# Solve the inequality : 14x^2 < 13x - 3 .

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

We have to solve the inequality 14x^2 < 13x - 3

Now 14x^2 < 13x - 3

subtract 13x - 3 from both the sides

=> 14x^2 - 13x + 3 < 0

=> 14x^2 - 7x - 6x + 3 < 0

=> 7x( 2x - 1) -3 (2x -1) < 0

=> (7x - 3) ( 2x -1) < 0

Now for the product of the factors to be less than 0 only one of them should be negative and the other positive.

Therefore first taking:

(7x - 3) <0  and ( 2x -1) >0

=> x < 3/7 and x > 1/2

But x cannot be less than 3/7 and greater than 1/2 at the same time, so we can't find valid values here.

Let's take (7x - 3) > 0  and ( 2x -1) <0

=> x > 3/7 and x < 1/2

This gives valid values in the range (3/7 , 1/2)

Therefore the set in which the values of x lie is (3/7 , 1/2).

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll solve the inequality in the following way. First, we'll move all terms to the left side:

14x^2 - 13x+3 < 0

We'll find the roots of the equation and we'll establish the rule: between the roots, the expression has the opposite sign to the sign of the coefficient of x^2. The expression will have the same sign with the sign of the coefficient of x^2, outside the roots.

Now, we'll calculate the roots of the quadratic:

14x^2 - 13x+3 = 0

x1 = [13+sqrt(169-168)]/2*14

x1 = 14/28

x1 = 1/2

x2 = 12/48

x2 = 1/4

Since the coefficient of x^2 is positive, the expression will be negative for the values found between the roots x1 and x2.

14x^2 - 13x+3 <0 for x belongs to the intervals:(1/4  , 1/2).