# Connexions

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Collection by: Sunil Kumar Singh. E-mail the author

Module by: Sunil Kumar Singh. E-mail the author

We are already acquainted with quadratic equation and its roots. In this module, we shall study quadratic expression from the point of view of a function. It is a polynomial function of degree 2. The general form of quadratic expression/ function is :

f x = a x 2 + b x + c ; a , b , c R , a > 0 f x = a x 2 + b x + c ; a , b , c R , a > 0

a x 2 + b x + c = 0 ; a , b , c R , a > 0 a x 2 + b x + c = 0 ; a , b , c R , a > 0

Nature of a given quadratic function is best understood in terms of discriminant, D, of corresponding quadratic equation. This is given as :

D = b 2 4 a c D = b 2 4 a c

Quadratic equation is obtained by equating quadratic function to zero. Quadratic equation has at most two roots. The roots are given by :

α = - b D 2 a = - b b 2 4 a c 2 a α = - b D 2 a = - b b 2 4 a c 2 a

β = - b + D 2 a = - b + b 2 4 a c 2 a β = - b + D 2 a = - b + b 2 4 a c 2 a

#### Properties of roots of quadratic equation

1 : If D>0, then roots are real and distinct.

2 : If D=0, then roots are real and equal.

3 : If D<0, then roots are complex conjugates with non-zero imaginary part.

4 : If D>0; a,b,c∈T (rational numbers) and D is a perfect square, then roots are rational.

5 : If D>0; a,b,c∈T (rational numbers) and D is not a perfect square, then roots are radical conjugates.

6 : If D>0; a=1;b,c∈Z (integer numbers) and roots are rational, then roots are integers.

7 : If a quadratic equation has more than two roots, then the function is an identity in x and a=b=c=0.

8 : If a quadratic equation has one real root and a,b,c∈R, then other root is also real.

The real roots of the quadratic equation are zeroes of quadratic function. The zeroes of quadratic function are real values of x for which value of quadratic function becomes zero. On graph, zeros are the points at which graph intersects y=0 i.e. x-axis.

Graph reveals important characteristics of quadratic function. The graph of quadratic function is a parabola. Working with the quadratic function, we have :

y = a x 2 + b x + c = a x 2 + b a x + c a y = a x 2 + b x + c = a x 2 + b a x + c a

In order to complete square, we add and subtract b 2 / 4 a 2 b 2 / 4 a 2 as :

y = a x 2 + b a x + b 2 4 a 2 + c a b 2 4 a 2 y = a x 2 + b a x + b 2 4 a 2 + c a b 2 4 a 2

y = a { x + b 2 a 2 - b 2 4 a c 4 a } y = a { x + b 2 a 2 - b 2 4 a c 4 a }

y + b 2 4 a c 4 a = a x + b 2 a 2 y + b 2 4 a c 4 a = a x + b 2 a 2

y + D 4 a = a x + b 2 a 2 y + D 4 a = a x + b 2 a 2

Y = a X 2 Y = a X 2

Where,

X = x + b 2 a and Y = y + D 4 a X = x + b 2 a and Y = y + D 4 a

Clearly, Y = a X 2 Y = a X 2 is an equation of parabola having its vertex given by (-b/2a, -D/4a). When a>0, parabola opens up and when a<0, parabola opens down. Further, parabola is symmetric about x=-b/2a.

#### Maximum and minimum values of quadratic function

The graph of quadratic function extends on either sides of x-axis. Its domain, therefore, is R. On the other hand, value of function extends from vertex to either positive or negative infinity, depending on whether “a” is positive or negative.

When a > 0, the graph of quadratic function is parabola opening up. The minimum and maximum values of the function are given by :

y min = - D 4 a at x = - b 2 a y min = - D 4 a at x = - b 2 a

y max y max

Clearly, range of the function is [-D/4a, ∞).

When a < 0, the graph of quadratic function is parabola opening down. The maximum and minimum values of the function are given by :

y max = - D 4 a at x = - b 2 a y max = - D 4 a at x = - b 2 a

y min y min

Clearly, range of the function is (-∞, -D/4a].

