\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus Precalculus - Quadratic Formula

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The quadratic formula is easy to remember and easy to use. Many students rely on it to solve lots of problems. It is very useful but it can become a crutch if you don't remember where it comes from, why you are using it and how to use it properly.

If you would like a complete lecture on this topic, we recommend this video from one our favorite instructors.

Prof Leonard - Solving Quadratic Equations with the Quadratic Formula

video by Prof Leonard

The Quadratic Formula

The values of x that solve \(ax^2 + bx + c = 0\) are

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

The quadratic formula is derived by completing the square. If you know how to complete the square, you do not need the quadratic formula and you can solve many more types of problems than with the quadratic formula alone. Everywhere you can use the quadratic formula, you can also complete the square. But there will be times when the quadratic formula cannot be used but completing the square will work.
We highly recommend that you completely understand how to derive the quadratic formula since some instructors may not let you use it. They may require you to complete the square and show your work. So make sure you understand this section.

Completing the square is a much important and versatile technique than using the quadratic formula.

Deriving The Quadratic Formula

Given a quadratic equation \(ax^2 + bx + c = 0\), we can solve this by completing the square as follows.

\( ax^2 + bx + c = 0 \)

Divide both sides by a.

\( x^2 + (b/a)x + (c/a) = 0 \)

\( x^2 + (b/a)x = - (c/a) \)

Take the term in front of the x and divide by 2 and square it. Then add that term to both sides of the equation.

\( x^2 + (b/a)x + [b/(2a)]^2 = - (c/a) + [b/(2a)]^2 \)

On the left, we now have a perfect square.

\( [x+b/(2a)]^2 = - (c/a) + [b/(2a)]^2 \)

Take the square root of both sides, remembering to have plus/minus on one side.

\( x+b/(2a) = \pm \sqrt{ - (c/a) + [b/(2a)]^2 } \)

\( x = -b/(2a) \pm \sqrt{ - (c/a) + [b/(2a)]^2 } \)

Let's look at the term under the square root.

\(\displaystyle{ \sqrt{ - (c/a) + [b/(2a)]^2 } }\)

\(\displaystyle{ \sqrt{ [b/(2a)]^2 - (c/a) } }\)

\(\displaystyle{ \sqrt{ \left[ \frac{b}{2a} \right]^2 - \frac{c}{a} } }\)

\(\displaystyle{ \sqrt{ \frac{b^2}{4a^2} - \frac{c}{a} } }\)

\(\displaystyle{ \sqrt{ \frac{b^2}{4a^2} - \frac{4ac}{4a^2} } }\)

\(\displaystyle{ \sqrt{ \frac{b^2-4ac}{4a^2} } }\)

\(\displaystyle{ \frac{\sqrt{b^2-4ac}}{\sqrt{4a^2}} }\)

\(\displaystyle{ \frac{\sqrt{b^2-4ac}}{2a} }\)

Now let's put this back into the equation and see what we have.

\(\displaystyle{ x = \frac{-b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a} }\)

\(\displaystyle{ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} }\)

This last equation is the quadratic formula.
Now, in order to learn this, we recommend that you start from \( ax^2 + bx + c = 0 \) and try to get the quadratic equation on your own without looking at this derivation. Do this several times until you know it. It will also help for you to watch a video or two of someone doing this derivation. The 17Calculus YouTube playlist below has several to choose from. Watch at least one or two of them before working the practice problems.

quadratic formula 17calculus youtube playlist

Here is a playlist of the videos on this page.

Practice

Unless otherwise instructed, use the quadratic formula to solve these equations. Check your answers by completing the square.

Unless otherwise instructed, use the quadratic formula to solve the equation \( x^2 + x - 12 = 0 \). Check your answer(s) by completing the square.

Problem Statement

Unless otherwise instructed, use the quadratic formula to solve the equation \( x^2 + x - 12 = 0 \). Check your answer(s) by completing the square.

Solution

3078 video

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Unless otherwise instructed, use the quadratic formula to solve the equation \( x^2 - 6x + 5 = 0 \). Check your answer(s) by completing the square.

Problem Statement

Unless otherwise instructed, use the quadratic formula to solve the equation \( x^2 - 6x + 5 = 0 \). Check your answer(s) by completing the square.

Solution

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Unless otherwise instructed, use the quadratic formula to solve the equation \( (1/4)x^2 - 2x + 3 = 0 \). Check your answer(s) by completing the square.

Problem Statement

Unless otherwise instructed, use the quadratic formula to solve the equation \( (1/4)x^2 - 2x + 3 = 0 \). Check your answer(s) by completing the square.

Solution

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Unless otherwise instructed, use the quadratic formula to solve the equation \( x^2 + 8x - 16 = 0 \). Check your answer(s) by completing the square.

Problem Statement

Unless otherwise instructed, use the quadratic formula to solve the equation \( x^2 + 8x - 16 = 0 \). Check your answer(s) by completing the square.

Solution

3081 video

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Unless otherwise instructed, use the quadratic formula to solve the equation \( 4x^2 - 12x + 9 = 0 \). Check your answer(s) by completing the square.

Problem Statement

Unless otherwise instructed, use the quadratic formula to solve the equation \( 4x^2 - 12x + 9 = 0 \). Check your answer(s) by completing the square.

Solution

3082 video

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Unless otherwise instructed, use the quadratic formula to solve the equation \( 3x^2 = 6x-1 \). Check your answer(s) by completing the square.

Problem Statement

Unless otherwise instructed, use the quadratic formula to solve the equation \( 3x^2 = 6x-1 \). Check your answer(s) by completing the square.

Solution

2563 video

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Unless otherwise instructed, use the quadratic formula to solve the equation \( x^2 + 5x + 2 = 0 \). Check your answer(s) by completing the square.

Problem Statement

Unless otherwise instructed, use the quadratic formula to solve the equation \( x^2 + 5x + 2 = 0 \). Check your answer(s) by completing the square.

Solution

2564 video

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Unless otherwise instructed, use the quadratic formula to solve the equation \( x^2 - 5x - 6 = 0 \). Check your answer(s) by completing the square.

Problem Statement

Unless otherwise instructed, use the quadratic formula to solve the equation \( x^2 - 5x - 6 = 0 \). Check your answer(s) by completing the square.

Solution

2565 video

video by PatrickJMT

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Unless otherwise instructed, use the quadratic formula to solve the equation \( 2x^2 - 4x + 7 = 0 \). Check your answer(s) by completing the square.

Problem Statement

Unless otherwise instructed, use the quadratic formula to solve the equation \( 2x^2 - 4x + 7 = 0 \). Check your answer(s) by completing the square.

Solution

2566 video

video by PatrickJMT

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Really UNDERSTAND Precalculus

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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