problems in quadratic equation with solution

• PRO Courses Guides New Tech Help Pro Expert Videos About wikiHow Pro Upgrade Sign In
• EDIT Edit this Article
• EXPLORE Tech Help Pro About Us Random Article Quizzes Request a New Article Community Dashboard This Or That Game Popular Categories Arts and Entertainment Artwork Books Movies Computers and Electronics Computers Phone Skills Technology Hacks Health Men's Health Mental Health Women's Health Relationships Dating Love Relationship Issues Hobbies and Crafts Crafts Drawing Games Education & Communication Communication Skills Personal Development Studying Personal Care and Style Fashion Hair Care Personal Hygiene Youth Personal Care School Stuff Dating All Categories Arts and Entertainment Finance and Business Home and Garden Relationship Quizzes Cars & Other Vehicles Food and Entertaining Personal Care and Style Sports and Fitness Computers and Electronics Health Pets and Animals Travel Education & Communication Hobbies and Crafts Philosophy and Religion Work World Family Life Holidays and Traditions Relationships Youth
• Browse Articles
• Learn Something New
• Quizzes Hot
• This Or That Game New
• Train Your Brain
• Explore More
• Support wikiHow
• About wikiHow
• Log in / Sign up
• Education and Communications
• Mathematics

How to Solve Quadratic Equations

Last Updated: February 10, 2023 Fact Checked

This article was co-authored by David Jia . David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. There are 9 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 1,367,045 times.

A quadratic equation is a polynomial equation in a single variable where the highest exponent of the variable is 2. [1] X Research source There are three main ways to solve quadratic equations: 1) to factor the quadratic equation if you can do so, 2) to use the quadratic formula, or 3) to complete the square. If you want to know how to master these three methods, just follow these steps.

Factoring the Equation

• Then, use the process of elimination to plug in the factors of 4 to find a combination that produces -11x when multiplied. You can either use a combination of 4 and 1, or 2 and 2, since both of those numbers multiply to get 4. Just remember that one of the terms should be negative, since the term is -4. [3] X Research source

• 3x = -1 ..... by subtracting
• 3x/3 = -1/3 ..... by dividing
• x = -1/3 ..... simplified
• x = 4 ..... by subtracting
• x = (-1/3, 4) ..... by making a set of possible, separate solutions, meaning x = -1/3, or x = 4 seem good.

• So, both solutions do "check" separately, and both are verified as working and correct for two different solutions.

Using the Quadratic Formula

• 4x 2 - 5x - 13 = x 2 -5
• 4x 2 - x 2 - 5x - 13 +5 = 0
• 3x 2 - 5x - 8 = 0

• {-b +/-√ (b 2 - 4ac)}/2
• {-(-5) +/-√ ((-5) 2 - 4(3)(-8))}/2(3) =
• {-(-5) +/-√ ((-5) 2 - (-96))}/2(3)

• {-(-5) +/-√ ((-5) 2 - (-96))}/2(3) =
• {5 +/-√(25 + 96)}/6
• {5 +/-√(121)}/6

• (5 + 11)/6 = 16/6
• (5-11)/6 = -6/6

• x = (-1, 8/3)

Completing the Square

• 2x 2 - 9 = 12x =
• In this equation, the a term is 2, the b term is -12, and the c term is -9.

• 2x 2 - 12x - 9 = 0
• 2x 2 - 12x = 9

• 2x 2 /2 - 12x/2 = 9/2 =
• x 2 - 6x = 9/2

• -6/2 = -3 =
• (-3) 2 = 9 =
• x 2 - 6x + 9 = 9/2 + 9

• x = 3 + 3(√6)/2
• x = 3 - 3(√6)/2)

Practice Problems and Answers

Expert Q&A

• If the number under the square root is not a perfect square, then the last few steps run a little differently. Here is an example: [14] X Research source Thanks Helpful 1 Not Helpful 0
• As you can see, the radical sign did not disappear completely. Therefore, the terms in the numerator cannot be combined (because they are not like terms). There is no purpose, then, to splitting up the plus-or-minus. Instead, we divide out any common factors --- but ONLY if the factor is common to both of the constants AND the radical's coefficient. Thanks Helpful 1 Not Helpful 0
• If the "b" is an even number, the formula is : {-(b/2) +/- √(b/2)-ac}/a. Thanks Helpful 2 Not Helpful 0

