Why Even Roots Restrict a Domain
A formula may look meaningful for many inputs but fail to produce a real number for some of them. For a real-valued function, every operation in its formula must be defined over the real numbers. An even root is a key example: the square root of a negative real number is not real.
In several variables, the radicand may depend on two or more inputs. The domain is then described by an inequality involving those variables. For instance, a condition such as \(x^2+y^2\le 9\) describes a region in the plane, not just an interval on the number line.
The equality case matters: if the radicand is zero, the root is zero and is defined. The restriction is nonnegativity, not strict positivity. Odd roots do not have this restriction over the real numbers, but this tutorial concerns even roots.
Turning a Radicand into an Inequality
To determine where an even-root formula is defined, identify the entire expression underneath the root. That expression is the radicand. Require it to be greater than or equal to zero, then solve the resulting inequality for the input variables.
Include every term under the root, respecting parentheses and signs.
Require the radicand to be at least zero.
Describe the input values or region that satisfies it.
Substitute points on both sides of the restriction to verify the sign and the equality case.
This method works because the real numbers have a nonnegative square root, and more generally a nonnegative even root, for every nonnegative radicand. Conversely, an even power of a real number cannot be negative. Thus an expression under a real even root must be nonnegative, and that condition is sufficient as well as necessary.
Worked Example: A Linear Radicand in Two Variables
Worked Example: A Linear Radicand in Two Variables
Find where \(f(x,y)=\sqrt{12-3x+2y}\) is real.
Step 1: Identify the radicand. The full radicand is \(12-3x+2y\).
Step 2: Require it to be nonnegative.
Step 3: Rearrange. Subtract \(12\), then add \(3x\), to obtain \(2y\ge 3x-12\). Dividing by the positive number \(2\) preserves the inequality:
The domain is the half-plane on or above the line \(y=\frac{3}{2}x-6\).
Check: At \((0,0)\), the radicand is \(12-3(0)+2(0)=12\), so the input is allowed. At \((4,0)\), it is \(12-3(4)+2(0)=0\), so the root is defined and equals \(0\). At \((5,0)\), it is \(12-3(5)+2(0)=-3\), so the root is not real. These checks agree with the inequality: the first two points satisfy \(y\ge \frac{3}{2}x-6\), while \((5,0)\) does not.
Quadratic Radicands Describe Curved Regions
When variables appear squared, the inequality may describe a disk, an ellipse, or a region outside a circle or ellipse. The sign of the inequality is important: a radicand that is a constant minus squared terms usually gives a bounded region, while squared terms minus a constant often gives an unbounded region.
Worked Example: A Circular and Elliptical Restriction
Worked Example: A Circular and Elliptical Restriction
Find where \(g(x,y)=\sqrt{16-x^2-4y^2}\) is real.
Step 1: Write the condition. The radicand must satisfy
Step 2: Rearrange. Moving the squared terms to the other side gives \(x^2+4y^2\le 16\). Dividing by \(16\), a positive number, gives
This is the filled ellipse centered at the origin, with horizontal semiaxis \(4\) and vertical semiaxis \(2\).
Check an interior point: At \((2,1)\), the radicand is \(16-2^2-4(1^2)=16-4-4=8\). Since \(8\ge 0\), the root is defined, and \(g(2,1)=\sqrt{8}=2\sqrt{2}\).
Check a point on the limiting curve: At \((0,2)\), the radicand is \(16-0^2-4(2^2)=16-16=0\), so the root is defined.
Check a point outside: At \((2,2)\), the radicand is \(16-2^2-4(2^2)=16-4-16=-4\), so the root is not real. Each result agrees with the inequality \(x^2+4y^2\le 16\).
Using Algebra to Simplify the Condition
A radicand may contain a quadratic expression that is easier to understand after factoring or rewriting it as a square. For example, \(x^2-2xy+y^2\) is exactly \((x-y)^2\). Recognizing that identity converts a two-variable quadratic inequality into a condition on the difference between the variables.
