Question

Solve the following inequality : $${\displaystyle \frac{x^2-6x+9}{5-4x-x^2}}⩾0$$

Answer 1

Hint: We have to find the value of $$x$$ from the given expression of inequality $${\displaystyle \frac{x^2-6x+9}{5-4x-x^2}}⩾0$$ . 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 :
$${\displaystyle \frac{x^2-6x+9}{5-4x-x^2}}⩾0$$
Splitting both the numerator and the denominator to form its factors , we can write the expression as :
$${\displaystyle \frac{x^2-3x-3x+9}{5-5x+x-x^2}}⩾0$$
Taking terms common so as to form its factor , we can write the expression as :
$${\displaystyle \frac{\left(x-3\right)x-3\left(x-3\right)}{5\left(1-x\right)+x\left(1-x\right)}}⩾0$$
$${\displaystyle \frac{\left(x-3\right)\left(x-3\right)}{\left(5+x\right)\left(1-x\right)}}⩾0$$
We can also write the expression as :
$${\displaystyle \frac{{\left(x-3\right)}^2}{\left(5+x\right)\left(1-x\right)}}⩾0$$
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 $${\displaystyle \frac{x^2-6x+9}{5-4x-x^2}}⩾0$$ 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.