You Might Also Like

• ↑ https://www.mathsisfun.com/definitions/quadratic-equation.html
• ↑ http://www.mathsisfun.com/algebra/factoring-quadratics.html
• ↑ https://www.mathportal.org/algebra/solving-system-of-linear-equations/elimination-method.php
• ↑ https://www.cuemath.com/algebra/quadratic-equations/
• ↑ https://www.purplemath.com/modules/solvquad4.htm
• ↑ http://www.purplemath.com/modules/quadform.htm
• ↑ https://uniskills.library.curtin.edu.au/numeracy/algebra/quadratic-equations/
• ↑ http://www.mathsisfun.com/algebra/completing-square.html
• ↑ http://www.umsl.edu/~defreeseca/intalg/ch7extra/quadmeth.htm

About This Article

To solve quadratic equations, start by combining all of the like terms and moving them to one side of the equation. Then, factor the expression, and set each set of parentheses equal to 0 as separate equations. Finally, solve each equation separately to find the 2 possible values for x. To learn how to solve quadratic equations using the quadratic formula, scroll down! Did this summary help you? Yes No

• Send fan mail to authors

Reader Success Stories

Did this article help you?

Sep 24, 2022

Kalo Morris

Mar 12, 2017

Matthew Mathers

Mar 25, 2017

Kapela Davis

Oct 10, 2017

Featured Articles

Trending Articles

Watch Articles

• Terms of Use
• Privacy Policy
• Do Not Sell or Share My Info
• Not Selling Info

Get all the best how-tos!

Sign up for wikiHow's weekly email newsletter

• My Preferences
• My Reading List
• Study Guides
• Solving Quadratic Equations
• Preliminaries
• Quiz: Preliminaries
• Properties of Basic Mathematical Operations
• Quiz: Properties of Basic Mathematical Operations
• Multiplying and Dividing Using Zero
• Quiz: Multiplying and Dividing Using Zero
• Powers and Exponents
• Quiz: Powers and Exponents
• Square Roots and Cube Roots
• Quiz: Square Roots and Cube Roots
• Grouping Symbols
• Quiz: Grouping Symbols
• Divisibility Rules
• Quiz: Divisibility Rules
• Signed Numbers (Positive Numbers and Negative Numbers)
• Quiz: Signed Numbers (Positive Numbers and Negative Numbers)
• Quiz: Fractions
• Simplifying Fractions and Complex Fractions
• Quiz: Simplifying Fractions and Complex Fractions
• Quiz: Decimals
• Quiz: Percent
• Scientific Notation
• Quiz: Scientific Notation
• Quiz: Set Theory
• Variables and Algebraic Expressions
• Quiz: Variables and Algebraic Expressions
• Evaluating Expressions
• Quiz: Evaluating Expressions
• Quiz: Equations
• Ratios and Proportions
• Quiz: Ratios and Proportions
• Solving Systems of Equations (Simultaneous Equations)
• Quiz: Solving Systems of Equations (Simultaneous Equations)
• Quiz: Monomials
• Polynomials
• Quiz: Polynomials
• Quiz: Factoring
• What Are Algebraic Fractions?
• Operations with Algebraic Fractions
• Quiz: Operations with Algebraic Fractions
• Inequalities
• Quiz: Inequalities
• Graphing on a Number Line
• Quiz: Graphing on a Number Line
• Absolute Value
• Quiz: Absolute Value
• Solving Equations Containing Absolute Value
• Coordinate Graphs
• Quiz: Coordinate Graphs
• Linear Inequalities and Half-Planes
• Quiz: Linear Inequalities and Half-Planes
• Quiz: Functions
• Quiz: Variations
• Introduction to Roots and Radicals
• Simplifying Square Roots
• Quiz: Simplifying Square Roots
• Operations with Square Roots
• Quiz: Operations with Square Roots
• Quiz: Solving Quadratic Equations
• Solving Technique
• Key Words and Phrases
• Simple Interest
• Compound Interest
• Ratio and Proportion
• Percent Change
• Number Problems
• Age Problems
• Motion Problems
• Coin Problems
• Mixture Problems
• Work Problems
• Number Problems with Two Variables
• Quiz: Word Problems

