −B. So squaring lets in answers where the two sides are opposites. These are false roots (also called extraneous roots).

Example: √(x+2) = x. Square: x + 2 = x², so x² − x − 2 = 0, giving x = 2 or x = −1. Check x = −1: left side √1 = 1, right side −1. No! Only x = 2 is true.

Shortcut: after isolating the root, the other side must be 0 or more. If your answer makes it negative, throw the answer out.

Another type uses two roots: √(x+3) + √(x−2) = 5. Isolate one root, square, isolate the second root, square again. You get x = 6; check: 3 + 2 = 5. True.

Irrational inequalities

For an inequality, we cannot just square, because squaring needs both sides to be non-negative. Use these two rules:

• √f < g is true when f ≥ 0, g > 0 and f < g². (All three at once.)
• √f > g is true in two cases: (1) g < 0 and f ≥ 0, or (2) g ≥ 0 and f > g².

Example: √(x−1) < 3. Need x − 1 ≥ 0 and x − 1 < 9. So 1 ≤ x < 10.

Example: √(x+2) > x. Case 1: x < 0 and x ≥ −2, so −2 ≤ x < 0. Case 2: x ≥ 0 and x + 2 > x², so 0 ≤ x < 2. Together: −2 ≤ x < 2. On the 3D graph this is exactly where the green curve is above the orange line.

Problems with a parameter

A parameter is a letter like a that stands for a fixed but unknown number. A typical question: "For which a does √(x+a) = x have solutions?"

Think of the graph. The line y = x is fixed. The curve √(x+a) slides left or right as a changes. Squaring: x² − x − a = 0, and we need x ≥ 0.

• a < −¼: the curve is too far right, no solution.
• −¼ ≤ a < 0: the curve cuts the line twice, two solutions (touching at a = −¼).
• a ≥ 0: exactly one solution.

Use the a slider in the 3D to see each case.

Try it: predict, then check

In the 3D set a = 2, b = 0 and look at the dots. Now set b = −2. Before you look, guess how many true roots there are. Then check the readout. Next, at home: take any number n, find its square root on a calculator, add 1, and call the result y. Can you now find n again from y by "undoing" the steps (subtract 1, then square)? That is exactly how you solve √n + 1 = y.

Key formulas and definitions

• √A = B ⟹ A = B², and B must be ≥ 0
• √f = √g ⟹ f = g, with f ≥ 0
• √f < g ⟺ f ≥ 0, g > 0, f < g²
• √f > g ⟺ (g < 0 and f ≥ 0) or (g ≥ 0 and f > g²)
• Domain: the part under a square root must be ≥ 0

Worked examples

Common mistakes

• Squaring each term separately: (√x + 1)² is NOT x + 1. It is x + 2√x + 1. Isolate the root first.
• Forgetting to check. Squaring can add false roots, so always put answers back in the original equation.
• Thinking √9 = ±3. The root sign means the non-negative root only: √9 = 3.
• Ignoring the domain. For √(x−3) we need x ≥ 3, so any answer below 3 is out.