Download lesson on quadratic inequalities

SOLVING QUADRATIC
INEQUALITIES
RULE 1:
If (x-a) (x-b) < 0 and a<b then
a<x<b
Rule 2:
If (x-a)(x-b) >0 and a<b then
x<a or x>b
Exercise 3.3
Find the range of values for each of the
following
1a) x2+5x-6 <0
Solution: The given inequality can be written as
(x+6)(x-1)< 0
-6<x<1
1f) (x+2)(x+3)  x+6
Solution: The given inequality can
be written as
2
x +3x+2x+6  x+6
x2+4x
0
x(x+4)  0
Solutions are
-4 x 0
1g) (x-1)(5x +4) > 2(x-1)
The given inequality can be written as
5x2 +4x -5x -4 > 2x -2
5x2 -3x -2 > 0
5x 2
2x
(5x+2)(x-1) >0
x - 1 -5x
5(x+2/5)(x-1)>0
___________
5x2
Solutions are
x< -2/5 or x >1
-2
-3x
4. There is no real value of x for which mx2+8x
+m =6. Find ‘m’
Since the equation has no real roots , the discriminant
D < 0 Rewrite the given equation as follows
mx2 +8x +m-6 =0
a=m
b=8
c = m- 6
D = b2 – 4ac
= (8) 2 – 4 (m) m-6)
= 64-4m2 +24m
= -4(m2-6m -16) < 0
ie m2-6m -16 >0
ie (m-8)(m+2) > 0
ie m<-2 or m>8
13. The curve y= (k-6)x2-8x +k cuts x-axis at two points and has a
minimum point. Find the range of values of ‘k’.
Solution .
(i)Since the curve has minimum value ,
k-6 > 0
i.e
k>6---------(1)
(ii) Since the curve cuts the x- axis at two points , its
D>0
a = k-6
b = -8
c=k
D = b2 – 4ac
= (-8) 2 – 4 (k-6)(k)
= 64-4k2 +24k
= -4(k2-6k -16)>0
ie
k2-6k -16 <0
ie
(k+2)(k-8) < 0
-2<k<8 -----------(2)
Eqns (1) and (2)  6 < k <8