Why Classify a Radicand?
A quadratic inside a radical can look unfamiliar even when it belongs to a standard form. Completing the square reveals the shape of the expression: a positive square plus a constant, a square minus a constant, or a constant minus a square. These patterns have different real domains, so identifying the right one is an important first step before choosing a method for an integral.
This tutorial focuses on rewriting and classification. The goal is not to integrate the resulting expressions, but to determine precisely which form is present and where the radicand is nonnegative.
Completing the Square and Normalizing
Let \(q(x)=ax^2+bx+c\), where \(a\ne 0\). Completing the square gives
Define the center \(h=-\frac{b}{2a}\) and the completed-square constant \(C=c-\frac{b^2}{4a}\). Then \(q(x)=a(x-h)^2+C\). To make the squared term have coefficient \(1\) or \(-1\), define \(u=\sqrt{|a|}(x-h)\). Thus \(a(x-h)^2\) is \(u^2\) when \(a>0\), and \(-u^2\) when \(a<0\).
The change from \(x\) to \(u\) is a rewriting of the expression, not a change to its sign or real domain: \(u\) is real for every real \(x\), and the relation between them can be reversed. The sign of \(C\), together with whether the square is added or subtracted, determines the classification.
The Three Main Sign Patterns
When the normalized expression is \(u^2+C\) with \(C>0\), it is a positive-sum form. Writing \(d=\sqrt C\), it is \(u^2+d^2\), which is positive for every real \(u\). If \(C<0\), write \(d=\sqrt{-C}\); the form becomes \(u^2-d^2\), a difference form. It is nonnegative only when \(u\le -d\) or \(u\ge d\).
When \(a<0\), the normalized expression is \(C-u^2\). If \(C>0\), writing \(d=\sqrt C\) gives \(d^2-u^2\), a bounded form. It is nonnegative exactly for \(-d\le u\le d\). If \(C<0\), both \(C\) and \(-u^2\) are nonpositive, and their sum is strictly negative for every real \(u\); there is no real square-root domain.
The case \(C=0\) is a boundary case rather than one of the three nonzero-constant patterns. The radicand is \(u^2\) if \(a>0\), and \(-u^2\) if \(a<0\). In the first case it is nonnegative for every real \(u\), but vanishes at \(u=0\). In the second case it is nonnegative only at \(u=0\).
A Sign Test for the Constant
The discriminant \(\Delta=b^2-4ac\) gives a useful check on the completed-square constant. From its definition,
Thus the sign of \(C\) is determined by the signs of \(\Delta\) and \(a\). This relation is a check, not a replacement for completing the square: the completed form still identifies the actual center and makes the domain conditions visible.
Worked Examples
Worked Example: A Positive-Sum Radicand
Classify \(q(x)=2x^2-8x+10\), and find the real domain of \(\sqrt{q(x)}\).
First factor the leading coefficient from the quadratic and linear terms:
Complete the square inside the parentheses. Half of \(-4\) is \(-2\), whose square is \(4\):
Here \(a=2>0\), \(h=2\), and \(C=2>0\). Set \(u=\sqrt{2}(x-2)\). Then \(2(x-2)^2=u^2\), so
This is the positive-sum form \(u^2+d^2\), with \(d=\sqrt2\). Since \(u^2+2\ge2>0\) for all real \(u\), the domain of the square root is all real \(x\). To check the rewrite, expand it:
This matches the original quadratic, and the constant positive term confirms that the radicand never vanishes.
Worked Example: A Difference Form and Two Exterior Intervals
Classify \(q(x)=2x^2+4x-6\), and find the real domain of \(\sqrt{q(x)}\).
Factor \(2\) from the quadratic and linear terms, then complete the square:
Here \(a=2>0\), \(h=-1\), and \(C=-8<0\). Define \(u=\sqrt2(x+1)\). The normalized form is
This is a difference form. The square root is real when \(u^2-8\ge0\), which is equivalent to \(|u|\ge2\sqrt2\). Substituting \(u=\sqrt2(x+1)\) gives
Therefore \(x\le-3\) or \(x\ge1\), so the domain is \((-\infty,-3]\cup[1,\infty)\). Check the result by factoring the original quadratic:
The roots are \(-3\) and \(1\). Since the leading coefficient is positive, the quadratic is nonnegative outside the roots, agreeing with the completed-square domain.
