Question

If $x + \dfrac{1}{x} = - 1$, then what is the value of ${x^{247}} + \dfrac{1}{{{x^{187}}}}$?

Answer 1

We use the concepts of quadratic equations and expressions and the ways of solving those equations. We will also use some complex number theories and formulas. Complex numbers deal with imaginary numbers which have no certain value.
We use quadratic formula which is $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete answer:
It is given that, $x + \dfrac{1}{x} = - 1$
Now, multiply the whole equation by $x$.
So, it implies ${x^2} + 1 = - x$
Now, on doing some transpositions, we get the equation as, ${x^2} + x + 1 = 0$
Now, we got a quadratic equation in $x$. So, now let us solve this and find the value of $x$.
So, to solve the equation ${x^2} + x + 1 = 0$, we will use the quadratic formula.
Consider a quadratic equation $a{x^2} + bx + c = 0$. The values of $x$ for which the equation gets satisfied are called roots of the equation. Generally, a quadratic equation has two roots.
And the roots of the equation $a{x^2} + bx + c = 0$ are given by quadratic formula which is,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Here, the term ${b^2} - 4ac$ is called the determinant of the quadratic equation.
If the determinant is less than 0, then the equation will have imaginary roots.
If the determinant is equal to 0, then the equation will have real and equal roots.
If the determinant is greater than 0, then the equation will have real and unique roots.
Now let us solve ${x^2} + x + 1 = 0$
If we compare this equation with $a{x^2} + bx + c = 0$, we get, $a = 1;b = 1;c = 1$
So, determinant is ${1^2} - 4(1)(1) = - 3 < 0$
Here, the determinant is less than 0, so the equation has imaginary roots.
And roots are, $x = \dfrac{{ - 1 \pm \sqrt { - 3} }}{2}$
That means, the roots are $x = \dfrac{{ - 1 + i\sqrt 3 }}{2}$ and $x = \dfrac{{ - 1 - i\sqrt 3 }}{2}$
Consider ${x^3}$. Let us find ${x^3}$.
$\Rightarrow {x^3} = {\left( {\dfrac{{ - 1 - i\sqrt 3 }}{2}} \right)^3} = \dfrac{{{{\left( { - 1 - i\sqrt 3 } \right)}^3}}}{{{2^3}}}$
$\Rightarrow {x^3} = \dfrac{{{{( - 1)}^3} + 3{{( - 1)}^2}( - i\sqrt 3 ) + 3( - 1){{( - i\sqrt 3 )}^2} + {{( - i\sqrt 3 )}^3}}}{8}$
Now, on simplifying by using ${i^2} = - 1$, we get,
$\Rightarrow {x^3} = \dfrac{{ - 1 + 3( - i\sqrt 3 ) - 3( - 3) + i3\sqrt 3 }}{8}$
$\Rightarrow {x^3} = \dfrac{{ - i3\sqrt 3 + 8 + i3\sqrt 3 }}{8} = 1$
So, we finally got ${x^3} = 1$
So, now let us evaluate ${x^{247}} + \dfrac{1}{{{x^{187}}}}$
$\Rightarrow {x^{247}} + \dfrac{1}{{{x^{187}}}} = {x^{246}}.x + \dfrac{1}{{{x^{186}}.x}}$
$\Rightarrow {\left( {{x^3}} \right)^{82}}.x + \dfrac{1}{{{{\left( {{x^3}} \right)}^{62}}.x}}$
And we know the value of $x$ and ${x^3}$. So, let us substitute those values here.
$\Rightarrow {\left( 1 \right)^{82}}.\dfrac{{ - 1 - i\sqrt 3 }}{2} + \dfrac{1}{{{{\left( 1 \right)}^{62}}.\left( {\dfrac{{ - 1 - i\sqrt 3 }}{2}} \right)}}$
$\Rightarrow \dfrac{{ - 1 - i\sqrt 3 }}{2} + \dfrac{2}{{ - 1 - i\sqrt 3 }}$
Now, on rationalizing the denominator of second factor, we get
$\Rightarrow \dfrac{{ - 1 - i\sqrt 3 }}{2} + \dfrac{2}{{ - 1 - i\sqrt 3 }}.\dfrac{{ - 1 + i\sqrt 3 }}{{ - 1 + i\sqrt 3 }}$
$\Rightarrow \dfrac{{ - 1 - i\sqrt 3 }}{2} + \dfrac{{2( - 1 + i\sqrt 3 )}}{{{{( - 1)}^2} - {{(i\sqrt 3 )}^2}}}$ (As $(a + b)(a - b) = {a^2} - {b^2}$ )
So, on simplification, we get,
$\Rightarrow \dfrac{{ - 1 - i\sqrt 3 }}{2} + \dfrac{{ - 2 + i2\sqrt 3 }}{{1 - ( - 3)}}$
$\Rightarrow \dfrac{{ - 2 - 2i\sqrt 3 }}{4} + \dfrac{{ - 2 + i2\sqrt 3 }}{4}$
So, on adding up, we get,
$\Rightarrow \dfrac{{ - 2 - 2i\sqrt 3 - 2 + 2i\sqrt 3 }}{4} = \dfrac{{ - 4}}{4} = - 1$
So, we can conclude that, ${x^{247}} + \dfrac{1}{{{x^{187}}}} = - 1$
This is the required value.

Note: Make a note that, the value $i$ is equal to $\sqrt { - 1}$ i.e., $i = \sqrt { - 1}$. All the rational numbers and irrational numbers come under complex numbers. And also remember the fact that the roots of equation ${x^3} = 1$ are $x = 1,\dfrac{{ - 1 + i\sqrt 3 }}{2},\dfrac{{ - 1 - i\sqrt 3 }}{2}$
Here, we calculated the ${x^3}$ by taking $x = \dfrac{{ - 1 - i\sqrt 3 }}{2}$
But instead, we can also take $x = \dfrac{{ - 1 + i\sqrt 3 }}{2}$ and we will get the same value.