Question

Find all rational values of x at which $y=\sqrt{{{x}^{2}}+x+3}$ is a rational number.

Answer 1

Hint: We start solving this question by first assuming x and y as rational numbers. Then we consider y-x, which is also a rational number and assume it to be d. Then we take $y=x+d$ and square it on both sides and substitute the given value of y, $y=\sqrt{{{x}^{2}}+x+3}$. Then we solve it to obtain the value of x in terms of d. Then we substitute the value of x in the given value of x and solve it to find the value of y in terms of d. Then we find the domain of d and then we can find the rational values of x that satisfy $y=\sqrt{{{x}^{2}}+x+3}$.

Complete step by step answer:
We are given that $y=\sqrt{{{x}^{2}}+x+3}$ is rational and let us assume that x is rational.
As both x and y are rational, we can say that y-x is rational. Let us say that $y-x=d$. So, d is rational.
$\begin{align}
  & \Rightarrow y-x=d \\
 & \Rightarrow y=d+x \\
 & \Rightarrow \sqrt{{{x}^{2}}+x+3}=d+x \\
\end{align}$
Squaring on both sides we get,
\[\begin{align}
  & \Rightarrow {{x}^{2}}+x+3={{\left( d+x \right)}^{2}} \\
 & \Rightarrow {{x}^{2}}+x+3={{x}^{2}}+2dx+{{d}^{2}} \\
 & \Rightarrow x-2dx={{d}^{2}}-3 \\
 & \Rightarrow x\left( 1-2d \right)={{d}^{2}}-3 \\
 & \Rightarrow x=\dfrac{{{d}^{2}}-3}{1-2d} \\
\end{align}\]
Now let us substitute this value in $y=\sqrt{{{x}^{2}}+x+3}$. Then we get,
\[\begin{align}
  & \Rightarrow y=\sqrt{{{\left( \dfrac{{{d}^{2}}-3}{1-2d} \right)}^{2}}+\dfrac{{{d}^{2}}-3}{1-2d}+3} \\
 & \Rightarrow y=\sqrt{\dfrac{{{d}^{4}}-6{{d}^{2}}+9}{{{\left( 1-2d \right)}^{2}}}+\dfrac{\left( {{d}^{2}}-3 \right)\left( 1-2d \right)}{{{\left( 1-2d \right)}^{2}}}+\dfrac{3{{\left( 1-2d \right)}^{2}}}{{{\left( 1-2d \right)}^{2}}}} \\
 & \Rightarrow y=\sqrt{\dfrac{{{d}^{4}}-6{{d}^{2}}+9}{{{\left( 1-2d \right)}^{2}}}+\dfrac{{{d}^{2}}-3-2{{d}^{3}}+6d}{{{\left( 1-2d \right)}^{2}}}+\dfrac{3\left( 4{{d}^{2}}-4d+1 \right)}{{{\left( 1-2d \right)}^{2}}}} \\
 & \Rightarrow y=\sqrt{\dfrac{{{d}^{4}}-6{{d}^{2}}+9}{{{\left( 1-2d \right)}^{2}}}+\dfrac{{{d}^{2}}-3-2{{d}^{3}}+6d}{{{\left( 1-2d \right)}^{2}}}+\dfrac{12{{d}^{2}}-12d+3}{{{\left( 1-2d \right)}^{2}}}} \\
 & \Rightarrow y=\sqrt{\dfrac{{{d}^{4}}+{{d}^{2}}+9-2{{d}^{3}}-6d+6{{d}^{2}}}{{{\left( 1-2d \right)}^{2}}}} \\
\end{align}\]
Now, let us consider the formula
${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca$
Using this formula, we can change the value of y as
\[\begin{align}
  & \Rightarrow y=\sqrt{\dfrac{{{\left( {{d}^{2}} \right)}^{2}}+{{d}^{2}}+{{3}^{2}}-2\left( {{d}^{2}} \right)\left( d \right)-2\left( 3 \right)\left( d \right)+2\left( 3 \right)\left( {{d}^{2}} \right)}{{{\left( 1-2d \right)}^{2}}}} \\
 & \Rightarrow y=\sqrt{\dfrac{{{\left( {{d}^{2}}-d+3 \right)}^{2}}}{{{\left( 1-2d \right)}^{2}}}} \\
 & \Rightarrow y=\dfrac{{{d}^{2}}-d+3}{1-2d} \\
\end{align}\]
So, we get the values of x and y in terms of d as \[x=\dfrac{{{d}^{2}}-3}{1-2d}\] and \[y=\dfrac{{{d}^{2}}-d+3}{1-2d}\].
As denominator cannot be zero in a fraction, we have
$\begin{align}
  & \Rightarrow 1-2d\ne 0 \\
 & \Rightarrow 2d\ne 1 \\
 & \Rightarrow d\ne \dfrac{1}{2} \\
\end{align}$
So, we get x and y as rational for any rational value of d except when $d=\dfrac{1}{2}$.
So, y is rational for infinitely many rational values of x.

So, the correct answer is “Infinitely many values.”.

Note: The common mistake that one does while solving this type of problem is one might write the equation inside the square root in $y=\sqrt{{{x}^{2}}+x+3}$ as perfect square as,
$\begin{align}
  & \Rightarrow y=\sqrt{{{x}^{2}}+x+3} \\
 & \Rightarrow y=\sqrt{{{x}^{2}}+2\left( \dfrac{1}{2} \right)x+{{\left( \dfrac{1}{2} \right)}^{2}}-{{\left( \dfrac{1}{2} \right)}^{2}}+3} \\
 & \Rightarrow y=\sqrt{{{\left( x+\dfrac{1}{2} \right)}^{2}}+3-\dfrac{1}{4}} \\
 & \Rightarrow y=\sqrt{{{\left( x+\dfrac{1}{2} \right)}^{2}}+\dfrac{11}{4}} \\
\end{align}$
and think that as it is not a perfect square there are no such x. But here we are not discussing the roots of ${{x}^{2}}+x+3=0$. So, one needs to carefully understand the question.