Modulus of Quadratic Expression

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We Know that,
| x | of a real number x is the non-negative value of x without regard to its sign.

 means,
if, y = |x|,
|x| \ge 0 
or we can say that , 
y \ge 0

 means no part of the graph of |x| will be below the x-axis.

 Actually, this is true for any f(x),
if y = |f(x)| ,

 means no part of curve will below the x - axis, since y \ge 0

 for, y = x

 means if ,\begin{array}{l} x = 1 \Rightarrow y = 1\\ x = - 2 \Rightarrow y = - 2\\ x = - 1 \Rightarrow y = - 1 \end{array}

 and for , y = |x|

 \begin{array}{l} x = 1 \Rightarrow y = 1\\ x = - 2 \Rightarrow y = 2\\ x = - 1 \Rightarrow y = 1 \end{array}

in |x| graph we can see from the figure that no part of curve will be on negative side of y- axis.

 In fact we take the mirror image or reflection of negative side of y - axis in respect of  x - axis and leave the part of negative side of y -axis.
Similarly if we have any y = |f(x)| , 
we will take the mirror image of part of negative side of y - axis about the x -axis and leave the part of negative side of y - axis or left the part of curve which is below the x -axis after taking the reflection,

 or , we can say that the part of curve which is below to the x - axis, we multiply with it negative sign to make the part positive......

 consider, y = |x|
 for x \le 0, part of  y = x is negative 

 therefore to make it positive we have to multiply it with -1

 therefore,
y = |x| = x, if x \ge 0

 and y = |x| = - x, if x \le 0
FROM ABOVE ARGUMENTS,

 if we have,

 y = |a{x^2} + bx + c|
i.e. f(x) is a quadratic expression,

 1st Condition,

 if, a > 0 and D < 0
concave upward and no real roots (imaginary roots)
( previous post http://iitmathseasy.blogspot.in/2013/03/quadratic-expression-concept-ii.html)

 therefore,
f(x) > 0 or, (a{x^2} + bx + c) > 0

 since a{x^2} + bx + c is always greater than '0'
 and mod of any positive number is that number,

 \Rightarrow  y = |a{x^2} + bx + c| = a{x^2} + bx + c


e.g. y = |2{x^2} + 5x + 7|

 since a > 0 and D = 25 - 56 = 31 < 0

 \Rightarrow y = 2{x^2} + 5x + 7 
2nd Condition,

 y = |a{x^2} + bx + c|

 if, a < 0 and D < 0

 curve is concave downwards and no real roots
( previous post http://iitmathseasy.blogspot.in/2013/03/quadratic-expression-concept-ii.html)

 \Rightarrow f(x) < 0 or, (a{x^2} + bx + c) < 0

 Since a{x^2} + bx + c is always less than zero that is negative.
So, we have to make it positive by multiplying with -1.

 \Rightarrow y = |a{x^2} + bx + c| = - (a{x^2} + bx + c)

3rd Condition,

 y = |a{x^2} + bx + c|

 a > 0 and D > 0


 curve is concave upwards and have real roots means it will cut x-axis on two points (suppose {x_1} and {x_2})

here use the method of solving modulus equation ( http://iitmathseasy.blogspot.in/2013/03/modulus-equality-solving-concept.html) step by step

 starting from left,

 i) x \le {x_1},
from figure,

 y = |a{x^2} + bx + c| = a{x^2} + bx + c

 because for x \le {x_1}, curve is positive

 ii){x_1} \le x \le {x_2}
 from figure,

 \Rightarrow y = |a{x^2} + bx + c| = - (a{x^2} + bx + c)

 in this interval curve is negative, so we have to multiply by -1.

 iii) x \ge {x_2}

 from figure,

 y = |a{x^2} + bx + c| = a{x^2} + bx + c

 curve is positive.
4th condition,

 y = |a{x^2} + bx + c|

a < 0,D > 0
 concave downwards two real roots,

 from the above arguments clear from figure.


i) x \le {x_1}


\Rightarrow y = |a{x^2} + bx + c| = - (a{x^2} + bx + c)

ii){x_1} \le x \le {x_2}


y = |a{x^2} + bx + c| = a{x^2} + bx + c

iii)x \ge {x_2}


\Rightarrow y = |a{x^2} + bx + c| = - (a{x^2} + bx + c)







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