How to Complete the Square to Solve a Quadratic Equation

Completing the square is a helpful technique that allows you to rearrange a quadratic equation into a neat form that makes it easier to visualize or even solve. It’s used to determine the vertex of a parabola and to find the roots of a quadratic equation. If you’re just starting out with completing the square, or if the math isn’t exactly adding up, follow along with these easy steps to become a quadratic whiz.

Formula for Completing the Square

To complete the square with $x^{2}+ax+C=y$ , subtract C from both sides, then add $\left({\frac {1}{2}}a\right)^{2}$ to both sides. Now the equation can be factored as $\left(x+{\frac {1}{2}}a\right)^{2}=y-C+{\frac {1}{2}}a$ , where the left side is a perfect square.

Section 1 of 3:

Completing the Square

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Let's say you're working with the following equation: $x^{2}+10x+5=0.$
To isolate your x terms, move the constant (number without an “x”) to the other side of the equation by subtracting it from both sides. $x^{2}+10x=-5.$ ^{[1]
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Take the number in front of the x (second) term in the equation, divide it by 2, and then square it.^{[2]
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Research source} In this example, the second term is $10x,$ so ${\frac {10}{2}}=5,$ and $5^{2}=25$ .
By adding this term to the left side of the equation, you create a polynomial that is a perfect square. To keep the equation balanced, add the number to the right side of the equation, as well.^{[3]
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Research source} $x^{2}+10x+25=-5+25.$
Now that the left side is a perfect square, you can rewrite it as (x + the halved second term)², or $\left(x+5\right)^{2}$ for this example. You can check your math by multiplying it out and seeing if it gives you the first three terms from the last step.^{[4]
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To solve your equation, undo the square on the left by taking the square root of both sides.^{[5]
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Research source} ${\sqrt {\left(x+5\right)^{2}}}=\pm {\sqrt {20}}$ .
When you take the square root of the right side of the equation, the result is both negative and positive, since negative numbers are also positive when they’re squared. To solve this equation, break it into two parts: $x=-5+{\sqrt {20}}$ and $x=-5-{\sqrt {20}}$ .^{[6]
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Section 2 of 3:

Writing Vertex Form by Completing the Square

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1
Write your equation. For this example, try an equation where the $x^{2}$ $x^{{2}}$ term has a coefficient, like $3x^{2}+4x-15=y$ $3x^{2}+4x-15=y$ .
- Vertex form is a way of writing a quadratic equation $\left(ax^{2}+bx+C\right)$ with the coordinates of the vertex of the parabola formed by that equation.
- Vertex form is written as $f(x)=a\left(x-h\right)^{2}+k$ where $(h,k)$ is the vertex of a parabola.^{[7]
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Isolate the x terms by adding 15 to both sides of the equation. $3x^{2}+4x-15+15=y+15$
To complete the square, the leading coefficient has to be 1, so factor 3 out of the left side of the equation.^{[8]
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Research source} $3\left(x^{2}+{\frac {4}{3}}x\right)=y+15$ .
4
Halve the second term and square it. The second term, also known as the b term, in the equation, is ${\frac {4}{3}}$ ${\frac {4}{3}}$ . First, divide it by two: ${\frac {4}{3}}*{\frac {1}{2}}={\frac {4}{6}}={\frac {2}{3}}$ ${\frac {4}{3}}*{\frac {1}{2}}={\frac {4}{6}}={\frac {2}{3}}$ . Then, square the term: ${\frac {2^{2}}{3^{2}}}={\frac {4}{9}}$ ${\frac {2^{2}}{3^{2}}}={\frac {4}{9}}$ .^{[9]
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- Completing the square refers to finding a constant “C” that you can add to the first two terms to make a perfect square trinomial, which can be factored into the expression $\left(x+a\right)^{2}$ . The number you get at this step is the constant that completes the square.
5
Add the term to both sides of the equation. Use the new term to make a perfect square trinomial by adding it into the parentheses.^{[10]
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Research source} Since you factored out a 3, multiply the new term by three before adding it to the right side of the equation with the y term.
- ${\begin{aligned}3\left(x^{2}+{\frac {4}{3}}x+{\frac {4}{9}}\right)&=y+15+3\left({\frac {4}{9}}\right)\\3\left(x^{2}+{\frac {4}{3}}x+{\frac {4}{9}}\right)&=y+15+{\frac {4}{3}}\\3\left(x^{2}+{\frac {4}{3}}x+{\frac {4}{9}}\right)&=y+{\frac {49}{3}}\end{aligned}}$
6
Factor the trinomial as a square and isolate “y.” Now that you have a perfect square in the parenthesis, factor it out using the halved b term from before, in this case ${\frac {2}{3}}$ ${\frac {2}{3}}$ . Write your equation as a factored square, subtract the constant from the y side, and add it to the x side.^{[11]
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- $3\left(x+{\frac {2}{3}}\right)^{2}-{\frac {49}{3}}=y+{\frac {49}{3}}-{\frac {49}{3}}$
- $3\left(x+{\frac {2}{3}}\right)^{2}-{\frac {49}{3}}=y$
- If you’re using the vertex form $f(x)=a(x-h)^{2}+k$ to find the vertex of a parabola (h, k), remember that “h” has to be negative and “k” has to be positive. In this example, the vertex is $\left(-{\frac {2}{3}},-{\frac {49}{3}}\right)$ .^{[12]
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Section 3 of 3:

