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

x2+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 -2 -3x

Solutions are

x< -2/5 or x >1

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