Quadratic Formula

Quadratic Formula - How it Works - Video

Definition

Definition of Quadratic Functions:

Here we have the standard form or general form of a quadratic function, f(x) = ax2 + bx + c, and a cannot be 0. If a is o, then we have f(x) = bx + c, which is a linear function.

Proof of the Quadratic Formula 

Proof of the Quadratic Formula - Complete the Square - Part 1:

Line 1 - Given

Line 2 - Subtracted c on both sides

Line 3 - Divided each term by a to make the coefficient for x2 equal to 1

Line 4 - Divided the coefficient, b/a, for x by 2 and inserted b/(2a)2 on both sides to complete the square

Line 5 - Factored the left side, x + b/a * x + (b / [2a] )2 to (x + b / [2a] )2 

Here is the first part of how to find the quadratic formula. We have to complete the square so we can solve for x.

Proof of the Quadratic Formula - Combine Like Terms - Part 2:

Line 1 - Last step of part 1

Line 2 - Took the power of (b/ [2a] )2

Line 3 - Multiplied -c/a by (4a) / (4a) to create a common denominator

Line 4 - Smashed the numbers together

Line 5 - Switched the ordered of the terms so the negative is 2nd

Line 6 - Combine terms since both have 4a2 in the denominator

Here is the second part of how to find the quadratic formula. We have to combine like terms to have one term on each side. 

Proof of the Quadratic Formula - Solve for x - Part 3:

Line 1 - Last step of part 2

Line 2 - Took the square root on both sides

Line 3 - Square and square root cancel for the left side and separated the fraction into two square roots for the right side

Line 4 - Dropped the parentheses for the left side and simplified the denominator for the right side

Line 5 - Subtracted -b / (2a) on both sides

Line 6 - Combine terms since both have 2a in the denominator

Here is the third part of how to find the quadratic formula. We have to solve for x so we can the zeros of any function

Interesting Tidbit

Interesting Tidbit:

Let's take a look at the step, x = -b / (2a) + √(b2 - 4ac) / (2a). (Line 5 for the Proof - Part 3.) We can find an interesting tidbit. The first part is the axis of symmetry, which goes through the vertex. Remember that quadratic functions are mirror images along the axis of symmetry. So we can the graph in half and the lines will match up.

So that means that the roots, zeros of the function, or the x-intercepts are equal distance. +√(b2 - 4ac) / (2a) is the distance to the right and -√(b2 - 4ac) / (2a) is the distance to the left.

So if we combine the axis of symmetry and the distance we can the x-intercepts for each graph, if there are any. Sometimes we have none, one, or two x-intercepts. We can look at the discriminant to determine that.

Example 1

Example 1:

Here we have our parabolic equation, 2x2 + x - 15 = 0.

Our first step is to solve for zero. Why zero? Well, we are solving for the zeros of the function, and another name for the zeros of the function are the x-intercepts or the roots of a function.

Luckily for us this time, zero is already on one side.

Let's rewrite our equation, so it is a bit easier to see the numbers, 2x2 + 1x - 15 = 0. Now, we can label the a, b, and c values of our equation, so a is 2, b is 1, and c is -15.  Now, we can put these numbers into the quadratic formula.

 [ -b ± {b2 - 4ac} ] / (2a)  

 [ -(1) ± √ {(1)2 - 4(2)(-15)} ] / ( 2[2] )  

[ -1 ± √ {1 + 120} ] / 4

[ -1 ± √ {121} ] / 4

[ -1 ± 11 ] / 4

Given

Substituted the a, b, and c values

Multiplied

Added the numbers in the radical

Square rooted the radicand

[ -1 + 11 ] / 4

[ 10 ] / 4

5 / 2

1st scenario

Add in the numerators

Simplified

[ -1 - 11 ] / 4

[ -12 ] / 4

-3

2nd scenario

Add in the numerators

Simplified

Now we have our two answers, x = 5/2 or x = -3.

Example 2

Example 2:

Here we have our parabolic equation, 2x2 + 6x = 7.

Our first step is to solve for zero. Why zero? Well, we are solving for the zeros of the function, and another name for the zeros of the function are the x-intercepts or the roots of a function.

This time we have one step to do before we find a, b, and c. Right now we don't have 0 on one side. So we have to subtract 9 on both sides to get 2x2 + 6x - 9 = 0.

Now, we can label the a, b, and c values of our equation, so a is 2, b is 6, and c is -7.  Now, we can put these numbers into the quadratic formula.

 [ -b ± √ {b2 - 4ac} ] / (2a)  

 [ -(6) ± √ {(6)2 - 4(2)(-7)} ] / ( 2[2] )  

[ -6 ± √ {36 + 56} ] / 4

[ -6 ± √ {92} ] / 4

[ -6 ± 2√ {23} ] / 4

Given

Substituted the a, b, and c values

Multiplied

Added the numbers in the radical

Square rooted the radicand

√ {92}

√ {2} * √ {2} * √ {23}

 √ {4} * √ {23}

 2√ {23}

Given

Found the prime factorization

Combined same radicands

Simplified

[ -6 + 2√ {23} ] / 4

 -6/4 + 2√ {23} ] / 4

-3/2 + √ {23} ] / 2

[ -3 + √ {23} ] / 2

1st scenario

Separated into two fractions

Simplified

Combined fractions

[ -6 - 2√ {23} ] / 4

 -6/4 - 2√ {23} ] / 4

-3/2 - √ {23} ] / 2

[ -3 - √ {23} ] / 2

2nd scenario

Separated into two fractions

Simplified

Combined fractions

Now we have our two answers, x = [ -3 ± √ {23} ] / 2

Example 3

Example 3:

Here we have our parabolic equation, 3x2 - 4x = -5.

Our first step is to solve for zero. Why zero? Well, we are solving for the zeros of the function, and another name for the zeros of the function are the x-intercepts or the roots of a function.

This time we have one step to do before we find a, b, and c. Right now we don't have 0 on one side. So we have to add 5 on both sides to get 3x2 - 4x + 5 = 0.

Now, we can label the a, b, and c values of our equation, so a is 3, b is -4, and c is 5.  Now, we can put these numbers into the quadratic formula.

 [ -b ± √ {b2 - 4ac} ] / (2a)  

 [ -(-4) ± √ {(-4)2 - 4(3)(5)} ] / ( 2[3] )  

[ 4 ± √ {16 - 60} ] / 6

[ 4 ± √ {-44} ] / 6

[ 4 ± 2i√ {11} ] / 6

Given

Substituted the a, b, and c values

Multiplied

Added the numbers in the radical

Square rooted the radicand

√ {44}

√ {4} * √ {11}

 2√ {11}

Given

Found a square number

Simplify

[ 4 + 2i√ {11} ] / 6

 4/6 + 2i√ {11} ] / 6

2/3 + i√ {23} ] / 3

[ 2 + i√ {23} ] / 3

1st scenario

Separated into two fractions

Simplified

Combined fractions

[ 4 - 2i√ {11} ] / 6

 4/6 - 2i√ {11} ] / 6

2/3 - i√ {23} ] / 3

[ 2 - i√ {23} ] / 3

2nd scenario

Separated into two fractions

Simplified

Combined fractions

Now we have our two answers, x = [ 2 ± i√ {11} ] / 3

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