##### Example 1

Problem : Determine range of f x = - 3 x 2 + 2 x 4 f x = - 3 x 2 + 2 x 4

Solution : The determinant of corresponding quadratic equation is :

D = b 2 4 a c = 4 4 X 3 X 4 = 4 48 = - 44 D < 0 D = b 2 4 a c = 4 4 X 3 X 4 = 4 48 = - 44 D < 0

a = - 3 a < 0 a = - 3 a < 0

The graph of function is parabola opening down. Its vertex represents the maximum function value. The maximum and minimum values of function are given by :

y max = - D 4 a = - - 44 4 X - 3 = - 44 12 = - 11 3 y max = - D 4 a = - - 44 4 X - 3 = - 44 12 = - 11 3

y min y min

Range = (-∞, -11/3)

The discriminant of corresponding quadratic equation and coefficient of term “ x 2 x 2 ” of quadratic function together determine nature of quadratic function and hence its graph. Graphs of quadratic function is intuitive and helpful to remember results. As a matter of fact, we can interpret all properties of quadratic function, if we can draw its graph.

### Case 1 : D<0

If D<0, then roots are complex conjugates. It means graph of function does not intersect x-axis. If a > 0, then parabola opens up. The value of quadratic function is positive for all values of x i.e.

D < 0, a > 0 f x > 0 for x R D < 0, a > 0 f x > 0 for x R

If a < 0, then parabola opens down. The value of quadratic function is negative for all values of x i.e.

D < 0, a < 0 f x < 0 for x R D < 0, a < 0 f x < 0 for x R

Sign rule : If D<0, then sign of function is same as that of “a” for all values of x in R.

### Case 2 : D=0

If D=0, then roots are equal and is given by –b/2a. It means graph of function just touches x-axis. If a > 0, then parabola opens up. The value of quadratic function is non-negative for all values of x i.e.

D = 0, a > 0 f x 0 for x R D = 0, a > 0 f x 0 for x R

If a < 0, then parabola opens down. The value of quadratic function is non-positive for all values of x i.e.

D = 0, a < 0 f x 0 for x R D = 0, a < 0 f x 0 for x R

Sign rule : If D=0, then sign of function is same as that of “a” for all values of x in R except at x=-b/2a, at which f(x)=0. We do not associate sign with zero.

### Case 3 : D>0

If D>0, then roots are unequal and are given by (–b±D)/2a. It means graph of function intersects x-axis at α and β (β>α). If a > 0, then parabola opens up. The value of quadratic function is positive for all values of x in the interval (-∞,α) U (β,∞).The values of quadratic function are zero for values of x ∈{α,β}. The value of quadratic function is negative for all values of x in the interval (α,β).

D > 0, a > 0 f x > 0 for x - , α β , Sign of function same as that of “a” D > 0, a > 0 f x > 0 for x - , α β , Sign of function same as that of “a”

D > 0, a > 0 f x = 0 for x { α , β } D > 0, a > 0 f x = 0 for x { α , β }

D > 0, a > 0 f x < 0 for x α , β Sign of function opposite to that of “a” D > 0, a > 0 f x < 0 for x α , β Sign of function opposite to that of “a”

If a < 0, then parabola opens down. The value of quadratic function is positive for all values of x in the interval (α,β).The values of quadratic function are zero for values of x ∈{α,β}. The value of quadratic function is negative for all values of x in the interval (-∞,α) U (β,∞).

D > 0, a < 0 f x < 0 for x α , β Sign of function same as that of “a” D > 0, a < 0 f x < 0 for x α , β Sign of function same as that of “a”

D > 0, a < 0 f x = 0 for x { α , β } D > 0, a < 0 f x = 0 for x { α , β }

D > 0, a < 0 f x > 0 for x - , α β , Sign of function opposite to that of “a” D > 0, a < 0 f x > 0 for x - , α β , Sign of function opposite to that of “a”

Sign rule : If D>0, then domain of function, which is R, is divided at root points in three intervals. The signs of function in side intervals are same as that of “a”, whereas sign of function in the middle interval is opposite to that of “a”.

## Examples

### Example 2

Problem : Determine interval of “a” for which graph of x 2 + a 1 x + 16 x 2 + a 1 x + 16 lie above x-axis.

Solution : Here coefficient of “ x 2 x 2 ” is positive. Now, the graph of quadratic function lie above x-axis when D>0 and a>0.

D = a 1 2 4 X 1 X 16 < 0 a 1 2 64 < 0 D = a 1 2 4 X 1 X 16 < 0 a 1 2 64 < 0

a 1 2 8 2 < 0 a 1 + 8 a 1 8 < 0 a 1 2 8 2 < 0 a 1 + 8 a 1 8 < 0

a + 7 a 9 < 0 a + 7 a 9 < 0

7 < a < 9 7 < a < 9

### Example 3

Problem : The graph of a quadratic expression f x = a x 2 + b x + c f x = a x 2 + b x + c is shown in the figure. Determine signs of a,b,c.

Solution : The parabola opens downward. It means a<0. Since both roots are positive, their sum is also positive.

