# 10th Class Mathematics Quadratic Inequation

Category : 10th Class

FUNDAMENTALS

Consider the quadratic equation$a{{x}^{2}}+bx+c=0.\,\,\,(a\ne 0)$where a, b, and c are real numbers.

The quadratic in equations related to $a{{x}^{2}}+bx+c=0$are $a{{x}^{2}}+bx+c<0$and $a{{x}^{2}}+bx+c>0$.

Assume that a > 0 (This assumption is always valid because if a<0, we can always multiply the in equation by (– 1) to get a > 0.)

For example, $-2{{x}^{2}}+3x+2<0$can be written as $-2{{x}^{2}}-3x-2>0$

Note: The change in the sign of the inequality, when it is multiplied by (– 1).

Then following cases arise:

Case – 1:    If${{b}^{2}}-4ac>0$, then the equation $a{{x}^{2}}+bx+c=0$has real and unequal roots. Let $\alpha$and $\beta (\alpha <\beta )$be the roots. Then,

$\therefore a{{x}^{2}}+bx+c=a(x-\alpha )(x-\beta )$

1. If$x<\alpha$, then $\left( x-\alpha \right)<0$ and $(x-\beta )<0$

$\therefore a{{x}^{2}}+bx+c>0$

1. If a$\alpha <x<\beta$, then $\left( x-\alpha \right)>0$ and $(x-\beta \text{)}<0$

$\therefore a{{x}^{2}}+bx+c<0$

1. If$x>\beta$, then$x-\alpha >0$and$x-\beta >0$.

$\therefore a{{x}^{2}}+bx+c>0$

Case – 2: If${{b}^{2}}-4ac=0$, then $a{{x}^{2}}+bx+c=0$has real and equal roots.

Let $\alpha$ be the equal roots.$\Rightarrow a{{x}^{2}}+bx+c=a(x-\alpha ')(x-\alpha ')$

If$x<\alpha '$. Then, $(x-\alpha ')<0$.    $\therefore a{{x}^{2}}+bx+c>0$.

However, if$x>\alpha '$, then,$\left( x-\alpha ' \right)>0$.

$\therefore a{{x}^{2}}+bx+c>0$.

Case – 3: If${{b}^{2}}-4ac<0$, then $a{{x}^{2}}+bx+c=0$has no real roots.

In this case, $a{{x}^{2}}+bx+c>0,\forall x\in R$.

The above three cases (case 1 to case 3) may be summarized as:

1. If$\alpha <x<\beta$, then $\left( x-\alpha \right)\left( x-\beta \right)<0$ and vice - versa.
2. If$~x<\alpha$ and$x>\beta \left( \alpha <\beta \right)$, then $\left( x-\alpha \right)\left( x-\beta \right)>0$ and vice - versa.

Sample Question: Solve the in equation${{x}^{2}}-7x+12<0$.

Solution: Given in equation is ${{x}^{2}}-7x+12<0\Rightarrow {{x}^{2}}-3x-4x+12<0$

$\Rightarrow \left( x-3 \right)\left( x-4 \right)<0;\Rightarrow \left( x-3 \right)<0$ and $\left( x-4 \right)>0$……………..(1)

Or.                                                       $\left( x-3 \right)>0$ and $\left( x-4 \right)<0$…………...(2)

In set notation, ‘and’ means intersection whereas ‘or’ means UNION.

From (1), x < 3 and x > 4

$\therefore$ There is not overlap and this is a NULL set $\phi$………. (1)

From $\left( 2 \right),x>3$ and $x<4\Rightarrow x\in (3,\infty )\cap x\in (-\infty ,4)\Rightarrow x\in (3,4)$

Clearly, the intersection set is (3, 4)

Finally, we have union of NULL set $\phi$ in equation (1) and (3, 4) in equation (2) $\Rightarrow$

Solution $=\phi \cup (3,4)=(3,4)$

Elementary Questions -1

Solve for $x:{{x}^{2}}-3x+2\ge 0$

Solution:

Given in equation is ${{x}^{2}}-3x+2\ge 0$.

$\Rightarrow (x-1)(x-2)\ge 0\Rightarrow x-1\ge 0$ and $x-2\ge 0$ or $x-1\le 0$and $x-2\le 0$

$\Rightarrow x\ge 1$and $x\ge 2\Rightarrow x\in [1,\infty ]\cap [2,\infty )\Rightarrow x\in [2,\infty )$…………..(1)

Or,

$x\le 1$and $x\le 2\Rightarrow x\in [-\infty ,1)\cap [-\infty ,2)\Rightarrow x\in (-\infty ,1]$.............(2)

(1) & (2) are combined as $x\in (-\infty ,1]\cup [2,\infty )$

Hence, the solution for the given in equation is $x\in (-\infty ,1]\cup [2,\infty )$.

This can be seen on number line as follows:

From $(1);x\in [2,\infty )$

OR,

From $(2);x\in (-\,\infty ,1]$

OR, means UNION$\Rightarrow x\in (-\infty ,1]\cup [2,\infty )$.

Mod Function

$y=f(x)=|x|$is called mod x or mod of x;

Mod of x is defined as, |x| = x when$x\ge 0$; And |x| = – x when$x\le 0$

Thus, mod of x is a always + ve value of x.

Elementary question

Find range of $y=|x+2|$when $x\in (-5,-1)$

Answer: $x>-5\therefore y=x+2>-3$ $(when\,\,|x+2|=x+2)$

Also $x<-1\Rightarrow x+2<+1;$           $\therefore y=x+2\in (-3,+1)$

Other case: When $y=|x+2|=-x-2;\,\,x>-5\Rightarrow -x<+5\Rightarrow -x-2<+3$

$x<-1\Rightarrow -x>+1\Rightarrow -x-2>-1;$   Thus $y=-x-2\in (-1,3)$

or,  $x+2\in (-3,0]\cup [0,1);$  $\therefore |x+2|\in (0,+3)\cup [0,1]=[0,3)$

Graph of |x|

Elementary question: Draw graph of |2x – 5|