A quadratic equation is an equation that could be written as

ax 2 + bx + c = 0 

There are three basic methods for solving quadratic equations: factoring, using the quadratic formula, and completing the square. 

To solve a quadratic equation by factoring,

• Put all terms on one side of the equal sign, leaving zero on the other side.
• Set each factor equal to zero.
• Solve each of these equations.
• Check by inserting your answer in the original equation.

Solve x 2 – 6 x = 16. 

Following the steps, 

x 2 – 6 x = 16 becomes x 2 – 6 x – 16 = 0 

( x – 8)( x + 2) = 0 

Both values, 8 and –2, are solutions to the original equation.

Solve y 2 = – 6 y – 5. 

Setting all terms equal to zero, 

y 2 + 6 y + 5 = 0 

( y + 5)( y + 1) = 0 

To check, y 2 = –6 y – 5 

A quadratic with a term missing is called an incomplete quadratic (as long as the ax 2 term isn't missing). 

Solve x 2 – 16 = 0. 

To check, x 2 – 16 = 0 

Solve x 2 + 6 x = 0. 

To check, x 2 + 6 x = 0 

Solve 2 x 2 + 2 x – 1 = x 2 + 6 x – 5. 

First, simplify by putting all terms on one side and combining like terms.

Now, factor.

To check, 2 x 2 + 2 x – 1 = x 2 + 6 x – 5 

The quadratic formula

a, b, and c are taken from the quadratic equation written in its general form of 

where a is the numeral that goes in front of x 2 , b is the numeral that goes in front of x , and c is the numeral with no variable next to it (a.k.a., “the constant”). 

When using the quadratic formula, you should be aware of three possibilities. These three possibilities are distinguished by a part of the formula called the discriminant. The discriminant is the value under the radical sign, b 2 – 4 ac . A quadratic equation with real numbers as coefficients can have the following:

• Two different real roots if the discriminant b 2 – 4 ac is a positive number. 
• One real root if the discriminant b 2 – 4 ac is equal to 0. 
• No real root if the discriminant b 2 – 4 ac is a negative number. 

Solve for x : x 2 – 5 x = –6. 

Setting all terms equal to 0, 

x 2 – 5 x + 6 = 0 

Then substitute 1 (which is understood to be in front of the x 2 ), –5, and 6 for a , b , and c, respectively, in the quadratic formula and simplify. 

Because the discriminant b 2 – 4 ac is positive, you get two different real roots. 

Example produces rational roots. In Example , the quadratic formula is used to solve an equation whose roots are not rational. 

Solve for y : y 2 = –2y + 2. 

y 2 + 2 y – 2 = 0 

Then substitute 1, 2, and –2 for a , b , and c, respectively, in the quadratic formula and simplify. 

Note that the two roots are irrational.

Solve for x : x 2 + 2 x + 1 = 0. 

Substituting in the quadratic formula,

Since the discriminant b 2 – 4 ac is 0, the equation has one root. 

The quadratic formula can also be used to solve quadratic equations whose roots are imaginary numbers, that is, they have no solution in the real number system.

Solve for x : x ( x + 2) + 2 = 0, or x 2 + 2 x + 2 = 0. 

Since the discriminant b 2 – 4 ac is negative, this equation has no solution in the real number system. 

Completing the square

A third method of solving quadratic equations that works with both real and imaginary roots is called completing the square.

• Put the equation into the form ax 2 + bx = – c . 

• Find the square root of both sides of the equation.
• Solve the resulting equation.

Solve for x : x 2 – 6 x + 5 = 0. 

