Completing The Square

Completing the square is a technique for manipulating a quadratic into a perfect square plus a constant. The most common use of completing the square is solving quadratic equations.

For a quadratic polynomial $f(x) = ax^2 + bx +c$, completing the square means finding an expression of the form

$$f(x) = A(x-B)^2 + C.$$

Let's consider the case where $a = 1$. For example, for $x^2 + 2x + 4$, we have the following:

$$x^2 + 2x +4 = (x^2 + 2x + 1) + 3 = (x+1)^2 + 3.$$

We find the necessary manipulations to complete the square on the basis of the perfect square identity:

$$(x+a)^2 = x^2 + 2ax + a^2.$$

This means that for a quadratic like $x^2 + bx + c$, we can make a perfect square by taking half of $b$ and squaring it:

$$\begin{aligned} x^2 + bx + c &= x^2 + bx + \left(\frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c \\ &= \left( x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 +c. \end{aligned}$$

Complete the square for the quadratic $x^2 + 8x + 10$.

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Since our middle term is $8x$, we know that we will want a perfect square with the form $(x+4)^2$, which expands to $x^2 + 8x + 16$.

Thus, we can do the following:

$$\begin{aligned} x^2 + 8x + 10 &= x^2 + 8x + 16 - 16 + 10\\ &= (x^2 + 8x + 16) - 6 \\ &= ( x+ 4 )^2 -6.\ _\square \end{aligned}$$

Which of the following is equivalent to $x^2 + 4x + 7$?

A. $\ (x+2)^2 + 3$
B. $\ (x-2)^2 + 3$
C. $\ (x+3)^2 + 3$
D. $\ (x-3)^2 + 3$

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Completing the square, we see that $x^2 + 4x + 7 = (x^2 + 4x + 4) + 3 = (x+2)^2 + 3$. $_\square$

Sometimes the leading coefficient is not $1,$ which means we must use a slightly more sophisticated approach. Note that we can proceed as follows:

$$\begin{array}{ll} ax^2 + bx + c & \text{} \\ =a\left(x^2 + \frac{b}{a}x\right) + c & (\text{factor out leading coefficient}) \\ =a\left[x^2 + \frac{b}{a}x + \left( \frac{b}{2a}\right)^2 \right] - a\left( \frac{b}{2a} \right)^2 + c & (\text{add the needed term to complete the square}) \\ =a\left(x + \frac{b}{2a} \right)^2 - a\left( \frac{b}{2a} \right)^2 + c & (\text{replace the perfect square term}) \\ =a\left(x + \frac{b}{2a} \right)^2 + \frac{4ac-b^2}{4a}. & (\text{simplify the constant terms}) \end{array}$$

Note: We would strongly recommend understanding the motivation for each step so you can reproduce this, rather than merely memorizing the formula.

Complete the square for the quadratic $2x^2 + 20x + 100$.

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The important term we need is $\left(\frac{b}{2a}\right)^2 = \left(\frac{20}{2\cdot 2}\right)^2 = 25.$

Thus, we have

$$\begin{aligned} 2x^2 + 20x + 100 &= 2(x^2 + 10x) + 100 \\ &= 2(x^2 + 10x + 25) + 100 - 2(25) \\ &= 2(x+5)^2 + 50.\ _\square \end{aligned}$$

Complete the square for $-x^2 + 4x + 10$

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We have

$$\begin{aligned} -x^2 + 4x + 10 &= -(x^2 -4x) + 10 \\ &= -(x^2 -4x +4) -(-4) + 10 \\ &= -(x-2)^2 + 14.\ _\square \end{aligned}$$

Note: When the leading coefficient is negative, be very careful with signs, expecially when adding the $\left( \frac{b}{2a}\right)^2$ term.

For a more extensive list, see Applications of Completing the Square

Finding maximum or minimum values:

What is the maximum value of $-x^2 + 4x + 10$?

