Find the set of values of xx for which 5x2+12x>3x+25x^2 + 12x > 3x + 2.

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Worked solution

6(a)

1. Rearrange the inequality into standard quadratic form

5x2+12x−3x−2>05x^2 + 12x - 3x - 2 > 0
5x2+9x−2>05x^2 + 9x - 2 > 0

2. Factorise the quadratic expression

(5x−1)(x+2)>0(5x - 1)(x + 2) > 0

3. Determine the range of values for x

The roots of the equation (5x−1)(x+2)=0(5x - 1)(x + 2) = 0 are x=15x = \frac{1}{5} and x=−2x = -2.
Since the inequality is >0> 0 and the coefficient of x2x^2 is positive, the solution is x<−2x < -2 or x>15x > \frac{1}{5}.

Answer

x<−2x < -2 or x>15x > \frac{1}{5}

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