Arrange in the form of

Take the square root of both sides.

x – 3 = ±2 

Solve for y : y 2 + 2 y – 4 = 0. 

Solve for x : 2 x 2 + 3 x + 2 = 0. 

There is no solution in the real number system. It may interest you to know that the completing the square process for solving quadratic equations was used on the equation ax 2 + bx + c = 0 to derive the quadratic formula. 

Previous Quiz: Operations with Square Roots

Next Quiz: Solving Quadratic Equations

• Online Quizzes for CliffsNotes Algebra I Quick Review, 2nd Edition

has been added to your

Reading List!

Removing #book# from your Reading List will also remove any bookmarked pages associated with this title.

Are you sure you want to remove #bookConfirmation# and any corresponding bookmarks?

Quadratic Equation Solver

What do you want to calculate.

• Solve for Variable
• Practice Mode
• Step-By-Step

Step-By-Step Example

Example (click to try), choose your method, solve by factoring.

Example: 3x^2-2x-1=0

Complete The Square

Example: 3x^2-2x-1=0 (After you click the example, change the Method to 'Solve By Completing the Square'.)

Take the Square Root

Example: 2x^2=18

Quadratic Formula

Example: 4x^2-2x-1=0

About quadratic equations

Need more problem types? Try MathPapa Algebra Calculator

Clear Quadratic Equation Solver »

Quadratic Equation Solver

We can help you solve an equation of the form " ax 2 + bx + c = 0 " Just enter the values of a, b and c below :

Is it Quadratic?

Only if it can be put in the form ax 2 + bx + c = 0 , and a is not zero .

The name comes from "quad" meaning square, as the variable is squared (in other words x 2 ).

These are all quadratic equations in disguise:

How Does this Work?

The solution(s) to a quadratic equation can be calculated using the Quadratic Formula :

The "±" means we need to do a plus AND a minus, so there are normally TWO solutions !

The blue part ( b 2 - 4ac ) is called the "discriminant", because it can "discriminate" between the possible types of answer:

• when it is positive, we get two real solutions,
• when it is zero we get just ONE solution,
• when it is negative we get complex solutions.

Learn more at Quadratic Equations

Quadratic Equation Questions

Quadratic equation questions are provided here for Class 10 students. A quadratic equation is a second-degree polynomial which is represented as ax 2 + bx + c = 0, where a is not equal to 0. Here, a, b and c are constants, also called coefficients and x is an unknown variable. Also, learn Quadratic Formula here.

Solving the problems based on quadratics will help students to understand the concept very well and also to score good marks in this section. All the questions are solved here step by step with a detailed explanation. In this article, we will give the definition and important formula for solving problems based on quadratic equations. The questions given here is in reference to the CBSE syllabus and NCERT curriculum. 

Definition of Quadratic Equation

Usually, the quadratic equation is represented in the form of ax 2 +bx+c=0, where x is the variable and a,b,c are the real numbers & a ≠ 0. Here, a and b are the coefficients of x 2 and x, respectively. So, basically, a quadratic equation is a polynomial whose highest degree is 2. Let us see some examples:

• 3x 2 +x+1, where a=3, b=1, c=1
• 9x 2 -11x+5, where a=9, b=-11, c=5

Roots of Quadratic Equations:

If we solve any quadratic equation, then the value we obtain are called the roots of the equation. Since the degree of the quadratic equation is two, therefore we get here two solutions and hence two roots.

There are different methods to find the roots of quadratic equation, such as:

• Factorisation
• Completing the square
• Using quadratic formula

Learn: Factorization of Quadratic equations

Quadratic Equation Formula:

The quadratic formula to find the roots of the quadratic equation is given by:

\(\begin{array}{l}x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\end{array} \)

Where b 2 -4ac is called the discriminant of the equation.

Based on the discriminant value, there are three possible conditions, which defines the nature of roots as follows:

• two distinct real roots, if b 2 – 4ac > 0
• two equal real roots, if b 2 – 4ac = 0
• no real roots, if b 2 – 4ac < 0

Also, learn quadratic equations for class 10 here.

