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Study Guides > College Algebra

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:
  1. First, move the constant term to the right side of the equal sign:
    [latex]a{x}^{2}+bx=-c[/latex]
  2. As we want the leading coefficient to equal 1, divide through by a:
    [latex]{x}^{2}+\frac{b}{a}x=-\frac{c}{a}[/latex]
  3. 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]
  4. 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]
  5. 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]
  6. 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

  1. Make sure the equation is in standard form: [latex]a{x}^{2}+bx+c=0[/latex].
  2. Make note of the values of the coefficients and constant term, [latex]a,b[/latex], and [latex]c[/latex].
  3. Carefully substitute the values noted in step 2 into the equation. To avoid needless errors, use parentheses around each number input into the formula.
  4. Calculate and solve.

Example 9: Solve the Quadratic Equation Using the Quadratic Formula

Solve the quadratic equation: [latex]{x}^{2}+5x+1=0[/latex].

Solution

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]

Example 10: Solving a Quadratic Equation with the Quadratic Formula

Use the quadratic formula to solve [latex]{x}^{2}+x+2=0[/latex].

Solution

First, we identify the coefficients: [latex]a=1,b=1[/latex], and [latex]c=2[/latex]. Substitute these values into the quadratic formula.
[latex]\begin{array}{l}x\hfill&=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}\hfill \\\hfill&=\frac{-\left(1\right)\pm \sqrt{{\left(1\right)}^{2}-\left(4\right)\cdot \left(1\right)\cdot \left(2\right)}}{2\cdot 1}\hfill \\\hfill&=\frac{-1\pm \sqrt{1 - 8}}{2}\hfill \\ \hfill&=\frac{-1\pm \sqrt{-7}}{2}\hfill \\\hfill&=\frac{-1\pm i\sqrt{7}}{2}\hfill \end{array}[/latex]
The solutions to the equation are [latex]x=\frac{-1+i\sqrt{7}}{2}[/latex] and [latex]x=\frac{-1-i\sqrt{7}}{2}[/latex] or [latex]x=\frac{-1}{2}+\frac{i\sqrt{7}}{2}[/latex] and [latex]x=\frac{-1}{2}-\frac{i\sqrt{7}}{2}[/latex].

Try It 8

Solve the quadratic equation using the quadratic formula: [latex]9{x}^{2}+3x - 2=0[/latex]. Solution  

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, 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 11: 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:
  1. [latex]{x}^{2}+4x+4=0[/latex]
  2. [latex]8{x}^{2}+14x+3=0[/latex]
  3. [latex]3{x}^{2}-5x - 2=0[/latex]
  4. [latex]3{x}^{2}-10x+15=0[/latex]

Solution

Calculate the discriminant [latex]{b}^{2}-4ac[/latex] for each equation and state the expected type of solutions.
  1. [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.
  2. [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.
  3. [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.
  4. [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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