Solving Quadratic Equations

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Let's say you're working with the following equation: $3x^{2}+4x+5=6$ .
The constant terms are any terms that aren't attached to a variable.^{[13]
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Research source} In this case, you have 5 on the left side and 6 on the right side, so subtract 6 from both sides of the equation. Now you have $3x^{2}+4x-1=0.$
In this case, 3 is the coefficient of the $x^{2}$ term. Divide each term by 3, then put the terms into parenthesis with a 3 in front. So, $3x^{2}\div 3=x^{2}$ , $4x\div 3={\frac {4}{3}}x$ , and $1\div 3={\frac {1}{3}}$ . Now the equation is: $3\left(x^{2}+{\frac {4}{3}}x-{\frac {1}{3}}\right)=0.$ ^{[14]
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Since you divided each term by 3, it can be removed without impacting the equation. Now you have $x^{2}+{\frac {4}{3}}x-{\frac {1}{3}}=0.$ ^{[15]
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Isolate the x terms by moving $-{\frac {1}{3}}$ to the other side of the equal sign. $x^{2}+{\frac {4}{3}}x-{\frac {1}{3}}+{\frac {1}{3}}=0+{\frac {1}{3}}$
6
Halve the second term and square it. Next, take the second term, ${\frac {4}{3}}$ ${\frac {4}{3}}$ , also known as the b term, and halve and square it to complete the square. When you’re done, add it to both sides of the equation to keep it balanced.^{[16]
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Research source} $x^{2}+{\frac {4}{3}}x+{\frac {4}{9}}={\frac {1}{3}}+{\frac {4}{9}}$ $x^{2}+{\frac {4}{3}}x+{\frac {4}{9}}={\frac {1}{3}}+{\frac {4}{9}}$
- ${\frac {4}{3}}*{\frac {1}{2}}={\frac {2}{3}}$
- ${\frac {2^{2}}{3^{2}}}={\frac {4}{9}}$
7
Write the left side of the equation as a perfect square. Since you've already used a formula to find the missing term, all you have to do is put x and half of the second coefficient in parentheses and square them, like so: $\left(x+{\frac {2}{3}}\right)^{2}$ $\left(x+{\frac {2}{3}}\right)^{2}$ . The equation should now read: $\left(x+{\frac {2}{3}}\right)^{2}={\frac {7}{9}}$ $\left(x+{\frac {2}{3}}\right)^{2}={\frac {7}{9}}$ .^{[17]
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- Note that factoring that perfect square will give you the three terms: $x^{2}+{\frac {4}{3}}x+{\frac {4}{9}}$ .
8
Take the square root of both sides. On the left side of the equation, the square root of $\left(x+{\frac {2}{3}}\right)^{2}$ $\left(x+{\frac {2}{3}}\right)^{2}$ is just $x+{\frac {2}{3}}$ $x+{\frac {2}{3}}$ . On the right side, the square root of the denominator, 9, is 3, and the square root of 7 is ${\sqrt {7}}$ ${\sqrt {7}}$ , so the square root is $\pm {\frac {\sqrt {7}}{3}}$ $\pm {\frac {\sqrt {7}}{3}}$ .^{[18]
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- Remember to write $\pm$ because a square root can be positive or negative.
To isolate the variable x, move the constant term ${\frac {2}{3}}$ over to the right side of the equation. You now have two possible answers for x: $\pm {\frac {\sqrt {7}}{3}}-{\frac {2}{3}}$ . You can leave it at that or find the actual square root of 7 if you need an answer without the radical sign.^{[19]
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Why do you halve the b value and then square it? It makes no sense to me.

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It does seem strange and arbitrary, but there is a reason for it. The power move is taking the square root of both sides, but you can't simplify the square root of most polynomials. The step you ask about is a setup move to make the power move work. If I have, for example, x^2 + 4x = 5, and take the square root of both sides, nothing happens, it just makes a mess. But if I add 4 to both sides first and take the square root of both sides of x^2 + 4x + 4 = 9, it simplifies to |x+2| = 3 and the quadratic equation is reduced to a linear equation.

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What's the completing the square formula if x > 1?

Donagan

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The value of x doesn't matter. The process remains as shown above.

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In Part 1 of 2, how did you get 11/9 in Step 8?

Donagan

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Both sides of that equation are being divided by 3 (to get rid of the coefficient of the first term). Dividing the second term (11/3) by 3 gives us 11/9.

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Even after you know the quadratic formula, regularly practice completing the square either by proving the quadratic formula or by doing some practice problems. That way you won't forget how to do it when you need it.

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Be sure to put the $\pm$ in front of square roots, otherwise you will only get one answer.

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