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From the previous example, $-x^2 + 4x + 10 = - (x-2)^2 + 14$. Since squares are non-negative,

$$- (x-2)^2 + 14 \leq 14.$$

Thus, the maximum value of the quadratic is $14$, which is achieved at $x=2$. $_\square$

Comparing graphs:

In what ways is the graph of $f(x)= x^2 - 2x + 4$ different from the graph of $g(x) = x^2?$

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Completing the square, we see that

$$\begin{aligned} f(x) &= x^2 - 2x + 4 \\ &= x^2 -2x + 1 -1 + 4 \\ &= (x^2 -2x + 1) + 3 \\ &= (x -1) ^2 + 3 \end{aligned}$$

Thus, the graph will be shifted up 3 units and right 1 unit, as shown below:

Note: You may want to refer to Graph Transformation.

Solving equations:

Solve the equation $x^2 + x - 10 = 0$.

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This equation appears difficult at first, but if we complete the square, the solution becomes apparent. Note that the needed term is $\frac{b^2}{4} = \frac{1}{4}$:

$$\begin{aligned} x^2 + x - 10 &= 0 \\ x^2 + x + \frac{1}{4} -10 &= \frac{1}{4} \\ x^2 + x + \frac{1}{4} &= 10 + \frac{1}{4} \\ \left(x+\frac{1}{2}\right)^2 &= \frac{41}{4}. \end{aligned}$$

Now we can simply take the root of both sides (remembering to find both the positive and negative roots) and simplify:

$$\begin{aligned} \left(x+\frac{1}{2}\right)^2 &= \frac{41}{4} \\ x + \frac{1}{2} &= \pm \sqrt{\frac{41}{4}} \\ x &= -\frac{1}{2} \pm \frac{\sqrt{41}}{2} \\ x &= \frac{-1 \pm \sqrt{41}}{2}.\ _\square \end{aligned}$$

We can use the perfect square identity to simplify polynomials even if they are of higher-degree than quadratics.

Solve $x^4 + 4x^2 -12 = 0$.

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By letting $u = x^2$ we can see that this is a quadratic in $u$ and thus we can complete the square:

$$\begin{aligned} u^2 + 4u - 12 &= 0 \\ u^2 + 4u + 4 - 16 &= 0 \\ (u+2)^2 - 16 &= 0. \end{aligned}$$

Treating the above as a difference of two squares, we have

$$\begin{aligned} (u+2)^2 - 16 &= 0 \\ (u+2 - 4)(u+2+4) &= 0 \\ (u-2)(u+6) &= 0. \end{aligned}$$

Remembering that $u = x^2$, we have either $x^2 = 2$ or $x^2 = -6$. Thus the possible values of $x$ are $\sqrt{2}, -\sqrt{2}, i\sqrt{6}, -i\sqrt{6}$. $_\square$

For what integer value $n$ is $n^2 + 6n + 10$ a perfect square?

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Completing the square, we see that $n^2 + 6n + 10 = (n+3)^2 +1$.

If $n^2 + 6n + 10 = m^2$ for some integer $m$, then $1 = m^2 - (n+3)^2$.

The only perfect squares that differ by $1$ are $0$ and $1$. Hence,

$$(n+3)^2 = 0,$$

which has the solution $n = -3$. $_\square$

If $x^2 + y^2 = 9$, what is the maximum value of $x^2 + 4y$?

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Since $x^2 = 9 - y^2 \geq 0$, we can see that the range of $y$ is given by

$$y^2 \leq 9 \implies -3 \leq y \leq 3.$$

Now, substituting $x^2 = 9 - y^2$ into $x^2 + 4y$ and completing the square, we have

$$\begin{aligned} x^2 + 4y &= (9 - y^2) + 4y \\ &= -(y^2 - 4y + 4) + 13 \\ &= -(y-2)^2 + 13. \end{aligned}$$

Thus the maximum value of $x^2 + 4y$ occurs when $y = 2$ and is equal to 13. $_\square$

• Quadratic Formula
• Applications of Completing the Square
• Completing the Square - Multiple Variables
• Parabolas