Question

Given that, x is a real number satisfying $\dfrac{{5{x^2} - 26x + 5}}{{3{x^2} - 10x + 3}} < 0$, then
$\left( a \right)x < \dfrac{1}{5}$
$\left( b \right)\dfrac{1}{5} < x < 3$
$\left( c \right)x > 5$
$\left( d \right)\dfrac{1}{5} < x < \dfrac{1}{3}{\text{ or }}3 < x < 5$

Answer 1

Hint: In this particular question use the concept that in order to get the solution less than zero, one of the factors from both of the equations is less than zero so that the complete solution is less than zero, so use this concept to reach the solution of the question.

Complete step-by-step solution:
Given expression
$\dfrac{{5{x^2} - 26x + 5}}{{3{x^2} - 10x + 3}} < 0$
As we see that in numerator and denominator the highest power of x is 2 so both are the quadratic equation.
So factorize both of the equation we have,
$ \Rightarrow \dfrac{{5{x^2} - 25x - x + 5}}{{3{x^2} - 9x - x + 3}} < 0$
Now take 5x common from first two terms of the numerator and 3x common from first two terms of the denominator we have,
$ \Rightarrow \dfrac{{5x\left( {x - 5} \right) - 1\left( {x - 5} \right)}}{{3x\left( {x - 3} \right) - 1\left( {x - 3} \right)}} < 0$
$ \Rightarrow \dfrac{{\left( {x - 5} \right)\left( {5x - 1} \right)}}{{\left( {x - 3} \right)\left( {3x - 1} \right)}} < 0$
Now as we see that the above function is less than zero, so in order to get the solution less than zero i.e. negative the required solution for x is
$ \Rightarrow x \in \left( {\dfrac{1}{5},\dfrac{1}{3}} \right){\text{ or }}x \in \left( {3,5} \right)$

$ \Rightarrow \dfrac{1}{5} < x < \dfrac{1}{3}{\text{ or }}3 < x < 5$
So this is the required solution
Hence option (d) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall that the quadratic equation is solved using factorization method or we can use the quadratic formula for complex quadratic equation which is given as $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ where, a is the coefficient of ${x^2}$, b is the coefficient of x and c is the constant term.