Question

Solve the following inequality : $\dfrac{{{x^2} - 6x + 9}}{{5 - 4x - {x^2}}} \geqslant 0$

Answer 1

We have to find the value of $x$ from the given expression of inequality $\dfrac{{{x^2} - 6x + 9}}{{5 - 4x - {x^2}}} \geqslant 0$ . We solve this question using the concept of solving linear equations of inequality . First we would simplify the terms of the numerator and the denominator in terms of the factors of the equation . Then we would solve the inequality obtained for the denominator to exist and for the condition of the inequality . On further solving the expression of the inequality we will get the range for the value of $x$ for which it satisfies the given expression .

Complete step-by-step solution:
Given :
$\dfrac{{{x^2} - 6x + 9}}{{5 - 4x - {x^2}}} \geqslant 0$
Splitting both the numerator and the denominator to form its factors , we can write the expression as :
$\dfrac{{{x^2} - 3x - 3x + 9}}{{5 - 5x + x - {x^2}}} \geqslant 0$
Taking terms common so as to form its factor , we can write the expression as :
$\dfrac{{\left( {x - 3} \right)x - 3\left( {x - 3} \right)}}{{5\left( {1 - x} \right) + x\left( {1 - x} \right)}} \geqslant 0$
$\dfrac{{\left( {x - 3} \right)\left( {x - 3} \right)}}{{\left( {5 + x} \right)\left( {1 - x} \right)}} \geqslant 0$
We can also write the expression as :
$\dfrac{{{{\left( {x - 3} \right)}^2}}}{{\left( {5 + x} \right)\left( {1 - x} \right)}} \geqslant 0$
Now , we will first solve the inequality of the numerator :
As , we know that the square of any number is always positive , so we can have any value of $x$ for the numerator to be positive .
Now , for the inequality to exist the denominator should exist and the value of the denominator should be positive . i.e. the value of the denominator should not be equal to zero .
So , according to the condition we can write the inequality of denominator as :
$\left( {5 + x} \right)\left( {1 - x} \right) > 0$
Thus , from here we get to points as :
$x = - 5$ and $x = 1$
Let us check the values of the denominator by putting various points as :
Put value of $x < \- 5$ , $x = - 6$ in the denominator , we get the value of denominator as :
$\left( {5 - 6} \right)\left( {1 + 6} \right) < 0$
The value obtained would be negative .
As , the value of the denominator should be positive so we can’t have a value of $x < \- 5$ .
Put value $x > 1$ , $x = 2$ in the denominator , we get the value of denominator as :
$\left( {5 + 2} \right)\left( {1 - 2} \right) < 0$
The value obtained would be negative .
As , the value of the denominator should be positive so we can’t have a value of $x > 1$ .
Put $x > - 5$ and $x < 1$ , $x = 0$ in the denominator , we get the value of denominator as :
$\left( {5 + 0} \right)\left( {1 - 0} \right) > 0$
The value obtained would be positive .
As , the value of the denominator should be positive and defined so we have values of $x > - 5$ and $x < 1$ .
Hence , the solution of $x$ from the inequality $\dfrac{{{x^2} - 6x + 9}}{{5 - 4x - {x^2}}} \geqslant 0$ is $\left( { - 5,1} \right)$ .

Note: We must take care about the sign and symbols of the inequality , as a slight change causes major errors in the solution . The solution of the range of the inequality states that each and every value which lies in that particular range satisfies the given equation . The round bracket $\left( {} \right)$ in the value of the range states that the end elements I.e. $- 5$ and $1$ in this question will not satisfy the given expression whereas the square bracket $\left[ {} \right]$ states that the end elements of the range will satisfy the given expression.