Solve the inequality: 17x + 5 >= -6x^2.  

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

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17x + 5 >= -6x^2

First we will move -6x^2 to the left side so the right side is 0.

==> 6x^2 + 17x + 5 >= 0

Now we have a quadratic function, we will use the quadratic formula to find the roots.

==> x1= ( -17 + sqrt(289-4*6*5) / 2*6

             = (-17 + 13) / 12 = -4/12 = -1/3

==> ( x+1/3 ) Or ( 3x+1) is a factor of the function.

==> x2= ( -17-13) /12 = -30/12 = -5/2

==> (x+5/2) OR (2x + 5) is a factor.

==> ( 2x+5) ( 3x+1) >=0

Then ( 2x+5) >= 0   and ( 3x+1) >=0

==> x >= -5/2  and   x>= -1/3

==> x = [ -1/3, inf)

OR:

 x =<-5/2   and  x=< =-1/3

==> x = (-inf, -5/2]

==> x = (-inf, -5/2] U [-1/3, inf)

 

Top Answer

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have the inequality 17x + 5 >= -6x^2

adding 6x^2 to both the sides

=> 6x^2 + 17x + 5 >= 0

=> 6x^2 + 15x + 2x + 5 >= 0

=> 3x (2x + 5) + 1(2x + 5) >= 0

=> (3x + 1) (2x + 5) >=0

We do not want the product of (3x + 1) and (2x + 5) to be negative. For this either both (3x + 1) and (2x + 5) are positive or both (3x + 1) and (2x + 5) are negative.

If both (3x + 1) and (2x + 5) are positive

=> 3x +1 >=0

=> x >= -1/3

and

2x + 5 >= 0

=> x >= -5/2

Both the conditions are satisfied with x >= -1/3

If both (3x + 1) and (2x + 5) are negative

(3x + 1) =< 0

=> x =< -1/3

2x + 5 =< 0

=> x =< -5/2

Both the conditions are satisfied by x < -5/2

Therefore we have x in the interval (- inf., -5/2] U [-1/3, + inf.)

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll subtract 17x + 5 both sides:

-6x^2 - 17x - 5 =< 0

We'll multiply by -1:

6x^2 + 17x + 5 > = 0

We'll find the roots of the expression:

6x^2 + 17x + 5  = 0

x1 = [-17 + sqrt(289 - 120)]/12

x1 = (-17 + 13)/12

x1 = -4/12

x1 = -1/3

x2 = -30/12

x2 = -15/6

The expression is positive over the intervals (-infinite ; -15/6) U (-1/3 ; +infinite).

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