# How do you solve the inequality 6x^2-5x>6?

Feb 4, 2015

I would start by solving it as a normal equation (2nd degree). So you write:
$6 {x}^{2} - 5 x - 6 > 0$ (I took the $6$ to the left);
Solving your equation you should get two values ${x}_{1} = \frac{3}{2}$ and ${x}_{2} = - \frac{2}{3}$.
Now the tricky bit:
your inequality asks you for values of $x$ that make your equation have a value bigger than zero.
They cannot be ${x}_{1} \mathmr{and} {x}_{2}$ because at these point your equation IS equal to zero.
Graphically your function gives the following parabola:

graph{6x^2-5x-6 [-11.96, 13.02, -7.14, 5.34]}

So, basically I have to choose values that are outside the boundaries formed by the two values ${x}_{1} \mathmr{and} {x}_{2}$ to get a value bigger than zero (in the graph the two bits that are ABOVE the $x$ axis).!!!
Consider ${x}_{1} = \frac{3}{2} = 1.5$ ok I cannot choose it but what about $2$ (which is bigger)?
If I put $x = 2$ in the equation I get: $6 \cdot 4 - 5 \cdot 2 - 6 = 8 > 0$ YES!
Consider now ${x}_{2} = - \frac{2}{3} = - 0.67$ again I cannot choose it but what about $- 1$?
If I put $x = - 1$ in the equation I get: $6 \cdot {\left(- 1\right)}^{2} - 5 \cdot \left(- 1\right) - 6 = 5 > 0$ YES!
So, OUTSIDE the interval bound by ${x}_{1} \mathmr{and} {x}_{2}$ you can choose $x$.
You express this by writing your solution as:
$- \frac{2}{3} > x > \frac{3}{2}$
Or graphically Hope it helps