Question

If $x$ is real, the maximum value of the expression $\dfrac{{\left( {{x^2} + 14x + 9} \right)}}{{\left( {{x^2} + 2x + 3} \right)}}$ will be?

Answer 1

Hint: In the question it is given that x is real, which means it takes the set of all real values, then we have to find out the maximum value of the given function. The standard representation of a quadratic equation is: $y\left( x \right) = a{x^2} + bx + c$ where $a,b,c$ are the coefficients of the quadratic equation. Since $x$ is real, the coefficients of the quadratic equation will also be real i.e. $a,b,c \in R$ .

Complete step by step answer:
 In the given rational expression, both the numerator and denominator are in the form of quadratic equations. So the first step is to convert it into a form of a single quadratic equation in terms of $x$ .
Let $y = \dfrac{{\left( {{x^2} + 14x + 9} \right)}}{{\left( {{x^2} + 2x + 3} \right)}}$
By cross multiplication, we get;
$ \Rightarrow \left( {{x^2} + 2x + 3} \right)y = {x^2} + 14x + 9$
On further simplification;
$ \Rightarrow {x^2}y + 2xy + 3y = {x^2} + 14x + 9$
Rearranging the equation according to coefficients, by taking all the terms to the L.H.S. ;
$ \Rightarrow {x^2}y - {x^2} + 2yx - 14x + 3y - 9 = 0$
Now , arrange the given equation to make it in the form of a standard quadratic equation in terms of $x$ ;
$ \Rightarrow {x^2}\left( {y - 1} \right) + 2x\left( {y - 7} \right) + \left( {3y - 9} \right) = 0$ $......\left( 1 \right)$
To find the maximum value of equation $\left( 1 \right)$ , the discriminant (D) of the equation must be;
$ \Rightarrow D \geqslant 0$
The formula for discriminant of a quadratic equation is given by;
$ \Rightarrow D = {b^2} - 4ac$
From equation $\left( 1 \right)$, $a = y - 1$ , $b = 2\left( {y - 7} \right)$ and $c = 3y - 9$
Putting the values of $a$ , $b$ , $c$ in the discriminant formula, we get;
$ \Rightarrow {\left[ {2\left( {y - 7} \right)} \right]^2} - 4\left( {y - 1} \right)\left( {3y - 9} \right) = 0$
Simplifying the above equation;
$ \Rightarrow 4{\left( {y - 7} \right)^2} - 4\left( {y - 1} \right)\left( {3y - 9} \right) = 0$
Taking 4 common and expand ${\left( {y - 7} \right)^2}$ by formula ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$
$ \Rightarrow {y^2} + 49 - 14y - 3{y^2} + 12y - 9 = 0$
Further simplifying the above equation ;
$ \Rightarrow - 2{y^2} - 2y + 40 = 0$
Dividing the above equation by $ - 2$ , we get;
$ \Rightarrow {y^2} + y - 20 = 0$ $......\left( 2 \right)$
Now solving equation $\left( 2 \right)$ to get the value of $y$;
$ \Rightarrow {y^2} + 5y - 4y - 20 = 0$
$ \Rightarrow \left( {y + 5} \right)\left( {y - 4} \right) = 0$
So, the values of $y$ are ;
$ \Rightarrow y = - 5$ and $y = 4$
Out of these two values, $y = 4$ is the maximum value of the given expression.
Therefore, the correct answer for this question is $y = 4$ .

Note: The concept of discriminant $\left( D \right)$ has been used to achieve the greatest value. The discriminant is a part of the quadratic equation formula $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ where $D = {b^2} - 4ac$ . The values of the determinant can be positive, negative or zero and these values are used to find out how many solutions are there for a given quadratic equation. Here $3$ cases are possible: $\left( 1 \right)$ If the $D$ value is positive then the quadratic equation has two distinct real number solutions. $\left( 2 \right)$ If the $D$ value is negative then neither of the solutions will be real, they will be imaginary numbers. $\left( 3 \right)$ If the $D$ value is zero , then the quadratic equation has repeated real number solutions.