Worked Example: A Bounded Form
Classify \(q(x)=-3x^2+12x-9\), and find the real domain of \(\sqrt{q(x)}\).
Factor out \(-3\) from the quadratic and linear terms, then complete the square:
Here \(a=-3<0\), \(h=2\), and \(C=3>0\). Set \(u=\sqrt3(x-2)\), so the radicand becomes
This is a bounded form. The condition for a real square root is \(3-u^2\ge0\), or \(|u|\le\sqrt3\). Substituting for \(u\),
The domain is \([1,3]\). Check by expansion:
The rewrite matches the original. At the endpoints, the radicand is zero: \(q(1)=0\) and \(q(3)=0\). At the center, \(q(2)=3>0\), as expected for a bounded form with its maximum at \(x=2\).
Why the Classification Works
The classification follows from elementary inequalities for squares. For any real \(u\), \(u^2\ge0\). Consequently, adding a positive constant to \(u^2\) keeps the result positive, while subtracting a positive constant makes nonnegativity depend on whether the square is large enough. A positive constant minus a square is nonnegative only while that square stays small enough.
Completing the square expresses the quadratic as \(a(x-h)^2+C\), without changing its value.
With \(u=\sqrt{|a|}(x-h)\), the squared term becomes \(u^2\) or \(-u^2\), according to the sign of \(a\).
Use \(u^2\ge0\) to decide whether the radicand is always positive, nonnegative on exterior intervals, nonnegative on a bounded interval, or nowhere nonnegative.
This argument also explains why endpoints must be handled with care. A square-root domain includes points where the radicand is zero. A reciprocal square root excludes them. For example, if \(q(x)=u^2-d^2\), the square-root domain includes \(u=\pm d\), but \(1/\sqrt{q(x)}\) is undefined at those points.
Common Pitfall: Ignoring the Leading Coefficient
A frequent error is to look only at the constant left after completing the square and label the form from that sign, without checking whether the squared term is added or subtracted. Consider \(-3(x-2)^2+3\). Its positive constant does not make it a positive-sum form. The negative coefficient of the square gives \(3-u^2\), a bounded form, not \(u^2+3\).
Another error is to divide by the coefficient of the square without accounting for how that affects the inequality. A safer procedure is to define \(u=\sqrt{|a|}(x-h)\), write the normalized expression explicitly, and then solve its nonnegativity condition. Finally, verify the completed-square identity by expanding it back to the original quadratic.
Why the Form Matters
Recognizing the radicand’s type helps organize later work with quadratic radicals. A positive-sum form has no real zeros; a difference form has two boundary points and is real outside them; a bounded form is real only between its boundary points. These distinctions affect which intervals are available and whether endpoints are included. They also prevent applying a formula intended for one sign pattern to a different one.
For a quadratic with \(a\ne0\), the essential checklist is: find \(h\) and \(C\), normalize with \(u=\sqrt{|a|}(x-h)\), identify the signs in \(u^2+C\) or \(C-u^2\), and solve the appropriate inequality. If \(C=0\), or if the radicand is negative throughout, state that case separately rather than forcing it into a nonzero-constant pattern.
Check Your Understanding
For each question, distinguish the algebraic form from the domain conclusion.
- Complete the square in \(x^2+6x+13\). Which sign pattern does it have, and where is its square root real?
- Rewrite \(4x^2-16x+12\) using a normalized variable \(u\). Determine the domain of its square root.
- Classify \(-2x^2+8x-8\). Is its square-root domain an interval, two exterior intervals, a single point, or empty?
- For a radicand of the form \(u^2-d^2\), with \(d>0\), explain why the square-root domain includes the two points where the radicand is zero.
- How does the domain change when \(\sqrt{q(x)}\) is replaced by \(1/\sqrt{q(x)}\)?
