Smallest Value of a Quadratic Without Calculus: Complete the Square

2026-10-26

What is the smallest value of \( x^2-6x+13 \)? You don't need calculus, a graphing calculator or lucky guessing. One idea does it: complete the square. The same idea also handles \( x+\tfrac9x \).

The problem

Step 0: try some values

The values fall, then rise, and they are symmetric around \( x=3 \). So the lowest point is in the middle. The table suggests 4, but a table never proves a minimum. The next steps do.

Step 1: make a square from the first two terms

Take the number in front of \( x \), which is \( -6 \). Half of it is \( -3 \). Now expand:

This matches the first two terms of our expression exactly.

Step 2: fix the leftover

Our expression ends in 13, not 9, and \( 13=9+4 \). So

Step 3: a square is never negative

For every real number, \( (x-3)^2\ge 0 \). So

with equality exactly when \( (x-3)^2=0 \), that is, when \( x=3 \).

The smallest value is 4, reached at x = 3.

Check it another way

For \( ax^2+bx+c \) with \( a>0 \), the lowest point is at \( x=-\tfrac{b}{2a} \). Here that is \( x=\tfrac{6}{2}=3 \), and the value there is \( 9-18+13=4 \). Same answer.

Want the recipe in one line? The minimum value is \( c-\tfrac{b^2}{4a} \). Here: \( 13-\tfrac{36}{4}=4 \).

The same trick on a different shape

Try \( x=1 \): 10. Try \( x=3 \): 6. Try \( x=9 \): 10. The lowest looks like it is at 3. To prove it, square the difference \( \sqrt x-\tfrac{3}{\sqrt x} \):

The square is zero when \( \sqrt x=\tfrac{3}{\sqrt x} \), that is, \( x=3 \).

The smallest value of x + 9/x is 6, reached at x = 3.

Notice the pattern: we wrote the expression as "a square plus a constant". The constant is the minimum. (If you know the AM-GM inequality, this is exactly why \( x+\tfrac9x\ge 2\sqrt9=6 \).)

Common mistakes

• Thinking the minimum is at \( x=0 \) because zero feels like the simplest number. At \( x=0 \) the value is 13.
• Forgetting the leftover number and answering 9 or 0.
• Using the wrong sign: half of \( -6 \) is \( -3 \), so the bracket is \( (x-3)^2 \), not \( (x+3)^2 \).
• If \( x^2 \) has a minus sign in front, the same trick finds the largest value instead.

What to remember

1. Take half of the number in front of \( x \), and square it.
2. Fix the leftover number.
3. A square is never negative, so the leftover number is the minimum.

Your turn