Lesson Introduction into Quadratic Equations

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Introduction into Quadratic Equations

graph%28+200%2C+200%2C+-5%2C+5%2C+-10%2C+10%2C+x%5E2%2B2x-3%29
Graph of x%5E2%2B2x-3

I am writing this lesson so that it contains everything you need to solve quadratic equations and do well on tests.
Other quadratic lessons in this module explain the fine points of quadratics if you are interested.
You can also try the quadratic equation solver that also shows you a graph.

What is a quadratic equation


A quadratic equation is an equation of form ax%5E2%2Bbx%2Bc=0 that involves only two things besides numbers: a variable and a square of this variable.

Examples: 2x%5E2-x-1=0, x%5E2-1=0, and so on.

Usually, they are arranged so that the square part goes first, then the part with the variable, and some constant, while the right side is equal to zero.
In your tests, a, b and c will be actual numbers.

Solving Quadratic Equations


There are two ways of solving quadratics: factoring and using the Quadratic formula (see solver). Of these methods, the Quadratic Formula is the most reliable method
that will give you the correct answer without guesswork. Here's how it works.

Let's say that you have an equation ax%5E2%2Bbx%2Bc=0. If your equation is in some other form, for example x%5E2=-2x%2B3, convert it to standard form with x%5E2 first, x part second,
and the number third. The previous example x%5E2=-2x%2B3, for example, converts to x%5E2%2B2x-3=0.

First, compute the discriminant. discriminant+=+b%5E2-4ac. You need to remember this formula. The discriminant can be positive, zero, or negative.


If the discriminant is positive, the quadratic equation has two different solutions (roots). One solution is x%5B1%5D+=+%28-b-sqrt%28+b%5E2-4ac+%29%29%2F2a. The other solution is x%5B2%5D+=+%28-b%2Bsqrt%28+b%5E2-4ac+%29%29%2F2a.
Click here to read the proof of these formulas.
These solutions are so similar, that they are often written as x+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F2a.
The +/- sign means that one solution comes with a "+" sign, and the other with a "-" sign. Last formula is called quadratic formula.


If the discriminant is equal to zero, the quadratic equation has one solution (root). This solution is x+=+%28-b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2a. You can still use the same formula, but because
the discriminant (which is an expression under the square root) is zero, it doesn't contribute, and +- doesn't make the difference. So, the root in this case is simply x+=+-b%2F2a.


If the discriminant is negative, the quadratic equation has no solution (root) in the area of real numbers.


Example 1.

Solve quadratic equation x%5E2-5x-6=0 by using the quadratic formula.
First calculate the discriminant: d=%28-5%29%5E2-4%2A1%2A%28-6%29+=+25%2B4%2A6+=+25%2B24+=+49.
Discriminant is positive, so the equation has two roots:
x%5B1%5D+=+%28-%28-5%29%2Bsqrt%2849%29%29%2F2+=+%285%2B7%29%2F2+=+12%2F2+=+6
and
x%5B2%5D+=+%28-%28-5%29-sqrt%2849%29%29%2F2+=+%285-7%29%2F2+=+-2%2F2+=+-1.
The two solutions are 6 and -1


Example 2.

Solve quadratic equation 2x%5E2%2B4x-3=0 by using the quadratic formula.
First calculate the discriminant: d=4%5E2-4%2A2%2A%28-3%29+=+16%2B4%2A2%2A3+=+16%2B24+=+40.
Discriminant is positive, so the equation has two roots:

and
x%5B2%5D+=+%28-4-sqrt%2840%29%29%2F4+=+%28-2-sqrt%2810%29%29%2F2+=+%28-2-sqrt%2810%29%29%2F2
The two solutions are %28-2%2Bsqrt%2810%29%29%2F2 and %28-2-sqrt%2810%29%29%2F2


Example 3.

Solve quadratic equation x%5E2-10x%2B25=0 by using the quadratic formula.
Calculating the discriminant: d=%28-10%29%5E2-4%2A1%2A25+=+100-100+=+0.
Discriminant is equal to zero, so the equation has only one root:
x+=+%28-10%2Bsqrt%280%29%29%2F2+=+-5
The solution is 5.


Example 4.

Solve quadratic equation x%5E2%2B6x%2B10=0 by using the quadratic formula.
Calculating the discriminant: d=6%5E2-4%2A1%2A10+=+36-40+=+-4.
Discriminant is negative, so the equation has no root in real numbers.
The equation has no solution in real numbers.


On solving quadratic equations see also other lessons in this site
    - PROOF of quadratic formula by completing the square

    - HOW TO complete the square - Learning by examples
    - HOW TO solve quadratic equation by completing the square - Learning by examples
    - Solving quadratic equations without quadratic formula
    - Who is who in quadratic equations
    - Using Vieta's theorem to solve quadratic equations and related problems

    - Find a number using quadratic equations
    - Find an equation of the parabola passing through given points
    - Problems on quadratic equations to solve them using discriminant
    - Relative position of a straight line and a parabola on a coordinate plane
    - Advanced minimax problems to solve them using the discriminant principle

    - Using quadratic equations to solve word problems
    - Word problems on engineering constructions of parabolic shapes
    - Challenging word problems solved using quadratic equations
    - Business-related problems to solve them using quadratic equations
    - Rare beauty investment problem to solve using quadratic equation
    - HOW TO solve the problem on quadratic equation mentally and avoid boring calculations
    - Entertainment problems on quadratic equations
    - Prime quadratic polynomials with real coefficients

    - Problem on a projectile moving vertically up and down
    - Problem on an arrow shot vertically upward
    - Problem on a ball thrown vertically up from the top of a tower
    - Problem on a toy rocket launched vertically up from a tall platform

    - Problems on the area and the dimensions of a rectangle
    - Problems on the area and the dimensions of a rectangle surrounded by a strip
    - Problems on a circular pool and a walkway around it

    - OVERVIEW of lessons on solving quadratic equations

Use this file/link  ALGEBRA-I - YOUR ONLINE TEXTBOOK  to navigate over all topics and lessons of the online textbook  ALGEBRA-I.



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