The Quadratic Formula

The quadratic formula can help you solve any quadratic equation of the form

ax² + bx + c

To find the solutions to this equation we can use the quadratic formula which is written as follows

(-b (+or-) sqrt(b²-4ac)) / 2a

Let us consider an example of

x² + 6x + 7

a=1
b=6
c=7

so

(-6 (+or-) sqrt(6²-4*1*7)) / 2*1

=

-6 (+or-) sqrt(36 - 28) / 2

=

-6 (+or-) sqrt(8) / 2

We can simplify the square root so we get

(-6 (+or-) 2sqrt(2)) / 2

=

-3 +or- sqrt(2)

and that is our final answer.

Solving Equations with a Square Root

When we solve an equation by taking a square root, we have to consider both a positive and negative outcome.

For example consider -3 and 3

now 3² = 9 and -3²=9

Thus when we have an equation such that

x² = 9 to find x we need to take the square root of 9, however, then we have to say x= + or - 3 that is to say positive or negative 3 since it could be either and we don't know.

Multiplying and dividing square roots (radicals)

Square roots act just like other numbers when you multiply and divide them, consider the example

3sqrt(5) * 5sqrt(6) = 15sqrt(30)

Similarly

15sqrt(30) / 5sqrt(6) = 3sqrt(5)

Adding and Subtracting Radicals

When adding and subtracting radicals you treat the radicands as variables.

Example

3sqrt(5) + 4sqrt(5) = 7sqrt(5)

However, we cannot add together a radicand that is different as in

3sqrt(5) + 4sqrt(2)

The square root of a fraction

By definition we cannot derive the square root of a fraction. Thus we must find a way to get a fraction out of the radical.

Consider the square root of 1/2

sqrt(1/2)

how can we get the fraction out? We have to multiply and make the denominator a perfect square. But what we multiply to the bottom we must also multiply to the top

sqrt((1/2)*(2/2))

=

1/2*sqrt(2)

Now we have got the fraction out of the radical and created a square root we can rationalize.

Another example

sqrt(3/5)

To get the denominator out we multiply by (5/5) (which is equal to 1)

sqrt((3/5)*(5/5))

=

(1/5)*sqrt(3*5)

=

(1/5)*sqrt(15)

Simplifying Radicals

Like so much of algebra, it is good to know how to simplify radicals for purposes of canceling out or combining like terms.

Consider the square root of 27 or sqrt(27) there is no whole number that equals sqrt(27) but we can write the radical as sqrt(9*3) and that is equal to 3*sqrt(3)

Thus we have simplified the radical for purposes of mathematical calculation or canceling.

Square roots are radical

It is true, taking the square root of an expression can also be called "The radical"

We will write square root as sqrt on this website, thus

sqrt(4) = 2

and

sqrt(9x²y¹⁰) ?

To solve this, it is good to factor under the radical ( or factor the square root)

So we get

sqrt(9x²y¹⁰)

=

(3xy⁵)

Solving quadractics by factoring

One way to solve a quadratic equation like:

(4x² - 4)=0 is by factoring and setting both factors equal to zero. Because a quadratic contains a x² they often have two solutions.

(4x² - 4)=0 can be factored to

(2x - 2)(2x + 2)=0

Now set both factors equal to zero

2x-2=0
2x+2=0

we get
x=1 and x=-1
substituting 1 or -1 for x will solve the quadratic equation (4x² + 4)=0

Quadratic Equation

A quadratic equation is described as an equation where the highest exponent is 2.
The graph of a quadratic is a smooth curve known as a parabola.

All of the following are quadratic equations.
x² + 4 = 0

x² + 4x + 3 = 0

3x² + 34x + 7 = 50

Factoring a trinomial

As I said yesterday, I really think factoring comes down to trial and error till the process is internalized. As an example today we will factor trinomials.

Consider

5x² - 8x - 21

This is quite a complicated trinomial to factor, lets start with a guess

First off, we know that to get 5x² we need to multiply 5x and x, so that gives us our first two terms:

(5x + ) (x - )

As a further guess I also alternated the signs.

Now we can try guess what two numbers can multiply to give us -21. How about 7 and -3?

(5x + 7) (x - 3)

Checking with the foil method we get
5x² - 8x - 21

Factoring with the difference of squares

I think factoring is something which becomes internal, you see a problem, make a guess, and then check. The best method is trial and error till it becomes intuitive.

Still the difference of squares method is often taught, and so I will show it here.

Basically the difference of squares is always factored in the following form:

(x+y)(x-y)

Which equals (x² - y²)

Example

x² - 9

9 is a perfect square so we can use the memorized formula

(x-3)(x+3)

Again, I prefer gaining an intuitive understanding of factoring, but memorizing a rule like this can help till you gain an intuitive understanding.

