Skip to main content
Mathematics LibreTexts

11.3: Quadratic formula

  • Page ID
    45112
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    The quadratic formula is derived from the method of completing the square. If we took a general quadratic equation

    \[ax^2+bx+c=0\nonumber\]

    and solved for \(x\) by completing the square, we would obtain the quadratic formula. Let’s try this.

    Example 11.3.1

    Solve for \(x\) by completing the square: \(ax^2+bx+c=0\)

    Solution

    First, we should rewrite the equation so that the leading coefficient is one and c is on the other side.

    \[\begin{aligned}ax^2+bx+c&=0 \\ ax^2+bx&=-c \\ a\left(x^2+\dfrac{b}{a}\right)&=-c \\ x^2+\dfrac{b}{a}x&=\dfrac{-c}{a}\end{aligned}\]

    Next, we take \(\dfrac{b}{a}\) and form \(\left(\dfrac{b}{2a}\right)^2\).

    \[\begin{aligned}x^2+\dfrac{b}{a}x+\color{blue}{\left(\dfrac{1}{2}\cdot\dfrac{b}{a}\right)^2}&\color{black}{=}\dfrac{-c}{a}+\color{blue}{\left(\dfrac{1}{2}\cdot\dfrac{b}{a}\right)^2} \\ x^2+\dfrac{b}{a}x+\color{blue}{\dfrac{b^2}{4a^2}}&\color{black}{=}\dfrac{-c}{a}+\color{blue}{\dfrac{b^2}{4a^2}}\end{aligned}\]

    Let’s factor the left side and combine fractions on the right:

    \[\begin{aligned}x^2+\dfrac{b}{a}x+\color{blue}{\dfrac{b^2}{4a^2}}&\color{black}{=}\dfrac{-c}{a}\cdot\color{blue}{\dfrac{4a}{4a}}\color{black}{+}\color{blue}{\dfrac{b^2}{4a^2}} \\ \left(x+\dfrac{b}{2a}\right)^2&=\dfrac{-4ac}{4a^2}+\dfrac{b^2}{4a^2}\end{aligned}\]

    We can apply the square root property and solve as usual:

    \[\begin{array}{rl}\left(x+\dfrac{b}{2a}\right)^2=\dfrac{b^2-4ac}{4a^2}&\text{Apply the square root property} \\ x+\dfrac{b}{2a}=\pm\sqrt{\dfrac{b^2-4ac}{4a^2}}&\text{Simplify} \\ x+\dfrac{b}{2a}=\pm\dfrac{\sqrt{b^2-4ac}}{2a}&\text{Isolate }x\\ x=-\dfrac{b}{2a}\pm\dfrac{\sqrt{b^2-4ac}}{2a}&\text{Same denominator; combine fractions} \\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}&\text{Solution}\end{array}\nonumber\]

    Quadratic Formula

    Let \(a\), \(b\) be coefficients of \(x^2\), \(x\), respectively, and \(c\) be the constant coefficient of the quadratic equation \(ax^2 + bx + c = 0\). Then

    \[x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\nonumber\]

    is called the quadratic formula. The quadratic formula is a formula for solving quadratic equations in terms of the coefficients.

    Note

    Indian mathematician, Brahmagupta, gave the first explicit formula for solving quadratics in 628 AD. However, at that time, mathematics was not written with variables and symbols, so the formula he gave was, “To the absolute number multiplied by four times the square, add the square of the middle term; the square root of the same, less the middle term, being divided by twice the square is the value.” Mathematically, this would translate to

    \[\dfrac{\sqrt{4ac+b^2}-b}{2a}\nonumber\]

    as the solution to the equation \(ax^2+bx=c\).