Quadratic Equations Problems and Solutions

1. Rahul and Rohan have 45 marbles together. After losing 5 marbles each, the product of the number of marbles they both have now is 124. How to find out how many marbles they had to start with.

Solution: Say, the number of marbles Rahul had be x.

Then the number of marbles Rohan had = 45 – x.

The number of marbles left with Rahul after losing 5 marbles = x – 5

The number of marbles left with Rohan after losing 5 marbles = 45 – x – 5 = 40 – x

The product of number of marbles = 124

(x – 5) (40 – x) = 124

40x – x 2 – 200 + 5x = 124

– x 2 + 45x – 200 = 124

x 2 – 45x + 324 = 0

This represents the quadratic equation. Hence by solving the given equation for x, we get;

x = 36 and x = 9

So, the number of marbles Rahul had is 36 and Rohan had is 9 or vice versa.

2. Check if x(x + 1) + 8 = (x + 2) (x – 2) is in the form of quadratic equation.

Solution: Given,

x(x + 1) + 8 = (x + 2) (x – 2)

Cancel x 2 both the sides.

Since, this expression is not in the form of ax 2 +bx+c, hence it is not a quadratic equation.

3. Find the roots of the equation 2x 2 – 5x + 3 = 0 using factorisation.

2x 2 – 5x + 3 = 0

2x 2 – 2x-3x+3 = 0

2x(x-1)-3(x-1) = 0

(2x-3) (x-1) = 0

2x-3 = 0; x = 3/2

(x-1) = 0; x=1

Therefore, 3/2 and 1 are the roots of the given equation.

4. Solve the quadratic equation 2x 2 + x – 300 = 0 using factorisation.

Solution: 2x 2 + x – 300 = 0

2x 2 – 24x + 25x – 300 = 0

2x (x – 12) + 25 (x – 12) = 0

(x – 12)(2x + 25) = 0

x-12=0; x=12

(2x+25) = 0; x=-25/2 = -12.5

Therefore, 12 and -12.5 are two roots of the given equation.

Also, read   Factorisation .

5. Solve the equation x 2 +4x-5=0.

x 2 + 4x – 5 = 0

x 2 -1x+5x-5 = 0

x(x-1)+5(x-1) =0

(x-1)(x+5) =0

Hence, (x-1) =0, and (x+5) =0

similarly, x+5 = 0

x=-5 & x=1

6. Solve the quadratic equation 2x 2 + x – 528 = 0, using quadratic formula.

Solution: If we compare it with standard equation, ax 2 +bx+c = 0

a=2, b=1 and c=-528

Hence, by using the quadratic formula:

Now putting the values of a,b and c.

x=64/4 or x=-66/4

x=16 or x=-33/2

7. Find the roots of x 2 + 4x + 5 = 0, if any exist, using quadratic formula.

Solution: To check whether there are real roots available for the quadratic equation, we need the find the discriminant value.

D = b 2 -4ac = 4 2 – 4(1)(5) = 16-20 = -4

Since the square root of -4 will not give a real number. Hence there is no real roots for the given equation.

8. Find the discriminant of the equation: 3x 2 -2x+⅓ = 0.

Solution: Here, a = 3, b=-2 and c=⅓

Hence, discriminant, D = b 2 – 4ac

D = (-2) 2 -4(3)(⅓)

Video Lesson

Quadratic equation worksheet.

Practice Questions

Solve these quadratic equations and find the roots. 

• x 2 -5x-14=0         [Answer: x=-2 & x=7]
• X 2  = 11x -28       [Answer: x=4 & x = 7]
• 6x 2 – x = 5            [Answer: x=-⅚ & x = 1]
• 12x 2 = 25x          [Answer: x=0 & x=25/12]

Frequently Asked Questions on Quadratic Equations

What is a quadratic equation, what are the examples of quadratic equations, what is the formula for quadratics, what are the methods to solve the quadratic equation, what are the roots of the quadratic equation, what are the zeroes of the quadratic equation, leave a comment cancel reply.

