Given that $x$ is a real number satisfying $\frac{5x^{2}-26x+5}{3x^{2}-10x+3} < 0 ,$ then
- $x
- $\frac{1}{5} < x < 3$
- $x > 5$
- $\frac{1}{5} < x < \frac{1}{3}$ or $3 < x < 5$
The Correct Option is D
Solution and Explanation
$\Rightarrow \frac{5 x^{2}-25 x-x+5}{3 x^{2}-9 x-x+3}<0$
$\Rightarrow \frac{5 x(x-5)-1(x-5)}{3 x(x-3)-1(x-3)}<0$
$\Rightarrow \frac{(x-5)(5 x-1)}{(x-3)(3 x-1)}< 0$
$\therefore x \in\left(\frac{1}{5}, \frac{1}{3}\right) \cup(3,5)$
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Concepts Used:
Complex Numbers and Quadratic Equations
Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.
Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.