    Apply the Quadratic Formula

    Example 11.3.2

    Solve: \(x^2+3x+2=0\)

    Solution

    We may note that we can solve this equation by factoring. However, we will use the quadratic formula and later compare.

    \[\begin{array}{rl}x^2+3x+2=0&\text{Apply quadratic formula} \\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}&\text{Plug-n-chug }a=1,\: b=3,\: c=2 \\ x=\dfrac{-\color{blue}{3}\color{black}{\pm}\sqrt{\color{blue}{3}\color{black}{^2}-4\color{blue}{(1)(2)}}}{2\color{blue}{(1)}}&\color{black}{\text{Simplify using order of operations}} \\ x=\dfrac{-3\pm\sqrt{9-8}}{2} \\ x=\dfrac{-3\pm\sqrt{1}}{2} \\ x=\dfrac{-3\pm 1}{2}&\text{Rewrite as two solutions} \\ x=\dfrac{-3+1}{2}\text{ or }\dfrac{-3-1}{2}&\text{Evaluate} \\ x=-1\text{ or }x=-2&\text{Solution}\end{array}\nonumber\]

    Let's compare with factoring the equation:

    \[\begin{aligned}x^2+3x+2&=0 \\ (x+1)(x+2)&=0 \\ x+1=0\text{ or }x+2&=0 \\ x=-1\text{ or }x&=-2\end{aligned}\]

    Notice, factoring would have been much quicker than using the quadratic formula.

    Note

    When we can factor the quadratic equation, we should, and when the equation isn’t factorable, we should use quadratic formula.

    Make Equal to Zero

    Example 11.3.3

    Solve: \(25x^2=30x+11\)

    Solution

    We first rewrite the equation so that zero is on one side of the equation. Then we can solve as usual.

    \[\begin{array}{rl}25x^2=30x+11&\text{Rewrite where zero is one side} \\ 25x^2-30x-11=0&\text{Apply the quadratic formula} \\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}&\text{Plug-n-chug }a=25,\: b=-30,\: c=-11 \\ x=\dfrac{-\color{blue}{(-30)}\color{black}{\pm}\sqrt{\color{blue}{(-30)}\color{black}{^2}-4\color{blue}{(25)(-11)}}}{2\color{blue}{(25)}}&\text{Simplify using order of operations} \\ x=\dfrac{30\pm\sqrt{900+1100}}{50} \\ x=\dfrac{30\pm\sqrt{2000}}{50}&\text{Rewrite }\sqrt{2000} \\ x=\dfrac{30\pm\sqrt{400\cdot 5}}{50}&\text{Apply product property of radicals} \\ x=\dfrac{30\pm 20\sqrt{5}}{50}&\text{Factor the numerator} \\ x=\dfrac{\cancel{10}(3\pm 2\sqrt{5})}{\cancel{50}^5}&\text{Reduce by a factor of }10 \\ x=\dfrac{3\pm 2\sqrt{5}}{5}&\text{Solution}\end{array}\nonumber\]

    We can see that the equation wasn’t factorable, so we applied the quadratic formula

    Example 11.3.4

    Solve: \(3x^2+4x+8=2x^2+6x-5\)

    Solution

    We first rewrite the equation so that zero is on one side of the equation. Then we can solve as usual.

    \[\begin{array}{rl}3x^2+4x+8=2x^2+6x-5 &\text{Rewrite where zero is on one side} \\ x^2-2x+13=0 &\text{Apply the quadratic formula} \\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}&\text{Plug-n-chug }a=1,\: b=-2,\: c=13 \\ x=\dfrac{2\pm\sqrt{\color{blue}{(-2)}\color{black}{^2}-4\color{blue}{(1)(13)}}}{2\color{blue}{(1)}}&\text{Simplify using order of operations} \\ x=\dfrac{2\pm\sqrt{4-52}}{2} \\ x=\dfrac{2\pm\sqrt{-48}}{2}&\text{Rewrite the radical using }i \\ x=\dfrac{2\pm i\sqrt{48}}{2}&\text{Simplify the radical} \\ x=\dfrac{2\pm 4i\sqrt{3}}{2}&\text{Factor a 2 from the numerator} \\ x=\dfrac{2(1\pm 2i\sqrt{3})}{2}&\text{Reduce by a factor of 2} \\ x=\dfrac{\color{blue}{\cancel{2}}\color{black}{(}1\pm 2i\sqrt{3})}{\color{blue}{\cancel{2}}}&\text{Rewrite} \\ x=1\pm 2i\sqrt{3}&\text{Solutions}\end{array}\nonumber\]

    When there is a negative value as the radicand, we rewrite the radical using the imaginary unit and the solutions are non-real numbers

    Example 11.3.5

    Solve: \(3x^2-7=0\)

    Solution

    If the term is missing from the quadratic equation, we solve the equation by using the quadratic formula and plug-n-chug zero for that term. If the linear term is missing, then \(b = 0\), and if the constant term is missing, then \(c = 0\).

