Why Even Roots Restrict the Domain
A formula can look simple but still fail to produce a real output for some inputs. For example, the real square root of a negative number is not defined. When a formula contains an even root, its radicand therefore limits which real inputs are allowed.
The radicand is the expression inside the root sign. To find a real domain, first require every even-root radicand to be nonnegative. If the formula also contains a denominator, require that denominator to be nonzero as well. The domain is the set of real inputs satisfying all the conditions at once.
The Nonnegative-Radicand Condition
For a square root \(\sqrt{A(x)}\), where \(A(x)\) is an expression in \(x\), the real-valued condition is
This same condition applies to any even root, such as \(\sqrt[4]{A(x)}\): its radicand must be nonnegative for the root to be real. In contrast, an odd root can be real for a negative radicand, so the rule in this tutorial is specifically about even roots.
Why is the condition necessary? If a real number \(r\) is the square root of a real number \(a\), then \(r^2=a\). Every real square is nonnegative, so \(a\geq 0\). Conversely, every nonnegative real number has a real square root. Thus, \(\sqrt{a}\) is real exactly when \(a\geq 0\). The corresponding existence property holds for every even root.
Worked Example: A Linear Radicand
Worked Example: A Linear Radicand
Find the real domain of \(f(x)=\sqrt{7-3x}\).
The radicand is \(7-3x\), so it must be nonnegative:
Subtract \(7\) from both sides:
Divide by \(-3\). Dividing an inequality by a negative number reverses its direction:
Therefore the domain is \((-\infty,\frac{7}{3}]\). To check the boundary, substitute \(x=\frac{7}{3}\) into the radicand:
The square root of \(0\) is defined, so the endpoint is included. If \(x=3\), which is outside the proposed domain, the radicand is \(7-9=-2\), and the real square root is not defined. This confirms both the boundary and the direction of the interval.
Combining Root and Denominator Restrictions
A denominator cannot equal zero. This is the denominator restriction used when finding domains: exclude every input that makes a denominator zero. With an even root, this restriction is applied in addition to the radicand condition, not instead of it.
If a square root itself appears in the denominator, two conditions apply. Its radicand must be nonnegative for the root to be real, and the root must not equal zero because it is a denominator. Together, those conditions mean that the radicand must be strictly positive.
For other formulas, write down each condition separately and take their intersection: retain only the real inputs that satisfy all the conditions.
Worked Example: A Root and a Separate Denominator
Worked Example: A Root and a Separate Denominator
Find the real domain of \(g(x)=\frac{\sqrt{x+2}}{x-3}\).
First, require the radicand to be nonnegative:
Next, require the denominator to be nonzero:
Both conditions must hold. Starting with \([-2,\infty)\), remove the one excluded point \(3\). The domain is \([-2,3)\cup(3,\infty)\).
Check the included endpoint \(x=-2\): the radicand is \(-2+2=0\), and the denominator is \(-2-3=-5\), which is nonzero. Thus \(g(-2)=0/(-5)=0\), so \(-2\) is allowed. At \(x=3\), the denominator is \(3-3=0\), so the formula is undefined. At \(x=4\), the radicand is \(4+2=6>0\) and the denominator is \(4-3=1\neq 0\), so an input greater than \(3\) is allowed. These checks agree with the stated domain.
When the Radicand Is a Fraction
Sometimes the radicand is itself a fraction. In that case, requiring the radicand to be nonnegative involves both the sign of its numerator and its denominator. The denominator of the fraction must also be nonzero. The resulting inequality can be solved by identifying where the numerator and denominator change sign, then checking the sign on each interval.
Worked Example: A Fraction Inside a Square Root
Worked Example: A Fraction Inside a Square Root
Find the real domain of \(h(x)=\sqrt{\frac{x+1}{x-2}}\).
The radicand must be nonnegative, and its denominator must not be zero:
The numerator is zero at \(x=-1\), and the denominator is zero at \(x=2\). These points divide the real line into three intervals. Check the sign of the fraction on each interval:
The fraction is positive on \((-\infty,-1)\) and \((2,\infty)\). At \(x=-1\), it equals \(0/(-3)=0\), so the square root is defined and \(-1\) is included. At \(x=2\), the fraction is undefined because its denominator is \(0\), so \(2\) is excluded. Therefore the domain is \((-\infty,-1]\cup(2,\infty)\).
Check one allowed input from each part: at \(x=-2\), the radicand is \(1/4\), so \(h(-2)=1/2\); at \(x=3\), the radicand is \(4\), so \(h(3)=2\). At \(x=0\), the radicand is \(-1/2\), so there is no real output. These substitutions verify the interval choices and the exclusion of the middle interval.
Worked Example: A Nonlinear Radicand and an Excluded Point
Worked Example: A Nonlinear Radicand and an Excluded Point
Find the real domain of \(p(x)=\frac{\sqrt{9-x^2}}{x+1}\).
The radicand condition is
Factor the expression:
Equivalently, \(x^2\leq 9\), which gives \(-3\leq x\leq 3\). The denominator must also be nonzero:
Removing \(-1\) from \([-3,3]\) gives the domain \([-3,-1)\cup(-1,3]\). At the endpoints, the radicand is \(9-(-3)^2=0\) and \(9-3^2=0\); the denominators are \(-3+1=-2\) and \(3+1=4\), both nonzero. Thus both endpoints are allowed. At \(x=-1\), the radicand is \(9-1=8>0\), but the denominator is \(-1+1=0\), so that input is excluded. At \(x=0\), the numerator is \(\sqrt{9}=3\) and the denominator is \(1\), so \(p(0)=3\), confirming that an input in the domain produces a real value.
A Common Mistake: Excluding a Zero Instead of Requiring a Nonnegative Radicand
Consider \(q(x)=\sqrt{x-5}\). A common incorrect calculation is: “The radicand cannot be zero, so \(x-5\neq 0\), and therefore the domain is all real numbers except \(5\).” This applies a denominator-style restriction to a square root, but a square root is allowed to equal zero.
The correct domain is \([5,\infty)\). Substituting \(x=5\) gives \(\sqrt{5-5}=\sqrt{0}=0\), so \(5\) must be included. Substituting \(x=4\) gives \(\sqrt{4-5}=\sqrt{-1}\), which is not real, so inputs below \(5\) cannot be included.
A Reliable Process
For a formula with even roots and possible denominators, use this sequence:
Write a separate condition requiring each one to be greater than or equal to zero.
Require each denominator to be nonzero. If an even root is in a denominator, combine the conditions to require its radicand to be strictly positive.
For a fraction used as a radicand, also account for where its denominator is zero and where the fraction is positive, zero, or negative.
Keep only inputs that satisfy every condition, then express the resulting set in interval notation.
Substitute boundary values into the formula’s conditions to decide whether each endpoint is included.
Finding a domain is not merely solving one inequality: it is collecting every restriction and applying them together. This habit prevents two especially common errors—excluding valid inputs where a radicand is zero, and including inputs that make a denominator zero.
Check Your Understanding
For each formula, find its real domain. Check any boundary values and excluded points by substituting them into the relevant conditions.
- Find the domain of \(f(x)=\sqrt{12-4x}\).
- Find the domain of \(g(x)=\frac{\sqrt{x-1}}{x-5}\).
- Find the domain of \(h(x)=\sqrt{\frac{x-2}{x+3}}\).
- For \(p(x)=\frac{1}{\sqrt{6-x}}\), explain why the radicand condition is strict rather than non-strict.
- A learner says that \(x=4\) must be excluded from the domain of \(\sqrt{x-4}\) because it makes the radicand zero. Explain the error and give the correct domain.
