Quadratic Equations

Summary

  • A quadratic equation is in the form  y\quad =\quad a{ x }^{ 2 }\quad +\quad bx\quad +\quad c\quad \quad \quad \quad \quad a\quad \neq \quad 0,
  • A quadratic function has a minimum value when a > 0
  • A quadratic function has a maximum value when a < 0
  • Vertex is a point on the where the curve is either maximum or minimum
  • To find the values of the maximum or minimum point we use the method called ”Completing the square method”.
  • (\frac { -b }{ 2a } ,\quad \frac { 4ac\quad -\quad { b }^{ 2 } }{ 4a } )  is the vertex, we get from the form obtained:

f\left( x \right) \quad =\quad a{ (x\quad +\quad \frac { b }{ 2a } ) }^{ 2 }\quad +\quad \frac { 4ac\quad -\quad { b }^{ 2 } }{ 4a }  after applying completing the square method.

An expression which is written in the form  f\left( x \right) \quad =\quad a{ x }^{ 2 }\quad +\quad bx\quad +\quad c  where a, b and c are constants and  a\quad \neq \quad 0,  is called a ”Quadratic Function”.

We know that a graph for the equation  f\left( x \right) \quad =\quad a{ x }^{ 2 }\quad +\quad bx\quad +\quad c,  when a > 0, is a parabola with a minimum value as shown in Fig 1.

However, when a < 0, the graph of the function has a maximum value as shown in Fig 2.

Here P and Q are called the vertex of the parabola.

Therefore, we now know that the maximum or minimum value of a quadratic function depends on the coefficient of  { x }^{ 2 }  in the function.

Completing the Square Method

To find the values of the maximum or minimum point we use the method called ”Completing the square method”.

Let’s look at a generic example to be able to understand how this method works.

Suppose we have a quadratic function  f\left( x \right) \quad =\quad a{ x }^{ 2 }\quad +\quad bx\quad +\quad c  where a, b and c are constants.

Firstly, we make 1 the coefficient of { x }^{ 2 }:

f\left( x \right) \quad =\quad a\left[ { x }^{ 2 }\quad +\quad \frac { b }{ a } x\quad +\quad \frac { c }{ a } \right] \\

We now add and subtract a term { (\frac { b }{ 2a } ) }^{ 2 }:

f\left( x \right) \quad=

=\quad a\left[ { x }^{ 2 }\quad +\quad \frac { b }{ a } x\quad +\quad { (\frac { b }{ 2a } ) }^{ 2 }\quad -\quad { (\frac { b }{ 2a } ) }^{ 2 }\quad +\quad \frac { c }{ a } \right]

The first three terms can be simplified to become { (x\quad +\quad \frac { b }{ 2a } ) }^{ 2 }:

f\left( x \right) \quad =\quad a\left[ { (x\quad +\quad \frac { b }{ 2a } ) }^{ 2 }\quad -\quad { (\frac { b }{ 2a } ) }^{ 2 }\quad +\quad \frac { c }{ a } \right]

Taking the LCM of the last two terms we get:

f\left( x \right) \quad =\quad a\left[ { (x\quad +\quad \frac { b }{ 2a } ) }^{ 2 }\quad +\quad \frac { 4ac\quad -\quad { b }^{ 2 } }{ 4{ a }^{ 2 } } \right]

Multiplying a outside the brackets with the terms inside, we get:

f\left( x \right) \quad =\quad a{ (x\quad +\quad \frac { b }{ 2a } ) }^{ 2 }\quad +\quad \frac { 4ac\quad -\quad { b }^{ 2 } }{ 4a }

From the expression obtained above, we know that a square quantity is always positive, therefore:

i) If a > 0, then the function has a minimum value  \frac { 4ac\quad -\quad { b }^{ 2 } }{ 4a }

ii) If a < 0, then the function has a maximum value  \frac { 4ac\quad -\quad { b }^{ 2 } }{ 4a }

iii) The graph of the function above is always symmetrical in shape about the line  x\quad =\quad -\frac { b }{ 2a }

iv) The point  (\frac { -b }{ 2a } ,\quad \frac { 4ac\quad -\quad { b }^{ 2 } }{ 4a } )  is called vertex of the parabola.

Graph of a quadratic function

To be able to sketch the graph of a quadratic function, we must remember the following four points:

1.     Check the value of coefficient of  { x }^{ 2 }  to know the shape of the curve, whether it has a minimum point (i.e Fig 1) or a maximum point (i.e Fig 2).

2.     Find the vertex (point where the curve is maximum or minimum) of the curve (\frac { -b }{ 2a } ,\quad \frac { 4ac\quad -\quad { b }^{ 2 } }{ 4a } )

3.     Substitute y = 0 in the equation of a function find x intercept.

4.     Substitute x = 0 in the equation of a function find y intercept.

Example #1

Q. Find the vertex of the curve  y\quad =\quad a{ x }^{ 2 }\quad +\quad bx\quad +\quad c

Solution:

To begin with we can see that since the coefficient of { x }^{ 2 } is < 0, this curve will have a maximum value and the shape of the curve will be like in Fig 2.

We know a = -3, b = -2 and c = -4

To find the vertex use the formula:  (\frac { -b }{ 2a } ,\quad \frac { 4ac\quad -\quad { b }^{ 2 } }{ 4a } )

(\frac { -(-2) }{ 2(-3) } ,\quad \frac { 4(-3)(-4)\quad -\quad { (-2) }^{ 2 } }{ 4(-3) } )

(\frac { -1 }{ 3 } ,\quad \frac { 48\quad -\quad 4 }{ -12 } )

(\frac { -1 }{ 3 } ,\quad \frac { -11}{ 3 } )      Ans

Example #2

Q. Solve  y\quad =\quad { x }^{ 2 }\quad +\quad 6x\quad +\quad 5  by completing the square method

Solution:

Since { x }^{ 2 } already has a coefficient 1, so we now add and subtract {(\frac { b }{ 2a } ) }^{ 2 }

a = 1, b = 6 and c = 5

{ (\frac { b }{ 2a } ) }^{ 2 }\quad =\quad { (\frac { 6 }{ 2 } ) }^{ 2 }\quad =\quad { 3 }^{ 2 }

y\quad =\quad { (x }^{ 2 }\quad +\quad 6x\quad +\quad { (3 })^{ 2 }\quad -\quad { (3) }^{ 2 }\quad +\quad 5)

y\quad =\quad { (x\quad +\quad 3 })^{ 2 }\quad -\quad { 9\quad +\quad 5 }

y\quad =\quad { (x\quad +\quad 3 })^{ 2 }\quad -\quad { 4 }

Equate the expression to 0 to find the values of x:

{ (x\quad +\quad 3 })^{ 2 }\quad -\quad { 4 }\quad =\quad 0

{ (x\quad +\quad 3 })^{ 2 }\quad =\quad 4

x\quad +\quad 3\quad =\quad \pm \sqrt { 4 }

This gives us two values of x:

x\quad +\quad 3\quad =\quad 2

x\quad =\quad -1       Ans

and

x\quad +\quad 3\quad =\quad -2

x\quad =\quad -5       Ans

 

Reference
  1. https://www.epcc.edu/CollegeReadiness/Documents/Quadratic_Review_20-50.pdf
Categories Algebra