Solve Quadratic Equations by Completing the Square
Completing the square refers to a process where we rewrite a quadratic expression in standard form to a quadratic expression that has an expression that is squared. This is helpful both in graphing as it puts the function in vertex form and in solving as it allows us to use square roots to solve. Any quadratic expression can be rewritten using completing the square, but it can get a little messy depending on the coefficients of the variables. We will first look at the process of completing the square and then how we can use it to solve.
To understand the process, let’s first look at quadratic function in vertex form and rewrite it in standard form. Completing the square will be the reverse of that process.
\(y=2\left(x-3\right)^2+7\)
\(y=2\left(x-3\right)\left(x-3\right)+7\)
\(y=2\left(x^2-6x+9\right)+7\)
\(y=2x^2-12x+18+7\)
\(y=2x^2-12x+25\)
If we were to work backwards in that process with a few adjustments, here’s what it would look like.
\(y=2x^2-12x+25\)
\(y=2\left(x^2-6x+ \right)+25\)
\(y=2\left(x^2-6x+9\right)+25-18\)
\(y=2\left(x-3\right)^2+7\)
The key in completing the square is to identify the number that must be added into the equation so that the expression can be factored as a perfect square. This value will always be half of the \(b\) value squared, \(\left(\frac{b}{2}\right)^2\). We will then need to make sure we account for this added value to keep an equivalent expression. In the example above, this is why we subtracted \(18\) at the end which came from twice the \(9\) that was added in the parentheses.
Completing the square refers to a process where we rewrite a quadratic expression in standard form to a quadratic expression that has an expression that is squared. This is helpful both in graphing as it puts the function in vertex form and in solving as it allows us to use square roots to solve. Any quadratic expression can be rewritten using completing the square, but it can get a little messy depending on the coefficients of the variables. We will first look at the process of completing the square and then how we can use it to solve.
To understand the process, let’s first look at quadratic function in vertex form and rewrite it in standard form. Completing the square will be the reverse of that process.
\(y=2\left(x-3\right)^2+7\)
\(y=2\left(x-3\right)\left(x-3\right)+7\)
\(y=2\left(x^2-6x+9\right)+7\)
\(y=2x^2-12x+18+7\)
\(y=2x^2-12x+25\)
If we were to work backwards in that process with a few adjustments, here’s what it would look like.
\(y=2x^2-12x+25\)
\(y=2\left(x^2-6x+ \right)+25\)
\(y=2\left(x^2-6x+9\right)+25-18\)
\(y=2\left(x-3\right)^2+7\)
The key in completing the square is to identify the number that must be added into the equation so that the expression can be factored as a perfect square. This value will always be half of the \(b\) value squared, \(\left(\frac{b}{2}\right)^2\). We will then need to make sure we account for this added value to keep an equivalent expression. In the example above, this is why we subtracted \(18\) at the end which came from twice the \(9\) that was added in the parentheses.
Steps to Completing the Square on a Quadratic Expression:
1) Make sure the quadratic expression is in standard form, \(ax^2+bx+c\).
2) If necessary, factor out a leading coefficient from the first two terms.
3) Add \(\left(\frac{b}{2}\right)^2\) in the parentheses to create an expression that can be factored as a perfect square.
4) Multiply \(\left(\frac{b}{2}\right)^2\) by the leading coefficient and subtract this value at the end to keep the expression equivalent to the original.
5) Factor the perfect square trinomial and simplify.
1) Make sure the quadratic expression is in standard form, \(ax^2+bx+c\).
2) If necessary, factor out a leading coefficient from the first two terms.
3) Add \(\left(\frac{b}{2}\right)^2\) in the parentheses to create an expression that can be factored as a perfect square.
4) Multiply \(\left(\frac{b}{2}\right)^2\) by the leading coefficient and subtract this value at the end to keep the expression equivalent to the original.
5) Factor the perfect square trinomial and simplify.
Here are the steps numbered with a specific example.
1) \(3x^2+24x+29\)
2) \(3\left(x^2+8x+ \right)+29\)
3) and 4) \(3\left(x^2+8x+16\right)+29-48\)
5) \(3\left(x+4\right)^2-19\)
Once you have completed the square, you can solve the equation using square roots.
Solve the following equations.
Example 1: Solve \(x^2-10x+7=0\).
\(\begin{align}\left(x^2-10x+ \right)+7&=0\ & \ &\text{1) Separate 1st two terms from constant...in this example, there is no leading}\\& \ & \ &\text{coefficient to factor out.}\\\left(x^2-10x+25\right)+7-25&=0\ & \ &\text{2) Take half of -10 and square it...add this value in the parentheses and subtract}\\& \ & \ &\text{it at the end.}\\\left(x-5\right)^2-18&=0\ & \ &\text{3) Factor.}\\\left(x-5\right)^2&=18\ & \ &\text{4)Add 18 to both sides.}\\x-5&=\pm\sqrt{18}\ & \ &\text{5) Take the square root of both sides of the equation. (Make sure to include}\ \pm)\\x&=5\pm3\sqrt{2}\ & \ &\text{6) Simplify the square root and add 5 to both sides.}\end{align}\)
Example 2: Solve \(2x^2+32=-8x\).
\(\begin{align}2x^2+8x+32&=0\ & \ &\text{1) Add 8x to both sides.}\\2\left(x^2+4x+ \right)+32&=0\ & \ &\text{2)Factor a 2 from the first two terms and separate the constant.}\\2\left(x^2+4x+4\right)+32-8&=0\ & \ &\text{3) Take half of 4 and square it...add this value in the parentheses.}\\
&\ & \ &\text{and multiply it by the leading coefficient and subtract at the end.}\\2\left(x+2\right)^2+24&=0\ & \ &\text{4) Factor.}\\2\left(x+2\right)^2&=-24\ & \ &\text{5) Subtract 24 from both sides.}\\\left(x+2\right)^2&=-12\ & \ &\text{6)
Divide both sides by 2.}\\x+2&=\pm\sqrt{-12}\ & \ &\text{7)
Take the square root of both sides. (Make sure to include}\ \pm)\\x&=-2\pm2i\sqrt{3}\ & \ &\text{8) Simplify the square root and subtract 2 from both sides.}\end{align}\)
&\ & \ &\text{and multiply it by the leading coefficient and subtract at the end.}\\2\left(x+2\right)^2+24&=0\ & \ &\text{4) Factor.}\\2\left(x+2\right)^2&=-24\ & \ &\text{5) Subtract 24 from both sides.}\\\left(x+2\right)^2&=-12\ & \ &\text{6)
Divide both sides by 2.}\\x+2&=\pm\sqrt{-12}\ & \ &\text{7)
Take the square root of both sides. (Make sure to include}\ \pm)\\x&=-2\pm2i\sqrt{3}\ & \ &\text{8) Simplify the square root and subtract 2 from both sides.}\end{align}\)
Example 3: Solve \(-3x^2-18x-11=0\).