Completing the Square and the Quadratic Formula
Learning Objectives
- Complete the square to solve a quadratic equation
- Use the quadratic formula to solve a quadratic equation
- Use the discriminant to determine the number and type of solutions to a quadratic equation
Not all quadratic equations can be factored or can be solved in their original form using the square root property. In these cases, we may use a method for solving a quadratic equation known as completing the square. Using this method, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property. To complete the square, the leading coefficient, a, must equal 1. If it does not, then divide the entire equation by a. Then, we can use the following procedures to solve a quadratic equation by completing the square.
We will use the example [latex]{x}^{2}+4x+1=0[/latex] to illustrate each step.
- Given a quadratic equation that cannot be factored, and with [latex]a=1[/latex], first add or subtract the constant term to the right sign of the equal sign.
[latex]{x}^{2}+4x=-1[/latex]
- Multiply the b term by [latex]\frac{1}{2}[/latex] and square it.
[latex]\begin{array}{l}\frac{1}{2}\left(4\right)=2\hfill \\ {2}^{2}=4\hfill \end{array}[/latex]
- Add [latex]{\left(\frac{1}{2}b\right)}^{2}[/latex] to both sides of the equal sign and simplify the right side. We have
[latex]\begin{array}{l}{x}^{2}+4x+4=-1+4\hfill \\ {x}^{2}+4x+4=3\hfill \end{array}[/latex]
- The left side of the equation can now be factored as a perfect square.
[latex]\begin{array}{l}{x}^{2}+4x+4=3\hfill \\ {\left(x+2\right)}^{2}=3\hfill \end{array}[/latex]
- Use the square root property and solve.
[latex]\begin{array}{l}\sqrt{{\left(x+2\right)}^{2}}=\pm \sqrt{3}\hfill \\ x+2=\pm \sqrt{3}\hfill \\ x=-2\pm \sqrt{3}\hfill \end{array}[/latex]
- The solutions are [latex]x=-2+\sqrt{3}[/latex], [latex]x=-2-\sqrt{3}[/latex].
Example: Solving a Quadratic by Completing the Square
Solve the quadratic equation by completing the square: [latex]{x}^{2}-3x - 5=0[/latex].Answer: First, move the constant term to the right side of the equal sign.
[latex]{x}^{2}-3x=5[/latex]
Then, take [latex]\frac{1}{2}[/latex] of the b term and square it.
[latex]\begin{array}{l}\frac{1}{2}\left(-3\right)=-\frac{3}{2}\hfill \\ {\left(-\frac{3}{2}\right)}^{2}=\frac{9}{4}\hfill \end{array}[/latex]
Add the result to both sides of the equal sign.
[latex]\begin{array}{l}\text{ }{x}^{2}-3x+{\left(-\frac{3}{2}\right)}^{2}=5+{\left(-\frac{3}{2}\right)}^{2}\hfill \\ {x}^{2}-3x+\frac{9}{4}=5+\frac{9}{4}\hfill \end{array}[/latex]
Factor the left side as a perfect square and simplify the right side.
[latex]{\left(x-\frac{3}{2}\right)}^{2}=\frac{29}{4}[/latex]
Use the square root property and solve.
[latex]\begin{array}{l}\sqrt{{\left(x-\frac{3}{2}\right)}^{2}}\hfill&=\pm \sqrt{\frac{29}{4}}\hfill \\ \left(x-\frac{3}{2}\right)\hfill&=\pm \frac{\sqrt{29}}{2}\hfill \\ x\hfill&=\frac{3}{2}\pm \frac{\sqrt{29}}{2}\hfill \end{array}[/latex]
The solutions are [latex]x=\frac{3}{2}+\frac{\sqrt{29}}{2}[/latex], [latex]x=\frac{3}{2}-\frac{\sqrt{29}}{2}[/latex].
Try It
Solve by completing the square: [latex]{x}^{2}-6x=13[/latex].Answer: [latex-display]x=3\pm \sqrt{22}[/latex-display]
Using the Quadratic Formula
The fourth method of solving a quadratic equation is by using the quadratic formula, a formula that will solve all quadratic equations. Although the quadratic formula works on any quadratic equation in standard form, it is easy to make errors in substituting the values into the formula. Pay close attention when substituting, and use parentheses when inserting a negative number. We can derive the quadratic formula by completing the square. We will assume that the leading coefficient is positive; if it is negative, we can multiply the equation by [latex]-1[/latex] and obtain a positive a. Given [latex]a{x}^{2}+bx+c=0[/latex], [latex]a\ne 0[/latex], we will complete the square as follows:- First, move the constant term to the right side of the equal sign:
[latex]a{x}^{2}+bx=-c[/latex]
- As we want the leading coefficient to equal 1, divide through by a:
[latex]{x}^{2}+\frac{b}{a}x=-\frac{c}{a}[/latex]
- Then, find [latex]\frac{1}{2}[/latex] of the middle term, and add [latex]{\left(\frac{1}{2}\frac{b}{a}\right)}^{2}=\frac{{b}^{2}}{4{a}^{2}}[/latex] to both sides of the equal sign:
[latex]{x}^{2}+\frac{b}{a}x+\frac{{b}^{2}}{4{a}^{2}}=\frac{{b}^{2}}{4{a}^{2}}-\frac{c}{a}[/latex]
- Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction:
[latex]{\left(x+\frac{b}{2a}\right)}^{2}=\frac{{b}^{2}-4ac}{4{a}^{2}}[/latex]
- Now, use the square root property, which gives
[latex]\begin{array}{l}x+\frac{b}{2a}=\pm \sqrt{\frac{{b}^{2}-4ac}{4{a}^{2}}}\hfill \\ x+\frac{b}{2a}=\frac{\pm \sqrt{{b}^{2}-4ac}}{2a}\hfill \end{array}[/latex]
- Finally, add [latex]-\frac{b}{2a}[/latex] to both sides of the equation and combine the terms on the right side. Thus,
[latex]x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}[/latex]
A General Note: The Quadratic Formula
Written in standard form, [latex]a{x}^{2}+bx+c=0[/latex], any quadratic equation can be solved using the quadratic formula:[latex]x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}[/latex]
where a, b, and c are real numbers and [latex]a\ne 0[/latex].
