Solve for x
x\in \left(-\infty,\frac{5}{3}\right)\cup \left(\frac{7}{4},\infty\right)
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12x^{2}-41x+35=0
To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.
x=\frac{-\left(-41\right)±\sqrt{\left(-41\right)^{2}-4\times 12\times 35}}{2\times 12}
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 12 for a, -41 for b, and 35 for c in the quadratic formula.
x=\frac{41±1}{24}
Do the calculations.
x=\frac{7}{4} x=\frac{5}{3}
Solve the equation x=\frac{41±1}{24} when ± is plus and when ± is minus.
12\left(x-\frac{7}{4}\right)\left(x-\frac{5}{3}\right)>0
Rewrite the inequality by using the obtained solutions.
x-\frac{7}{4}<0 x-\frac{5}{3}<0
For the product to be positive, x-\frac{7}{4} and x-\frac{5}{3} have to be both negative or both positive. Consider the case when x-\frac{7}{4} and x-\frac{5}{3} are both negative.
x<\frac{5}{3}
The solution satisfying both inequalities is x<\frac{5}{3}.
x-\frac{5}{3}>0 x-\frac{7}{4}>0
Consider the case when x-\frac{7}{4} and x-\frac{5}{3} are both positive.
x>\frac{7}{4}
The solution satisfying both inequalities is x>\frac{7}{4}.
x<\frac{5}{3}\text{; }x>\frac{7}{4}
The final solution is the union of the obtained solutions.
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