Question

If ${x^2} - 5x + 1 = 0$, then what is the value of ${x^5} + \dfrac{1}{{{x^5}}}$?
A) $3125$
B) $2625$
C) $3025$
D) $2525$

Answer 1

Hint: In this question, we are given a quadratic equation whose roots are not real. Hence, we cannot find the value of x and solve directly. We have been asked the value of ${x^5} + \dfrac{1}{{{x^5}}}$. At first, we will find the value of $x + \dfrac{1}{x}$ by shifting the given equation. Then we will square this equation. It will give us the value of ${x^2} + \dfrac{1}{{{x^2}}}$. Next, we will multiply both the equations to get the value of ${x^3} + \dfrac{1}{{{x^3}}}$. Then we will again square the equation of ${x^2} + \dfrac{1}{{{x^2}}}$. This will give us an equation in ${x^4} + \dfrac{1}{{{x^4}}}$. Multiply this equation with the equation in $x + \dfrac{1}{x}$. It will give us the equation in ${x^5} + \dfrac{1}{{{x^5}}}$ and ${x^3} + \dfrac{1}{{{x^3}}}$. Put the value of ${x^3} + \dfrac{1}{{{x^3}}}$ as already found above. This will give us the value of ${x^5} + \dfrac{1}{{{x^5}}}$

Complete step-by-step answer:
We are given ${x^2} - 5x + 1 = 0$. Let’s rearrange it.
$ \Rightarrow {x^2} + 1 = 5x$
Dividing both the sides by $x$,
$ \Rightarrow \dfrac{{{x^2}}}{x} + \dfrac{1}{x} = \dfrac{{5x}}{x}$
We will have,
$ \Rightarrow x + \dfrac{1}{x} = 5$ …. (1)
Squaring both the sides,
$ \Rightarrow {\left( {x + \dfrac{1}{x}} \right)^2} = {5^2} = 25$
Using the algebraic formula ${(a + b)^2} = {a^2} + 2ab + {b^2}$,
$ \Rightarrow {x^2} + \dfrac{1}{{{x^2}}} + 2 = 25$
$ \Rightarrow {x^2} + \dfrac{1}{{{x^2}}} = 25 - 2 = 23$ …. (2)
We will once again square both the sides,
$ \Rightarrow {\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right)^2} = {(23)^2}$
Using the algebraic formula ${(a + b)^2} = {a^2} + 2ab + {b^2}$,
$ \Rightarrow {x^4} + \dfrac{1}{{{x^4}}} + 2 = 529$
$ \Rightarrow {x^4} + \dfrac{1}{{{x^4}}} + = 527$ …. (3)
Now, since we want ${x^5} + \dfrac{1}{{{x^5}}}$, we will multiply equation (1) with equation (3),
$ \Rightarrow \left( {{x^4} + \dfrac{1}{{{x^4}}}} \right)\left( {x + \dfrac{1}{x}} \right) = 527 \times 5$
On solving we get,
$ \Rightarrow {x^5} + {x^3} + \dfrac{1}{{{x^3}}} + \dfrac{1}{{{x^5}}} = 2635$
Grouping like terms,
$ \Rightarrow {x^5} + \dfrac{1}{{{x^5}}} + \left( {{x^3} + \dfrac{1}{{{x^3}}}} \right) = 2635$ …. (4)
Now we will find the value of ${x^3} + \dfrac{1}{{{x^3}}}$ by multiplying equation (1) and (2),
$ \Rightarrow \left( {x + \dfrac{1}{x}} \right)\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right) = 5 \times 23$
On solving we get,
$ \Rightarrow {x^3} + \dfrac{1}{x} + x + \dfrac{1}{{{x^3}}} = 115$
Putting the value,
$ \Rightarrow x + \dfrac{1}{x} = 5$,
$ \Rightarrow {x^3} + \dfrac{1}{{{x^3}}} + 5 = 115$
$ \Rightarrow {x^3} + \dfrac{1}{{{x^3}}} = 110$
Putting this value in equation (4),
$ \Rightarrow {x^5} + \dfrac{1}{{{x^5}}} + 110 = 2635$
$ \Rightarrow {x^5} + \dfrac{1}{{{x^5}}} = 2635 - 110$
$ \Rightarrow {x^5} + \dfrac{1}{{{x^5}}} = 2525$

Hence, the value of ${x^5} + \dfrac{1}{{{x^5}}}$ is option (D) 2525.

