1 / 8

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

abra-weaver
Télécharger la présentation

SOLVING QUADRATIC INEQUALITIES

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related