Your Mobile number and Email id will not be published. Required fields are marked *

Request OTP on Voice Call

Post My Comment

Thank you so much for providing this

• Share Share

Solver Title

Generating PDF...

• Pre Algebra Order of Operations Factors & Primes Fractions Long Arithmetic Decimals Exponents & Radicals Ratios & Proportions Percent Modulo Mean, Median & Mode Scientific Notation Arithmetics
• Algebra Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power Sums Interval Notation Pi (Product) Notation Induction Logical Sets Word Problems
• Pre Calculus Equations Inequalities Simultaneous Equations System of Inequalities Polynomials Rationales Complex Numbers Polar/Cartesian Functions Arithmetic & Comp. Coordinate Geometry Plane Geometry Solid Geometry Conic Sections Trigonometry
• Calculus Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series Fourier Transform
• Functions Line Equations Functions Arithmetic & Comp. Conic Sections Transformation
• Linear Algebra Matrices Vectors
• Trigonometry Identities Proving Identities Trig Equations Trig Inequalities Evaluate Functions Simplify
• Statistics Mean Geometric Mean Quadratic Mean Average Median Mode Order Minimum Maximum Probability Mid-Range Range Standard Deviation Variance Lower Quartile Upper Quartile Interquartile Range Midhinge Standard Normal Distribution
• Physics Mechanics
• Chemistry Chemical Reactions Chemical Properties
• Finance Simple Interest Compound Interest Present Value Future Value
• Economics Point of Diminishing Return
• Conversions Radical to Exponent Exponent to Radical To Fraction To Decimal To Mixed Number To Improper Fraction Radians to Degrees Degrees to Radians Hexadecimal Scientific Notation Distance Weight Time
• Pre Algebra
• One-Step Addition
• One-Step Subtraction
• One-Step Multiplication
• One-Step Division
• One-Step Decimals
• Two-Step Integers
• Two-Step Add/Subtract
• Two-Step Multiply/Divide
• Two-Step Fractions
• Two-Step Decimals
• Multi-Step Integers
• Multi-Step with Parentheses
• Multi-Step Rational
• Multi-Step Fractions
• Multi-Step Decimals
• Solve by Factoring
• Completing the Square
• Quadratic Formula
• Biquadratic
• Logarithmic
• Exponential
• Rational Roots
• Floor/Ceiling
• Equation Given Roots
• Newton Raphson
• Substitution
• Elimination
• Cramer's Rule
• Gaussian Elimination
• System of Inequalities
• Perfect Squares
• Difference of Squares
• Difference of Cubes
• Sum of Cubes
• Polynomials
• Distributive Property
• FOIL method
• Perfect Cubes
• Binomial Expansion
• Negative Rule
• Product Rule
• Quotient Rule
• Expand Power Rule
• Fraction Exponent
• Exponent Rules
• Exponential Form
• Logarithmic Form
• Absolute Value
• Rational Number
• Powers of i
• Partial Fractions
• Is Polynomial
• Leading Coefficient
• Leading Term
• Standard Form
• Complete the Square
• Synthetic Division
• Linear Factors
• Rationalize Denominator
• Rationalize Numerator
• Identify Type
• Convergence
• Interval Notation
• Pi (Product) Notation
• Boolean Algebra
• Truth Table
• Mutual Exclusive
• Cardinality
• Caretesian Product
• Age Problems
• Distance Problems
• Cost Problems
• Investment Problems
• Number Problems
• Percent Problems
• Addition/Subtraction
• Multiplication/Division
• Dice Problems
• Coin Problems
• Card Problems
• Pre Calculus
• Linear Algebra
• Trigonometry
• Conversions

Most Used Actions

Number line.

