Quadratic/Fraction Inequalities?

Question

Solve each of the following inequalities:
(a) (16-x^2)/(x^2-9)>=0
and
(b) (x^2-9)*(x^2-16)<=0

Answer 1

When solving quadratic and fraction inequalities it is important to remember to _never_ multiply or divide by a term whose sign is not known.

For (a), first factor [(4-x)(4+x)]/[(x-3)(x+3)]>=0. Notice that the terms are 0 at -4,-3,3, and 4. Now, draw a number line and place those numbers on it. Pick a number less than -4. Then (4-x) is positive, (4+x) is negative, (x-3) is negative, and (x+3) is negative. Since this is a product/quotient with an odd number of negatives, the end result is negative. That is

[(4-x)(4+x)]/[(x-3)(x+3)]<0 if x<-4

As we pass through the -4, the 4+x changes sign from negative to postive so

[(4-x)(4+x)]/[(x-3)(x+3)]>0 if -4
As we pass through the -3, the x+3 changes sign from negative to postive so

[(4-x)(4+x)]/[(x-3)(x+3)]<0 if -3
As we pass through the 3, the x-3 changes sign from negative to postive so

[(4-x)(4+x)]/[(x-3)(x+3)]>0 if 3
As we pass through the 4, the 4-x changes sign from positive to negative so

[(4-x)(4+x)]/[(x-3)(x+3)]<0 if x>4.

Remember, we can't divide by 0 so +/-3 are not in our solution sets. In this case, using interval notation, we have [-4,-3) union (3,4].

(b) is similar but in this case we don't have to worry about dividing by 0:

(x+3)(x-3)(x+4)(x-4) is 0 at -4,-3,3,4.

In the same way as before,

(x+3)(x-3)(x+4)(x-4)>0 if x<-4

(x+3)(x-3)(x+4)(x-4)<0 if -4
(x+3)(x-3)(x+4)(x-4)>0 if -3
(x+3)(x-3)(x+4)(x-4)<0 if 3
(x+3)(x-3)(x+4)(x-4)>0 if x>4

so in interval notation our answer is

(-infinity,-4] union [-3,3] union [4,infinity)

Challenge: Solve (x^2-2x+1)/(x+1) >=0

Answer 2

Fraction Inequalities

Answer 3

i can write the first one as-
( 4+x)(4-x)/(x+3)(x-3)>=0
=> (x+4)(x-4)/(x+3)(x-3)<=0
this is because, when an inequality is multiplied by -1, it changes direction.....
then, obtain 'critical points', by equating each to zero.....like if x+4=0, then x=-4......the critical points will be-
-4, -3, 3, 4,
then plot these on a number line, and start drawing a wave from the extreme right number (4 in this case) the ways goes from up to down the no. line at 4, from down to up at 3, from up to down at -3 and from down to up at -4. NOTE that the upper part of number line represents positive and lower represents negative.
then, as the expression, is <= 0, therefore, takes intervals of the wave which are below the number line,,,,, also as there is an equality sign in the expression, the critical points obtained from the numerator will be included in the answer, hence, the answer becomes,
x belongs to [-4,-3) union (3,4]

similarly, solve for b part......its easy if u get it......
NOTE: i have explained the whole procedure quite properly, so if u dont understand it the first time, read once or thrice, and i bet u'll get it.

Answer 4

a) (16- x^2)/(x^2-9)>=0

<=> (16-x^2)>=0
<=> 16>= x^2
<=> x^2 <=16
<=> x <= (+ or -)4

b)(x^2-9)*(x^2-16)<=0

<=> (x^2-9)*(x^2-16)<=0
<=> x^2-9<=0 or x^2-16<=0
<=> x^2<=9 or x^2 <=16
<=> x <= (+ or -)3 or x <= (+ or -)4

Answer 5

(16-x^2)/(x^2-9)>=0
(4 -x) (4 +x)/ (x -3) (x+3) >=0....................i
x cannot be equal to -4,-3,3 and 4 being critical points
CASE I
If x is greater than -4 and less than -3 say x= -3.5 then the equation i is satisfied
CASE II
If x is greater than -3 and less than 3 say x= 0 then the equation i is not satisfied
CASE III
If x is greater than 3 and less than 4 say x=3.5 then the equation i is satisfied
CASE IV
If x is greater than 4 say x= 5 then the equation i is satisfied
CASE V
If x is less than -4 say x= -5 then the equation i is satisfie
It means x < -4, x>5, -4 Similarly you can try the second question

Answer 6

(16- x^2)(x^2 - 9) >= 0
(4-x)(4+x)(x-3)(x+3) >= 0

when 4-x >=0 , 4+x>= 0 , x-3>=0 , x+3>= 0
-x>= -4 x>= -4 x > 3 x>= -3
x<= 4

using the number line....

- - + + +
--------------------------->
- - - + +
-------------->
- + + + +
----------------------------------->
+ + + + -
<--------------------------------------
-----------------------------------------
(-) -4 (+) -3 (-) 3 (+) 4 (-)

since they are asking for the values >=0, just consider the signs showing "+ve"

therefore the answer is {x | -4 <=x <= -3, 3<=x<=4 }

try it out for b. it's just the same. Good Luck !!

Answer 7

(a)
case 1:
16-x^2>=0
16>=x^2
x is between [-4,4](1)
x^2-9>0 (not equal)
x^2>9
x is between (-infinity,-3) and (3,infinity)(2)
(1)+(2) we have that x is between [-4,-3) and (3,4]

case 2:
16-x^2<=0
16<=x^2
x is between (-infinity,-4] and [4,infinity)(1)
x^2-9<0 (not equal)
x^2<9
x is between (-3,3) (2)
(1)+(2) x has no common values
So for the first equation the answer is: x is between [-4,-3) and (3,4]

So for the second equation the answer will be :x is between [-4,-3] and [3,4]

Answer 8

a.)
(16 - x^2)/(x^2 - 9) >= 0

-4 <= x < -3
or
3 < x <= 4

----------------------------------------

b.)
(x^2 - 9)(x^2 - 16) <= 0

-4 <= x <= -3
or
3 <= x <= 4

Answer 9

Do your own damn homework!

Answer 10

huh