How To: Given a quadratic equation, solve it using the quadratic formula
- Make sure the equation is in standard form: [latex]a{x}^{2}+bx+c=0[/latex].
- Make note of the values of the coefficients and constant term, [latex]a,b[/latex], and [latex]c[/latex].
- Carefully substitute the values noted in step 2 into the equation. To avoid needless errors, use parentheses around each number input into the formula.
- Calculate and solve.
Example : Solve A Quadratic Equation Using the Quadratic Formula
Solve the quadratic equation: [latex]{x}^{2}+5x+1=0[/latex].Answer: Identify the coefficients: [latex]a=1,b=5,c=1[/latex]. Then use the quadratic formula.
[latex]\begin{array}{l}x\hfill&=\frac{-\left(5\right)\pm \sqrt{{\left(5\right)}^{2}-4\left(1\right)\left(1\right)}}{2\left(1\right)}\hfill \\ \hfill&=\frac{-5\pm \sqrt{25 - 4}}{2}\hfill \\ \hfill&=\frac{-5\pm \sqrt{21}}{2}\hfill \end{array}[/latex]
Try It
Solve the quadratic equation using the quadratic formula: [latex]9{x}^{2}+3x - 2=0[/latex].Answer: [latex]x=-\frac{2}{3}[/latex], [latex]x=\frac{1}{3}[/latex]
The Discriminant
The quadratic formula not only generates the solutions to a quadratic equation, it tells us about the nature of the solutions when we consider the discriminant, or the expression under the radical, [latex]{b}^{2}-4ac[/latex]. The discriminant tells us whether the solutions are real numbers or complex numbers (which we will learn about in more depth later in the course), and how many solutions of each type to expect. The table below relates the value of the discriminant to the solutions of a quadratic equation.| Value of Discriminant | Results |
|---|---|
| [latex]{b}^{2}-4ac=0[/latex] | One rational solution (double solution) |
| [latex]{b}^{2}-4ac>0[/latex], perfect square | Two rational solutions |
| [latex]{b}^{2}-4ac>0[/latex], not a perfect square | Two irrational solutions |
| [latex]{b}^{2}-4ac<0[/latex] | Two complex solutions |
A General Note: The Discriminant
For [latex]a{x}^{2}+bx+c=0[/latex], where [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex] are real numbers, the discriminant is the expression under the radical in the quadratic formula: [latex]{b}^{2}-4ac[/latex]. It tells us whether the solutions are real numbers or complex numbers and how many solutions of each type to expect.Example: Using the Discriminant to Find the Nature of the Solutions to a Quadratic Equation
Use the discriminant to find the nature of the solutions to the following quadratic equations:- [latex]{x}^{2}+4x+4=0[/latex]
- [latex]8{x}^{2}+14x+3=0[/latex]
- [latex]3{x}^{2}-5x - 2=0[/latex]
- [latex]3{x}^{2}-10x+15=0[/latex]
Answer: Calculate the discriminant [latex]{b}^{2}-4ac[/latex] for each equation and state the expected type of solutions.
- [latex]{x}^{2}+4x+4=0[/latex][latex]{b}^{2}-4ac={\left(4\right)}^{2}-4\left(1\right)\left(4\right)=0[/latex]. There will be one rational double solution.
- [latex]8{x}^{2}+14x+3=0[/latex][latex]{b}^{2}-4ac={\left(14\right)}^{2}-4\left(8\right)\left(3\right)=100[/latex]. As [latex]100[/latex] is a perfect square, there will be two rational solutions.
- [latex]3{x}^{2}-5x - 2=0[/latex][latex]{b}^{2}-4ac={\left(-5\right)}^{2}-4\left(3\right)\left(-2\right)=49[/latex]. As [latex]49[/latex] is a perfect square, there will be two rational solutions.
- [latex]3{x}^{2}-10x+15=0[/latex][latex]{b}^{2}-4ac={\left(-10\right)}^{2}-4\left(3\right)\left(15\right)=-80[/latex]. There will be two complex solutions.
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