How do I solve: $6(x^2+2)<17x$

Question

How do I solve this kinds of inequality. I can do it if all the 'x' is in one side. However, this one have x at both sides of the equation. And we don't know whether it's a positive or negative value.

Answer 1

Start by simplifying the left side.

$6(x^2+2) < 17x$

$6x^2+12 < 17x$

Then subtract the $17x$ from both sides.

$6x^2-17x+12<0$

Use the quadratic formula to find the $x$ values when the expression equals zero.

$\dfrac{17\pm \sqrt{17^2-4(6)(12)}}{2(6)}$

$\dfrac{17 \pm 1}{12}$

$\dfrac{4}{3},\dfrac{3}{2}$

Since the expression is $\textit{less}$ than zero, and the inequality is an upward opening parabola, the solution is between those two points.

$\dfrac{4}{3} < x < \dfrac{3}{2}$

Answer 2

We have $\displaystyle6x^2-17x+12<0$

Now, if $\displaystyle(x-a)(x-b)<0$ where $a,b$ are real such that $a<b,$

If $x-a<0\iff x<a,$ we need $x-b>0\iff x>b\implies a>x>b $ which is impossible as $a<b$

Or if $x-a>0\iff x>a,$ we need $x-b<0\iff x<b\implies a<x<b$

Answer 3

Hint: after expanding and simplifying, you should find that $$6x^2-17x+12<0 \iff (3x-4)(2x-3)<0 \iff \left(x-\frac{4}{3}\right)\left(x-\frac{3}{2}\right)<0.$$

Now use lab bhattacharjee's last line to solve this inequality.