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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).
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
Since it is a quadratic, we'll apply the quadratic formula.
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).
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