Solution Set in interval notation for inequality

[datafiend]

HI all,
I have the equation, 6x^2-2>9x for which I'm to find the solution set in interval notation.

I've rewritten the inequalty as 6X^2-9x-2=0. I tried to factor, but no go. Then I used the quadratic and got 9+/- rad(129)/-18. The answers I get for x are -1.1309 and .1309. The calculator gives. -.1964 and 1.6965. Is this just some rounding error?

Thanks,

 

[Evgeny.Makarov]

I've rewritten the inequalty as 6X^2-9x-2=0. I tried to factor, but no go. Then I used the quadratic and got 9+/- rad(129)/-18.

It should be $\dfrac{9\pm\sqrt{129}}{12}$. Your calculator is correct for approximate values.

 

[MarkFL]

You did well to move the $9x$ to the left so that you have:

\(\displaystyle 6x^2-9x-2>0\)

Now, observing that expression is a quadratic, and that its graph will be parabolic, we see that it is an upward opening parabola since the coefficient of the squared term is positive. Thus, the expression must be negative in between the roots. Having the correct roots of the expression now, can you state the solution in interval notation?

 

[datafiend]

I think -9+/-\sqrt{129}/-12 to negative infinity/positive infinity. Basically, the area between the pos and negative is excluded. Yes?

 

[MarkFL]

I think -9+/-\sqrt{129}/-12 to negative infinity/positive infinity. Basically, the area between the pos and negative is excluded. Yes?

The region between and including the roots is excluded because we have a strict inequality. To express this using interval notation, we would write:

\(\displaystyle \left(-\infty,\frac{9-\sqrt{129}}{12}\right)\,\bigcup\,\left(\frac{9+\sqrt{129}}{12},\infty\right)\)