Question

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

Answer 1

Hint: 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.