Answer
Verified
406.2k+ views
Hint: Use the fact that if for some integer x, $\sqrt{{{x}^{2}}+x+1}$ is rational, we have $\sqrt{{{x}^{2}}+x+1}$ is a non-negative integer. Hence let $m\in {{\mathbb{Z}}^{+}}$ be such that $\sqrt{{{x}^{2}}+x+1}=m$. Square both sides and hence form a quadratic equation in x. Use the fact that if the quadratic equation $a{{x}^{2}}+bx+c=0,a,b,c\in \mathbb{Q}$ has integral roots, then $D={{b}^{2}}-4ac$ is a perfect square. Simplify the resulting expression and use the fact that since 3 is prime, the only factors of 3 are 1 and 3. Hence find the number of integral values of x for which $\sqrt{{{x}^{2}}+x+1}$ is rational.
Complete step-by-step answer:
Let for some integer x, $\sqrt{{{x}^{2}}+x+1}$ is rational. Since ${{x}^{2}}+x+1$ is an integer, we have $\sqrt{{{x}^{2}}+x+1}$ is rational if and only if $\sqrt{{{x}^{2}}+x+1}$ is an integer. So let $m\in {{\mathbb{Z}}^{+}}$ be such that $m=\sqrt{{{x}^{2}}+x+1}$.
Squaring both sides of the equation, we get
${{m}^{2}}={{x}^{2}}+x+1$
Subtracting ${{m}^{2}}$ on both sides, we get
${{x}^{2}}+x+\left( 1-{{m}^{2}} \right)=0$, which is a quadratic equation in x.
We know that the roots of the quadratic equation $a{{x}^{2}}+bx+c=0,a,b,c\in \mathbb{Q}$ are given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
Since x is an integer, we have
${{b}^{2}}-4ac$ must be a perfect square.
Here $a=1,b=1$ and $c=1-{{m}^{2}}$
Hence, we have
$1-4\left( 1 \right)\left( 1-{{m}^{2}} \right)={{k}^{2}},k\in \mathbb{Z}$
Simplifying, we get
$4{{m}^{2}}={{k}^{2}}+3$
Subtracting ${{k}^{2}}$ from both sides, we get
$4{{m}^{2}}-{{k}^{2}}=3$
Using ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right),$ we get
$\left( 2m-k \right)\left( 2m+k \right)=3$
Since 3 is prime, the only factors of 3 are 1 and 3.
Hence, we have
$2m-k=3,2m+k=1$ or $2m-k=1,2m+k=3$or $2m-k=-1,2m+k=-3$ or $2m-k=-3,2m+k=-1$
Solving the first system:
We have
$\begin{align}
& 2m-k=3\text{ (i)} \\
& 2m+k=1\text{ (ii)} \\
\end{align}$
Adding equation (i) and equation (ii), we get
$4m=4\Rightarrow m=1$
Substituting the value of m in equation (i), we get
$2-k=3\Rightarrow k=-1$
Hence m = 1 and k = -1 is the solution of the system.
Solving the second system:
$\begin{align}
& 2m+k=3\text{ (iii)} \\
& 2m-k=1\text{ (iv)} \\
\end{align}$
Adding equation (iv) and equation (v), we get
$4m=4\Rightarrow m=1$
Substituting the value of m in equation (iv), we get
$2-k=1\Rightarrow k=1$
Similarly solving the third system yields m=-1 and k= -1 and solving the fourth system yields m = -1 and k = 1.
Since m is non-negative, we have
(m=-1,k=1) and (m=-1, k =1) are rejected.
Hence m = 1 is the only possible integral value.
When m = 1, we have
${{x}^{2}}+x+1=1\Rightarrow {{x}^{2}}+x=0\Rightarrow \left( x \right)\left( x+1 \right)=0$
Hence x = 0 or x = -1 are the only integral values of x for which $\sqrt{{{x}^{2}}+x+1}$ is a rational number.
Hence the number of integral values of x for which $\sqrt{{{x}^{2}}+x+1}$ is rational is two.
Hence option [c] is correct.
Note: Verification: In the above question it is important to check the values for which we suppose $\sqrt{{{x}^{2}}+x+1}$ is rational.
