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SOLVING 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) x 2 +5x-6 <0

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SOLVING QUADRATIC INEQUALITIES

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  1. 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

  2. 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

  3. 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

  4. 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

  5. 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

  6. 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

  7. 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

  8. 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

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