    \[\begin{array}{rl}3x^2-7=0&\text{Apply the quadratic formula} \\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}&\text{Plug-n-chug }a=3,\: b=0,\: c=-7 \\ x=\dfrac{-\color{blue}{0}\color{black}{\pm}\sqrt{\color{blue}{0}\color{black}{^2}-4\color{blue}{(3)(-7)}}}{2\color{blue}{(3)}}&\text{Simplify using order of operations} \\ x=\dfrac{\pm\sqrt{84}}{6}&\text{Rewrite }\sqrt{84} \\ x=\dfrac{\pm\sqrt{4\cdot 21}}{6}&\text{Apply product property of radicals} \\ x=\dfrac{\pm\cancel{2}\sqrt{21}}{\cancel{6}^3}&\text{Reduce by a factor of }2 \\ x=\dfrac{\pm\sqrt{21}}{3}&\text{Solution}\end{array}\nonumber\]

    Note

    When we can factor the quadratic equation, we should, and when the equation isn’t factorable, we should use the quadratic formula.

    If the linear term is missing, then \(b = 0\), and we can solve by the square root property. If the constant term is missing, then \(c = 0\), then we can solve by factoring.

    Quadratic Formula Homework

    Solve each equation by applying the quadratic formula.

    Exercise 11.3.1

    \(4a^2-6=0\)

    Exercise 11.3.2

    \(2x^2-8x-2=0\)

    Exercise 11.3.3

    \(2m^2-3=0\)

    Exercise 11.3.4

    \(3r^2-2r-1=0\)

    Exercise 11.3.5

    \(4n^2-36=0\)

    Exercise 11.3.6

    \(v^2-4v-5=-8\)

    Exercise 11.3.7

    \(2a^2+3a+14=6\)

    Exercise 11.3.8

    \(3k^2+3k-4=7\)

    Exercise 11.3.9

    \(7x^2+3x-16=-2\)

    Exercise 11.3.10

    \(2p^2+6p-16=4\)

    Exercise 11.3.11

    \(3n^2+3n=-3\)

    Exercise 11.3.12

    \(2x^2=-7x+49\)

    Exercise 11.3.13

    \(5x^2=7x+7\)

    Exercise 11.3.14

    \(8n^2=-3n-8\)

    Exercise 11.3.15

    \(2x^2+5x=-3\)

    Exercise 11.3.16

    \(4a^2-64=0\)

    Exercise 11.3.17

    \(4p^2+5p-36=3p^2\)

    Exercise 11.3.18

    \(-5n^2-3n-52=2-7n^2\)

    Exercise 11.3.19

    \(7r^2-12=-3r\)

    Exercise 11.3.20

    \(2n^2-9=4\)

    Exercise 11.3.21

    \(3k^2+2=0\)

    Exercise 11.3.22

    \(6n^2-1=0\)

    Exercise 11.3.23

    \(5p^2+2p+6=0\)

    Exercise 11.3.24

    \(2x^2-2x-15=0\)

    Exercise 11.3.25

    \(3b^2+6=0\)

    Exercise 11.3.26

    \(2x^2+4x+12=8\)

    Exercise 11.3.27

    \(6n^2-3n+3=-4\)

    Exercise 11.3.28

    \(4x^2-14=-2\)

    Exercise 11.3.29

    \(4n^2+5n=7\)

    Exercise 11.3.30

    \(m^2+4m-48=-3\)

    Exercise 11.3.31

    \(3b^2-3=8b\)

    Exercise 11.3.32

    \(3r^2+4=-6r\)

    Exercise 11.3.33

    \(6a^2=-5a+13\)

    Exercise 11.3.34

    \(6v^2=4+6v\)

    Exercise 11.3.35

    \(x^2=8\)

    Exercise 11.3.36

    \(2k^2+6k-16=2k\)

    Exercise 11.3.37

    \(12x^2+x+7=5x^2+5x\)

    Exercise 11.3.38

    \(7m^2-6m+6=-m\)

    Exercise 11.3.39

    \(3x^2-3=x^2\)

    Exercise 11.3.40

    \(6b^2=b^2+7-b\)


    This page titled 11.3: Quadratic formula is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Darlene Diaz (ASCCC Open Educational Resources Initiative) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.