• ax^2+bx+c=0
• x^2+2x+1=3x-10
• 2x^2+4x-6=0
• How do you calculate a quadratic equation?
• To solve a quadratic equation, use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a).
• What is the quadratic formula?
• The quadratic formula gives solutions to the quadratic equation ax^2+bx+c=0 and is written in the form of x = (-b ± √(b^2 - 4ac)) / (2a)
• Does any quadratic equation have two solutions?
• There can be 0, 1 or 2 solutions to a quadratic equation. If the discriminant is positive there are two solutions, if negative there is no solution, if equlas 0 there is 1 solution.
• What is quadratic equation in math?
• In math, a quadratic equation is a second-order polynomial equation in a single variable. It is written in the form: ax^2 + bx + c = 0 where x is the variable, and a, b, and c are constants, a ≠ 0.
• How do you know if a quadratic equation has two solutions?
• A quadratic equation has two solutions if the discriminant b^2 - 4ac is positive.

quadratic-equation-calculator

• High School Math Solutions – Quadratic Equations Calculator, Part 1 A quadratic equation is a second degree polynomial having the general form ax^2 + bx + c = 0, where a, b, and c... Read More

• First-order ODE
• Introduction Differential equations Order Linearity
• Direction field method Implicit/explicit form Method
• Separation of variables Separable Reduction to separable form
• Equilibrium Characterization Stability
• Linear 1st order ODE technique Standard form General solution Reduction to linear form
• ODE existence and uniqueness Existence theorem Uniqueness theorem
• Initial value numerical methods Error Taylor series Stability Euler methods Runge-Kutta methods
• System of linear ODEs Definition Characteristic equation Eigenvector/eigenvalue System of homogeneous ODEs
• View PDF version on GitHub

First-order Ordinary Differential Equations cheatsheet Star

By Afshine Amidi and Shervine Amidi

Introduction

Differential Equations A differential equation is an equation containing derivatives of a dependent variable $y$ with respect to independent variables $x$. In particular,

- Ordinary Differential Equations ( ODE ) are differential equations having one independent variable.

- Partial Differential Equations ( PDE ) are differential equations having two or more independent variables.

Order An ODE is said to be of order $n$ if the highest derivative of the unknown function in the equation is the $n^{th}$ derivative with respect to the independent variable.

Linearity An ODE is said to be linear only if the function $y$ and all of its derivatives appear by themselves. Thus, it is of the form:

Direction Field Method

Implicit form The implicit form of an ODE is where $y'$ is not separated from the remaining terms of the ODE. It is of the form:

Remark: Sometimes, $y'$ cannot be separated from the other terms and the implicit form is the only one that we can write.

Explicit form The explicit form of an ODE is where $y'$ is separated from the remaining terms of the ODE. It is of the form:

Direction field method The direction field method is a graphical representation for the solution of ODE $y'=f(x,y)$ without actually solving for $y(x)$. Here is the procedure:

Separation of Variables

Separable An ODE is said to be separable if it can be written in the form:

Reduction to separable form The following table sums up the variable changes that allow us to change the ODE $y'=f(x,y)$ to $u'=g(x,u)$ that is separable.

Equilibrium

Characterization In order for an ODE to have equilibrium solutions, it must be (1) autonomous and (2) have a value $y^*$ that makes the derivative equal to 0, i.e:

Stability Equilibrium solutions can be classified into 3 categories:

- Unstable : solutions run away with any small change to the initial conditions. - Stable : any small perturbation leads the solutions back to that solution. - Semi-stable : a small perturbation is stable on one side and unstable on the other.

Linear first-order ODE technique

Standard form The standard form of a first-order linear ODE is expressed with $p(x), r(x)$ known functions of $x$, such that:

Remark: If $r=0$, then the ODE is homogenous, and if $r\neq0$, then the ODE is inhomogeneous.

General solution The general solution $y$ of the standard form can be decomposed into a homogenous part $y_h$ and a particular part $y_p$ and is expressed in terms of $p(x), r(x)$ such that:

Remark 1: Here, for any function $p$, the notation $\displaystyle\int pdx$ denotes the primitive of $p$ without additive constant.

Remark 2: The term $e^{-\int pdx}$ is called the basis of the ODE and $e^{\int pdx}$ is called the integrating factor.

Reduction to linear form The one-line table below sums up the change of variables that we apply in order to have a linear form:

Existence and uniqueness of an ODE

Here, we are given an ODE $y'=f(x,y)$ with initial conditions $y(x_0)=y_0$.

