Completing the square for quadratic polynomials

Process for completing the square

The zeroes of a single-variable polynomial are the values of that variable at which the polynomial is equal to $0$.

Completing the square is a method we can use to find the zeroes of a quadratic polynomial.

Another way to say this is that completing the square is a method we can use to solve the corresponding quadratic equation (the equation that has the quadratic polynomial on one side and $0$ on the other side).

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The solutions of any polynomial equation are called the roots of that equation. So the zeroes of a quadratic polynomial are numerically equal to the roots of the corresponding quadratic equation.

Completing the square is a useful method when it’s not possible to solve for the roots by factoring, because completing the square creates a trinomial that we can factor as the square of a binomial.

The formal way to write a quadratic polynomial is $ax^2+bx+c$, where $a$ is the coefficient of the $x^2$ term, $b$ is the coefficient of the $x$ term, and $c$ is the constant term.

These are the steps we’ll follow every time we want to complete the square in order to find the roots of a quadratic equation $ax^2+bx+c=0$.

Before we go through the steps, however, we’ll first divide both sides of the equation by $a$ (if $a\ne1$), because it will be easier to solve the equation if the coefficient of the $x^2$ term is $1$. If we have to do that division, we won’t define $b$ and $c$ until after we do it. That is, $b$ will be the coefficient of the new $x$ term, and $c$ will be the new constant term. So we’ll actually be starting with an equation of the form $x^2+bx+c=0$.

1. Move the constant term to the right side of the equation by subtracting $c$ from both sides.

$x^2+bx+c-c=0-c$

$x^2+bx=-c$

2. Find $(b/2)^2$. Take the coefficient of the $x$ term, divide it by $2$, and then square the result.

$x^2+bx+\left(\frac{b}{2}\right)^2=-c+\left(\frac{b}{2}\right)^2$

3. Factor the left side, which is now

$x^2+bx+\left(\frac{b}{2}\right)^2$

To factor this quadratic polynomial, we have to find a pair of factors of $(b/2)^2$ whose sum is $b$. The only pair of factors with this property is $b/2$ and $b/2$. So the left side of the equation becomes

$\left(x+\frac{b}{2}\right)\left(x+\frac{b}{2}\right)$

Notice that the trinomial

$x^2+bx+\left(\frac{b}{2}\right)^2$

factors as the square of the binomial $x+(b/2)$:

$x^2+bx+\left(\frac{b}{2}\right)^2=\left(x+\frac{b}{2}\right)\left(x+\frac{b}{2}\right)=\left(x+\frac{b}{2}\right)^2$

So the equation we have to solve is

$\left(x+\frac{b}{2}\right)^2=-c+\left(\frac{b}{2}\right)^2$

4. Square root both sides of the equation. Remember that the right side will now include a $\pm$ sign.

$\sqrt{\left(x+\frac{b}{2}\right)^2}=\sqrt{-c+\left(\frac{b}{2}\right)^2}$

$x+\frac{b}{2}=\pm\sqrt{-c+\left(\frac{b}{2}\right)^2}$

5. Solve for $x$ to get the roots of the original quadratic equation, by subtracting $b/2$ from both sides.

$x=-\frac{b}{2}\pm\sqrt{-c+\left(\frac{b}{2}\right)^2}$

How to complete the square on a quadratic polynomial

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Find the roots of the quadratic by completing the square

Example

Solve for $x$ by completing the square.

$x^2+6x+4=0$

There is no pair of factors of $4$ whose sum is $6$, so we’ll need to solve by completing the square. Start by moving the constant constant term, $4$, to the right side of the equation by subtracting $4$ from both sides.

$x^2+6x+4-4=0-4$

$x^2+6x=-4$

Find $(b/2)^2$. In this case, $b=6$.

$\left(\frac{b}{2}\right)^2=\left(\frac{6}{2}\right)^2=3^2=9$

Add $9$ to both sides of the equation.