Worked Example: A Quadratic Condition on a Difference
Worked Example: A Quadratic Condition on a Difference
Find where \(h(x,y)=\sqrt{x^2-2xy+y^2-9}\) is real.
Step 1: Require a nonnegative radicand.
Step 2: Factor the quadratic part. Since \(x^2-2xy+y^2=(x-y)^2\), the condition becomes
Step 3: Solve the square inequality. Here \(9=3^2\), so \(|x-y|\ge 3\). Therefore either \(x-y\ge 3\) or \(x-y\le -3\). The domain consists of the two regions on either side of the strip \(-3<x-y<3\), including the two boundary lines.
Check an allowed point: At \((4,0)\), the radicand is \(4^2-2(4)(0)+0^2-9=16-0+0-9=7\), which is nonnegative. Also, \(x-y=4\ge 3\).
Check a boundary point: At \((3,0)\), the radicand is \(3^2-2(3)(0)+0^2-9=9-0+0-9=0\). The root is defined and equals \(0\).
Check a disallowed point: At \((1,0)\), the radicand is \(1^2-2(1)(0)+0^2-9=1-0+0-9=-8\). Also, \(-3<1-0<3\), so this point does not satisfy either allowed inequality. The algebraic condition and the direct substitution agree.
Why the Boundary Is Included
For an even root, a zero radicand is valid: \(\sqrt{0}=0\). Consequently, when solving the radicand inequality, replacing \(\ge 0\) with \(>0\) would incorrectly discard inputs where the radicand is zero. In geometric descriptions, this is why the limiting curve or line is included when it makes the radicand exactly zero.
This is not a special geometric rule. It follows directly from the definition of the real square root. The same reasoning applies to any even root: \(\sqrt[n]{0}=0\) for an even positive integer \(n\).
Common Mistake: Treating Variables Separately
A common error is to inspect each variable separately and assume that individual bounds are enough to describe the full condition. For the radicand \(9-x^2-y^2\), the correct requirement is
A mistaken calculation might reason that \(x^2\le 9\) and \(y^2\le 9\), and therefore accept every point with \(-3\le x\le 3\) and \(-3\le y\le 3\). But the point \((3,3)\) meets both individual bounds while its radicand is \(9-3^2-3^2=9-9-9=-9\). The root is not real there.
The correction is to keep the terms together in the original inequality. At \((3,0)\), the radicand is \(9-9-0=0\), so the point is allowed. At \((3,3)\), the combined squared terms exceed \(9\), so it is not. The condition describes a disk, not the entire square given by separate coordinate bounds.
A Reliable Checklist
Before accepting a proposed domain for a formula with an even root, check the following:
- Did you include the entire expression under the radical, with its signs and parentheses?
- Did you require the radicand to be nonnegative rather than strictly positive?
- When dividing or multiplying an inequality, did you account for the sign of the factor?
- If the radicand contains several variables, did you preserve their combined condition?
- Can you test one point that satisfies the inequality and one that does not by substituting into the original radicand?
These checks help distinguish an algebra error from a mistaken interpretation of the region. They also verify that the description follows from the original formula, not merely from a visual guess.
Check Your Understanding
For each question, begin by writing the radicand inequality. Then solve it and, where requested, verify the result by substitution.
- For \(p(x,y)=\sqrt{5-x+2y}\), what inequality describes its real domain?
- For \(q(x,y)=\sqrt{25-x^2-y^2}\), is \((3,4)\) allowed? Is \((4,4)\) allowed? Show the radicand calculations.
- Find the condition on \(x\) and \(y\) that makes \(\sqrt{(x+y)^2-4}\) real.
- Why is it incorrect to replace the condition \(g(x,y)\ge 0\) with \(g(x,y)>0\) for a real square root?
- For the radicand \(16-x^2-4y^2\), explain why separate bounds on \(x\) and \(y\) do not by themselves give the full domain condition.