# Quadratic Equations

Contents

## 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