Existence theorem If $f(x,y)$ is continuous at all points in a rectangular region containing $(x_0,y_0)$, then $y'=f(x,y)$ has at least one solution $y(x)$ passing through $(x_0,y_0)$.

Remark: If the condition does not apply, then we cannot say anything about existence.

Uniqueness theorem If both $f(x,y)$ and $\frac{\partial f}{\partial y}(x,y)$ are continuous at all points in a rectangular region containing $(x_0,y_0)$, then $y'=f(x,y)$ has a unique solution $y(x)$ passing through $(x_0,y_0)$.

Remark: If the condition does not apply, then we cannot say anything about uniqueness.

Numerical methods for ODE - Initial value problems

In this section, we would like to find $y(t)$ for the interval $[0,t_f]$ that we divide into $N+1$ equally-spaced points $t_0< t_1 < ... < t_N = t_f$, such that:

Error In order to assess the accuracy of a numerical method, we define its local and global errors $\epsilon_{\textrm{local}}, \epsilon_{\textrm{global}}$ as follows:

Remark 1: If $\epsilon_{\textrm{local}}=O(h^k)$, then $\epsilon_{\textrm{global}}=O(h^{k-1})$.

Remark 2: When we talk about the 'error' of a method, we refer to its global error.

Taylor series The Taylor series giving the exact expression of $y_{n+1}$ in terms of $y_n$ and its derivatives is:

We can also have an expression of $y_n$ in terms of $y_{n+1}$ and its derivatives:

Stability The stability analysis of any ODE solver algorithm is performed on the model problem , defined by:

which gives $y_n=y_0\sigma^n$, for which $h$ verifies the condition $|\sigma(h)| < 1$.

Euler methods The Euler methods are numerical methods that aim at estimating the solution of an ODE:

Runge-Kutta methods The table below sums up the most commonly used Runge-Kutta methods:

System of Linear ODEs

Definition A system of $n$ first order linear ODEs

can be written in matrix form as:

where $A=\left(\begin{array}{ccc}a_{11}& \cdots& a_{1n}\\\vdots& \ddots & \vdots\\a_{n1}& \cdots& a_{nn}\end{array}\right)$ and $\vec{y}=\left(\begin{array}{c}y_1\\\vdots\\y_n\end{array}\right)$

Characteristic equation The characteristic equation of a linear system of $n$ equations represented by $A$ is given by:

For $n=2$, this equation can be written as:

Eigenvector, eigenvalue The roots $\lambda$ of the characteristic equation are the eigenvalues of $A$. The solutions $\vec{v}$ of the equation $A\vec{v}=\lambda I$ are called the eigenvectors associated with the eigenvalue $\lambda$.

System of homogeneous ODEs The resolution of the system of 2 homogeneous linear ODEs $\vec{y}'=A\vec{y}$ is detailed in the following table:

Take the annual global survey

Help | Advanced Search

Computer Science > Computational Complexity

Title: the complexity of computing kkt solutions of quadratic programs.

Abstract: It is well known that solving a (non-convex) quadratic program is NP-hard. We show that the problem remains hard even if we are only looking for a Karush-Kuhn-Tucker (KKT) point, instead of a global optimum. Namely, we prove that computing a KKT point of a quadratic polynomial over the domain $[0,1]^n$ is complete for the class CLS = PPAD$\cap$PLS.

Submission history

Access paper:.

• Download PDF
• Other Formats

References & Citations

• Google Scholar
• Semantic Scholar

BibTeX formatted citation

Bibliographic and Citation Tools

Code, data and media associated with this article, recommenders and search tools.

• Institution

arXivLabs: experimental projects with community collaborators

arXivLabs is a framework that allows collaborators to develop and share new arXiv features directly on our website.

Both individuals and organizations that work with arXivLabs have embraced and accepted our values of openness, community, excellence, and user data privacy. arXiv is committed to these values and only works with partners that adhere to them.

Have an idea for a project that will add value for arXiv's community? Learn more about arXivLabs .