Note: We can solve the problem in another way of method,
Given equation ${x^2} - 5x + 1 = 0$
Here equate the given to ${a^2} + bx + c = 0$
We get, $a = 1$, $b = - 5$, $c = 1$
Let a and b are the roots of the given equation,
By using property of relation between roots and coefficients,
$x + y = \dfrac{{ - b}}{a}$ and $x \cdot y = \dfrac{c}{a}$
Since, $x \cdot y = 1$
$ \Rightarrow y = \dfrac{1}{x}$
Hence, ${x^5} + {\left( {\dfrac{1}{x}} \right)^5}$ is same as ${x^5} + {y^5}$
Now, let us consider,
${x^2} - 5x + 1 = 0 - - - (1)$
${y^2} - 5y + 1 = 0 - - - (2)$
Multiply by ${x^{n - 2}}$ in equation (1),
${x^2}\left( {{x^{n - 2}}} \right) - 5x\left( {{x^{n - 2}}} \right) + 1\left( {{x^{n - 2}}} \right) = 0$
Multiplying the powers we get,
$ \Rightarrow \left( {{x^{n - 2 + 2}}} \right) - 5\left( {{x^{n - 2 + 1}}} \right) + 1\left( {{x^{n - 2}}} \right) = 0$
Hence,
$ \Rightarrow \left( {{x^n}} \right) - 5\left( {{x^{n - 1}}} \right) + \left( {{x^{n - 2}}} \right) = 0$
Multiply by ${y^{n - 2}}$ in equation (2) we get,
${y^2}\left( {{y^{n - 2}}} \right) - 5y\left( {{y^{n - 2}}} \right) + 1\left( {{y^{n - 2}}} \right) = 0$
Multiplying the powers we get,
$ \Rightarrow \left( {{y^{n - 2 + 2}}} \right) - 5\left( {{y^{n - 2 + 1}}} \right) + 1\left( {{y^{n - 2}}} \right) = 0$
Hence,
$ \Rightarrow \left( {{y^n}} \right) - 5\left( {{y^{n - 1}}} \right) + \left( {{y^{n - 2}}} \right) = 0$
Rearranging the terms on both equations,
$ \Rightarrow \left( {{x^n}} \right) = 5\left( {{x^{n - 1}}} \right) - \left( {{x^{n - 2}}} \right)$
$ \Rightarrow \left( {{y^n}} \right) = 5\left( {{y^{n - 1}}} \right) - \left( {{y^{n - 2}}} \right)$
Adding the both we get,
$ \Rightarrow \left( {{x^n}} \right) + \left( {{y^n}} \right) = = 5\left( {{x^{n - 1}}} \right) - \left( {{x^{n - 2}}} \right) + 5\left( {{y^{n - 1}}} \right) - \left( {{y^{n - 2}}} \right)$
Let consider the term $\left( {{x^n}} \right) + \left( {{y^n}} \right)$ as ${A_n}$,
Hence,
${A_n} = 5{A_{n - 1}} - {A_{n - 2}} - - - \left( * \right)$
Then, we can substituting the values for n as $n = 0,{\text{ }}1,{\text{ }}2,{\text{ }}3,{\text{ }}4,{\text{ }}5,{\text{ }}......$
Since by the property of relation between roots and coefficients,
${A_0} = 2$
${A_1} = 5$ Sum of the roots,
Hence,
${A_2} = 5{A_1} - {A_0} = 25 - 2 = 23$
${A_3} = 5{A_2} - {A_1} = 115 - 5 = 110$
${A_4} = 5{A_3} - {A_2} = 550 - 23 = 527$
${A_5} = 5{A_4} - {A_3} = 2635 - 110 = 2525$
Hence we get,
${A_5} = \left( {{x^5}} \right) + \left( {{y^5}} \right)$
$ \Rightarrow \left( {{x^5}} \right) + \left( {{{\left( {\dfrac{1}{x}} \right)}^5}} \right)$
$ \Rightarrow {x^5} + \dfrac{1}{{{x^5}}} = 2525$