Solve for all real $x$: $(3x^2-8x+5)/(4x^2-3x+7) > 0$

Question

I know how to solve quadratic inequalities but I can't figure out how to solve $$\frac{3x^2-8x+5}{4x^2-3x+7} > 0$$ for all real $x$. Help would be appreciated

Answer 1

$4x²-3x+7>0$ since the discriminant is negative and if we evaluate it at 0 for example the result is 7 which is positive. Therefore your inequality is equivalent to $3x^2-8x+5>0$ which is easy to solve. Note that $$3x^2-8x+5=(x-1)(3x-5)>0\iff x\in(-\infty,1)\cup(\frac53,\infty)$$

Answer 2

Solve the pair of systems $$\begin{cases}3x^2-8x+5>0\\ 4x^2-3x+7>0\end{cases}\lor \begin{cases}3x^2-8x+5<0\\ 4x^2-3x+7<0\end{cases}$$ with your favourite method.

Answer 3

Hint: Consider two cases: $$3x^2-8x+5\geq 0$$ and $$4x^2-3x+7>0$$ or $$3x^2-8x+5\le 0$$ and $$4x^2-3x+7<0$$ The result is given by $$x<1$$ or $$x>\frac{5}{3}$$

Answer 4

Note the denominator has constant sign (for real $x$) since its discriminant is negative and therefore has no real zeroes; by inspection, it is always positive.

So you only need to find where the numerator is positive.