α + β = b a > 0 Signs of a and b are opposite. α + β = b a > 0 Signs of a and b are opposite.

It means b>0. Now, putting x=0 in the function, we have value of function as :

f 0 = a X 0 2 + b X 0 + c = c f 0 = a X 0 2 + b X 0 + c = c From figure, graph can intersects y-axis only at negative y-value. Hence, “c” is negative.

Value of quadratic function is positive, zero or negative depending on the nature of coefficient of “ x 2 x 2 ” and discriminant, D. Accordingly, a quadratic function will hold true for particular interval(s) depending upon these parameters. Consider an example :

2 x 2 5 x 3 > 0 2 x 2 5 x 3 > 0

2 x 2 6 x + x 3 > 0 2 x 2 6 x + x 3 > 0

2 x x 3 + x 3 > 0 2 x x 3 + x 3 > 0

2 x + 1 x 3 > 0 2 x + 1 x 3 > 0

x = - 1 2 , 3 x = - 1 2 , 3

Here,

a = 2 a > 0 a = 2 a > 0

D = 25 4 X 2 X 3 = 1 D > 0 D = 25 4 X 2 X 3 = 1 D > 0

Under these conditions, given quadratic function is positive for - , - 1 / 2 3, - , - 1 / 2 3, . On the real number line, the intervals are shown as :

Let us consider slightly changed inequality involving less than equal sign,

2 x 2 5 x 3 0 2 x 2 5 x 3 0

Again quadratic function is positive for - , - 1 / 2 3, - , - 1 / 2 3, . However, equality is also allowed. It means valid intervals should also include root points. The modified valid interval corresponding to "less than equal to inequality" is : - , - 1 / 2 ] [ 3, - , - 1 / 2 ] [ 3, .

From this illustration, it is clear that we can determine valid interval(s) of x, provided we know the signs of quadratic function in different intervals. If equality is also allowed as in the case of “less than equal to” or “greater then equal to”, then we need to include root points also.

### Example 4

Problem : Determine the interval of x for which f x = 2 x 2 5 x 3 f x = 2 x 2 5 x 3 is non-positive and negative.

Solution : As already determined earlier, roots of function are -1/2, 3. Also a> 0 and D > 0. Sign rule for the function is shown here :

From the figure, it is clear that function is non-positive i.e. f(x) ≤0 in the interval [-1/2, 3]. Also, function is negative i.e. f(x)<0 in the interval (-1/2, 3).

### Example 5

Problem : A quadratic function is given by f x = x 2 4 x + 4 f x = x 2 4 x + 4 . Find solution for each of four inequalities viz f(x)<0, f(x) ≤ 0, f(x) > 0 and f(x) ≥ 0.

Solution : Here, coefficient of “ x 2 x 2 ” is 1. Thus,

a > 0 a > 0

Determinant of corresponding quadratic equation is :

D = - 4 2 4 X 1 X 4 = 0 D = - 4 2 4 X 1 X 4 = 0

For D=0 and a>0, f(x) ≥0. Further, since D=0, it means that corresponding quadratic equation has one real root. Now, the root is :

α = b 2 a = - - 4 2 X 1 = 2 α = b 2 a = - - 4 2 X 1 = 2

This means that f(x) is positive for all values of x except for x=2. At x=2, f(x)=0. It follows then that :

f x 0 ; x R f x 0 ; x R f x > 0 ; x R - { 2 } f x > 0 ; x R - { 2 } f x 0 ; x { 2 } f x 0 ; x { 2 } f x < 0 ; No solution f x < 0 ; No solution

### Exercise 1

A quadratic function is given by f x = - x 2 + 2 x 4 f x = - x 2 + 2 x 4 . Find solution for each of four inequalities viz f(x)<0, f(x) ≤ 0, f(x) > 0 and f(x) ≥ 0.

#### Solution

Here, coefficient of “ x 2 x 2 ” is -1. Thus,

a < 0 a < 0

Determinant of corresponding quadratic equation is :

D = 2 2 4 X 1 X 4 = - 8 D < 0 D = 2 2 4 X 1 X 4 = - 8 D < 0

For D<0 and a<0, f(x) < 0. Further, since D<0, it means that corresponding quadratic equation has no real root. This means that f(x) is negative for all values of x. It follows then that

f x < 0 ; x R f x < 0 ; x R f x 0 ; x R f x 0 ; x R f x > 0 ; No solution f x > 0 ; No solution f x 0 ; No solution f x 0 ; No solution

## Acknowledgment

Author wishes to thank Mr. Ralph David for helping to correct a mistake in the module.

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