$x^2+6x+9=-4+9$

$x^2+6x+9=5$

Factor $x^2+6x+9$, and notice that it factors as the square of a binomial, $(x+3)^2$, so our equation becomes

$(x+3)^2=5$

Take the square root of each side of the equation.

$\sqrt{(x+3)^2}=\sqrt{5}$

Since $x+3$ could be either positive or negative (because $(x+3)^2$ is equal to $5$, which is positive), we get

$x+3=\pm\sqrt{5}$

Solve for $x$ by subtracting $3$ from both sides. To avoid confusion, put the $-3$ in front of the $\pm\sqrt5$.

$x+3-3=-3\pm\sqrt{5}$

$x=-3\pm\sqrt{5}$

The roots of the original quadratic equation are

$x=-3+\sqrt{5}$ and $x=-3-\sqrt{5}$

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Let’s try another example of completing the square.

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Completing the square is a useful method when it’s not possible to solve for the roots by factoring, because completing the square creates a trinomial that we can factor as the square of a binomial.

Example

Solve for the variable by completing the square.

$v^3-5v^2+4v=0$

In this case we could solve by factoring since we can first factor out a $v$ and then factor the quadratic polynomial that remains.

$v(v^2-5v+4)=0$

$(-1)(-4)=4$ and $-1+(-4)=-5$, so $v^2-5v+4$ can be factored as $(v-1)(v-4)$. Therefore, our original equation can be rewritten as

$v(v-1)(v-4)=0$

One solution is $v=0$, and we’ll set each of the other two factors ($v-1$ and $v-4$) equal to $0$ and solve each of the resulting equations for $v$.

$v-1=0$

$v-1+1=0+1$

$v=1$

and

$v-4=0$

$v-4+4=0+4$

$v=4$

The solutions are therefore

$v=0$, $v=1$, and $v=4$

However, we were not asked to solve by factoring, so let’s look at how this problem can be solved by completing the square. Start by factoring out aso that we’ll have a quadratic polynomial inside parentheses.

$v(v^2-5v+4)=0$

Again, one solution is $v=0$, so now we’ll find the other solutions, that is, the solutions of the equation

$v^2-5v+4=0$

Now that we have a quadratic polynomial on the left side, we can start by moving the constant term, $4$, to the right side of the equation by subtracting $4$ from both sides.

$v^2-5v+4-4=0-4$

$v^2-5v=-4$

Find $(b/2)^2$. In this case, $b=-5$.

$\left(\frac{b}{2}\right)^2=\left(\frac{-5}{2}\right)^2=\frac{25}{4}$

Add $25/4$ to both sides of the equation.

$v^2-5v+\frac{25}{4}=-4+\frac{25}{4}$

$v^2-5v+\frac{25}{4}=\frac{9}{4}$

We see that the trinomial on the left side,

$v^2-5v+\frac{25}{4}$

factors as the square of a binomial, $[v-(5/2)]^2$, so our equation becomes

$\left(v-\frac{5}{2}\right)^2=\frac{9}{4}$

Take the square root of each side of the equation.

$\sqrt{\left(v-\frac{5}{2}\right)^2}=\sqrt{\frac{9}{4}}$

So we get

$v-\frac{5}{2}=\pm\frac{3}{2}$

Solve for $v$ by adding $5/2$ to both sides. To avoid confusion, put the $5/2$ in front of the $\pm(3/2)$.

$v-\frac{5}{2}+\frac{5}{2}=\frac{5}{2}\pm\frac{3}{2}$

$v=\frac{5}{2}\pm\frac{3}{2}$

If we solve this both ways (adding and subtracting), we get

$v=\frac{5}{2}+\frac{3}{2}=\frac{8}{2}=4$

$v=\frac{5}{2}-\frac{3}{2}=\frac{2}{2}=1$

So the solutions are

$v=0$, $v=1$, and $v=4$

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