Question

The value of parameters “a” for which $\dfrac{a{{x}^{2}}+3x-4}{a+3x-4{{x}^{2}}}$ takes all real values for $x\in R$, then a\in \\_\\_\\_?

Answer 1

This question is from the topic of quadratic equations. We will first assume that the given term is equal to y. After that, we will equate and solve that equation. From there, we will get the equation in quadratic form of x. From there, we will find the discriminant and equate to greater than or equal to zero. From there, we will get the quadratic equation in terms of y. Then, we will solve and find the value of a.

Complete step-by-step solution:
Let us solve this question.
In this question, we have given an expression as $\dfrac{a{{x}^{2}}+3x-4}{a+3x-4{{x}^{2}}}$ and say that it takes all real values for all real values of x.
So, firstly let us assume
$y=\dfrac{a{{x}^{2}}+3x-4}{a+3x-4{{x}^{2}}}$
The above can also be written as
$\Rightarrow y\left( a+3x-4{{x}^{2}} \right)=a{{x}^{2}}+3x-4$
The above equation can also be written as
$\Rightarrow ay+3xy-4{{x}^{2}}y=a{{x}^{2}}+3x-4$
The above equation can also be written as
$\Rightarrow ay+3xy-4{{x}^{2}}y-a{{x}^{2}}-3x+4=0$
After taking common as ${{x}^{2}}$ and $x$ in the above equation, we can write
$\Rightarrow -{{x}^{2}}\left( 4y+a \right)-x\left( 3-3y \right)+4+ay=0$
Now, multiplying -1 to the both side of the equation, we get
$\Rightarrow {{x}^{2}}\left( 4y+a \right)+x\left( 3-3y \right)-4-ay=0$
As it is given that the value of expression takes only for real value of x. So, the discriminant of the above quadratic equation will be greater than equal to zero.
Hence, we can write
$D\ge 0$
From the equation ${{x}^{2}}\left( 4y+a \right)+x\left( 3-3y \right)-4-ay=0$. So, we can write
$\Rightarrow {{\left( 3-3y \right)}^{2}}-4\left( 4y+a \right)\left( -4-ay \right)\ge 0$
We can write the above equation as
$\Rightarrow 9+9{{y}^{2}}-18y-4\left( -16y-4a{{y}^{2}}-4a-{{a}^{2}}y \right)\ge 0$
The above equation can also be written as
$\Rightarrow 9+9{{y}^{2}}-18y+64y+16a{{y}^{2}}+16a+4{{a}^{2}}y\ge 0$
We can write the above equation as
$\Rightarrow {{y}^{2}}\left( 9+16a \right)+y\left( -18+64+4{{a}^{2}} \right)+9+16a\ge 0$
As it is given that the expression $\dfrac{a{{x}^{2}}+3x-4}{a+3x-4{{x}^{2}}}$ which we have taken as y takes all real values, so for taking all real values of y, we can say that the discriminant of the above equation will be greater than zero.
And, also 9+16a will not be zero. We can say from here that a cannot be $-\dfrac{9}{16}$.
The discriminant will be
$D={{\left( -18+64+4{{a}^{2}} \right)}^{2}}-4\left( 9+16a \right)\left( 9+16a \right)$
The term y has taken all real values. So, we can write
$D={{\left( -18+64+4{{a}^{2}} \right)}^{2}}-4\left( 9+16a \right)\left( 9+16a \right)\ge 0$
The above equation can also be written as
$D={{\left( 4{{a}^{2}}+46 \right)}^{2}}-{{2}^{2}}{{\left( 9+16a \right)}^{2}}\ge 0$
The above can also be written as
$\Rightarrow {{\left( 4{{a}^{2}}+46 \right)}^{2}}-{{\left( 18+32a \right)}^{2}}\ge 0$
Using the formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$, we can write
$\Rightarrow \left( 4{{a}^{2}}+46-18-32a \right)\left( 4{{a}^{2}}+46+18+32a \right)\ge 0$
The above can also be written as
$\Rightarrow \left( 4{{a}^{2}}-32a+28 \right)\left( 4{{a}^{2}}+32a+64 \right)\ge 0$
We can write the above equation as
$\Rightarrow 4\left( {{a}^{2}}-8a+7 \right)4\left( {{a}^{2}}+8a+16 \right)\ge 0$
We can write the above equation as
$\Rightarrow \left( {{a}^{2}}-8a+7 \right)\left( {{a}^{2}}+8a+16 \right)\ge 0$
We can write the above equation as
$\Rightarrow \left( {{a}^{2}}-a-7a+7 \right)\left( {{a}^{2}}+2.4.a+{{4}^{2}} \right)\ge 0$
$\Rightarrow \left( a-1 \right)\left( a-7 \right){{\left( a+4 \right)}^{2}}\ge 0$
As a+4 will always be positive because it is in the square form. Let's understand the range of ‘a’ from the following figure.

From here, we can say that the range is
$a\in \text{(-}\infty \text{,1 }\\!\\!]\\!\\!\text{ }\cup [\text{7,+}\infty \text{)- }\\!\\!\\{\\!\\!\text{ }\dfrac{-9}{16}\text{ }\\!\\!\\}\\!\\!\text{ }$
As we have seen above that the value of ‘a’ cannot be taken as $\dfrac{-9}{16}$.

Note: We should have better knowledge in the topic of quadratic equations to solve this type of question easily. Always remember that the discriminant of the general equation $a{{y}^{2}}+by+c=0$ is $D={{b}^{2}}-4ac$. Always remember that whenever we have the equation in the form of $a{{y}^{2}}+by+c$, and it is given that it takes all real values of y. So, for taking all real values of y, ‘a’ should not be equal to zero and discriminant should be greater than equal to zero.