When x = 0, we have $\sqrt{{{x}^{2}}+x+1}=\sqrt{1}=1$
When x = -1, we have $\sqrt{{{x}^{2}}+x+1}=\sqrt{1-1+1}=1$
Hence at both x= 0 and x = -1 $\sqrt{{{x}^{2}}+x+1}$ is a rational number.
Hence our solution is correct
Complete step-by-step answer:
Let for some integer x, $\sqrt{{{x}^{2}}+x+1}$ is rational. Since ${{x}^{2}}+x+1$ is an integer, we have $\sqrt{{{x}^{2}}+x+1}$ is rational if and only if $\sqrt{{{x}^{2}}+x+1}$ is an integer. So let $m\in {{\mathbb{Z}}^{+}}$ be such that $m=\sqrt{{{x}^{2}}+x+1}$.
Squaring both sides of the equation, we get
${{m}^{2}}={{x}^{2}}+x+1$
Subtracting ${{m}^{2}}$ on both sides, we get
${{x}^{2}}+x+\left( 1-{{m}^{2}} \right)=0$, which is a quadratic equation in x.
We know that the roots of the quadratic equation $a{{x}^{2}}+bx+c=0,a,b,c\in \mathbb{Q}$ are given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
Since x is an integer, we have
${{b}^{2}}-4ac$ must be a perfect square.
Here $a=1,b=1$ and $c=1-{{m}^{2}}$
Hence, we have
$1-4\left( 1 \right)\left( 1-{{m}^{2}} \right)={{k}^{2}},k\in \mathbb{Z}$
Simplifying, we get
$4{{m}^{2}}={{k}^{2}}+3$
Subtracting ${{k}^{2}}$ from both sides, we get
$4{{m}^{2}}-{{k}^{2}}=3$
Using ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right),$ we get
$\left( 2m-k \right)\left( 2m+k \right)=3$
Since 3 is prime, the only factors of 3 are 1 and 3.
Hence, we have
$2m-k=3,2m+k=1$ or $2m-k=1,2m+k=3$or $2m-k=-1,2m+k=-3$ or $2m-k=-3,2m+k=-1$
Solving the first system:
We have
$\begin{align}
& 2m-k=3\text{ (i)} \\
& 2m+k=1\text{ (ii)} \\
\end{align}$
Adding equation (i) and equation (ii), we get
$4m=4\Rightarrow m=1$
Substituting the value of m in equation (i), we get
$2-k=3\Rightarrow k=-1$
Hence m = 1 and k = -1 is the solution of the system.
Solving the second system:
$\begin{align}
& 2m+k=3\text{ (iii)} \\
& 2m-k=1\text{ (iv)} \\
\end{align}$
Adding equation (iv) and equation (v), we get
$4m=4\Rightarrow m=1$
Substituting the value of m in equation (iv), we get
$2-k=1\Rightarrow k=1$
Similarly solving the third system yields m=-1 and k= -1 and solving the fourth system yields m = -1 and k = 1.
Since m is non-negative, we have
(m=-1,k=1) and (m=-1, k =1) are rejected.
Hence m = 1 is the only possible integral value.
When m = 1, we have
${{x}^{2}}+x+1=1\Rightarrow {{x}^{2}}+x=0\Rightarrow \left( x \right)\left( x+1 \right)=0$
Hence x = 0 or x = -1 are the only integral values of x for which $\sqrt{{{x}^{2}}+x+1}$ is a rational number.
Hence the number of integral values of x for which $\sqrt{{{x}^{2}}+x+1}$ is rational is two.
Hence option [c] is correct.
Note: Verification: In the above question it is important to check the values for which we suppose $\sqrt{{{x}^{2}}+x+1}$ is rational.
When x = 0, we have $\sqrt{{{x}^{2}}+x+1}=\sqrt{1}=1$
When x = -1, we have $\sqrt{{{x}^{2}}+x+1}=\sqrt{1-1+1}=1$
Hence at both x= 0 and x = -1 $\sqrt{{{x}^{2}}+x+1}$ is a rational number.
Hence our solution is correct
Recently Updated Pages
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Find the values of other five trigonometric functions class 10 maths CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Select the correct plural noun from the given singular class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
The sum of three consecutive multiples of 11 is 363 class 7 maths CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How many squares are there in a chess board A 1296 class 11 maths CBSE