# How do you solve 7+13x - 2x^2>0?

May 24, 2015

$y = - 2 {x}^{2} + 13 x + 7 > 0$. First, find the 2 real roots.

D = d^2 = 169 + 56 = 225 -> d = +- 15.
x = -13/-4 +- 15/-4
x = 13/4 +- 15/4
x1 = 28/4 = 7
x2 = -2/4 = -1/2

---------------|-1/2==|0==========|7-------------------

Test point method . Plot $\left(- \frac{1}{2}\right) \mathmr{and} \left(7\right)$on the number line. Use O as test point -> x = 0 -> f(x) = 7 > 0. OK. Then, O is located on the solution set, that is the open interval $\left(- \frac{1}{2} , 7\right)$
Algebraic method . Between the 2 real roots $\left(- \frac{1}{2}\right) \mathmr{and} \left(7\right)$, f(x) is positive as opposite in sign to a = -2. Then the solution set is the open interval $\left(- \frac{1}{2} , 7\right)$

NOTE. When a is negative (< 0), the parabola opens downward. Between the 2 real roots, a part of the parabola is above the x-axis, meaning, f(x) is positive (> 0) in this segment, as opposite in sign to a(< 0).
The algebraic method is very fast, because you don't have to draw the number line each time. Just look at the side of a, you would easily know where is the solution set.