Factoring binomials using the greatest common factor

One way to factor binomials is by searching for the greatest common factor.

Consider

(5x * 25)

In this case the greatest common factor is 5 and the phrase can be written as

5(x*5)

Multiplying a trinomial by a binomial

Multiplying a trinomial by a binomial is a lot like multiplying a binomial by a binomial. You multiply the first term by all the factors of the second term, then multiply the outer(last) term by all the factors of the second term, then simplify.

Consider the example:

(x+5) (5x² + 3x + 6)

First we multiply our first term (x) by every term in the trinomial (5x² + 3x + 6)

this gives us:

(5x³ + 3x² + 6x)

next we multiply our outer term (5) by every term in the trinomial (5x² + 3x + 6)

this gives us:
(25x² + 15x + 30)

so we have
(5x³ + 3x² + 6x) + (25x² + 15x + 30)

we can simplify by adding like terms to get:
(5x³ + 28x² + 21x + 30)

Multiplying a binomial by a binomial

The most common way to multiply binomials is what is called the FOIL method

First
Outer
Inner
Last

Let us look at an example

(x+4) (x+1)

These are both binomials, to multiply them we first multiply the first two terms to get x²

Then we still take the first x and multiply it by the outer number: 1, to get x.

So far we have
x² + x

Now we do the inner number: 4

4 times x is 4x

and finally the last number 4 times 1 is 4

So in total we have
x² + x + 4x + 4

which can be simplified by combining the like x terms to

x² + 5x + 4

Polynomials

Polynomials can be anything from a single number to a variable to a combination of numbers and variables

Monomials have one term.
Such as... 8x⁴ , 6 , or 2xy

Binomials have 2 terms which are not like.
Such as... 2wz - 4dt , 4x² - 3x , 4c - 2d

Trinomials have 3 terms which are not like.
Such as... 4bt - 5yu + 9o , 3x² - 2x + 9 , 5t + 7y - 8u

Dividing exponents

Yesterday we learned that when you multiply exponents you add the number in the exponent, today we see that when you divide exponent you subtract the number in the exponent.

Consider

x⁷ / ³

What is this equal to?

x*x*x*x*x*x*x / x*x*x = x*x*x*x or x⁴

x⁷ / ³ = x^7-3 = x⁴

what about

x³ / ⁷

= 1 / x^7-3 = 1 / x⁴

We take the reciprocal because the exponent is greater in the divisor, or denominator.

Exponent

Exponents tell you how many times a factor is multiplied.

x * x * x (x times x times x)

Can be written as x³ or x^3 , when we write the multiplication in this way, we call it an exponent.

To multiply exponents we add them, for example, consider we have

x² * x³

what is this equal to?

x⁵

why is this? Well if we write it out, it becomes obvious


x² * x³

=


(x*x) * (x*x*x) or x⁵

if we have

5x² * 2x³

Then the bottom numbers (or base numbers) are multiplied, while the exponents are added


5x² * 2x³
=
10x⁵

Using Substitution to Solve a System of Equations

Suppose we had a system of equations

2x + y = 4
and
3x + 2y = 5

How can we solve for x and y?

The good thing is that we have two equations for two variables.

One way is to solve on equation for y and substitute. Let us start with

2x + y = 4

subtract 2x from both sides

y = 4 - 2x we can use this informaiton to solve for x by substituting y into the other equation

3x + 2y = 5 becomes

3x + 2(4-2x) = 5

3x + 8 - 4x = 5

-x = -3
x=3

so x = 3

Now we can substitute x into our first equation to find y

2x + y = 4

6 + y = 4

y = -2

To check let us substitute our answers into the equations and see if we get the same answer:


2x + y = 4
and
3x + 2y = 5

x=3 y= -2

2(3) - 2 = 4
6-2 =4 Correct.

Next

3(3) + 2(-2) =5
9 - 4 = 5 Correct.

So our solutions to the system check OK and are correct.

Systems of Linear Equations

A system of linear equations is a composed of two or more equations with the same variables.

If you have two variables then you need two equations
three variables - three equations, and so on.

Let say you have a system of two equations, if you were to graph the two equations
we would say the system has a solution where the two lines intersect.
If the two lines run parallel then there are no solutions.
If the two lines coincide, then they are the same, and there is an infinite number of solutions.

Linear vs non-linear equations

A linear equation is any equation which graphs a straight line and is of the form

Ax + By = C where A and B are not equal to zero.

examples:
3x + 5y = 8
(4/3)x + 6y = 0
x = 19

These are all linear equations.

Non-linear equations will not be a straight line, and are generally less intuitive, examples are

x^3 + 4y = 8 (this is exponential)

(5/x) + 3y = 9 (contains a variable in the denominator)

2xy = 8 (is multiplicative)