Question

The solution set of the equation $${x^4} - 5{x^3} - 4{x^2} - 5x + 1 = 0$$ is
A.$$\left\{ {3 \pm 2\sqrt 2 ,\dfrac{{ - 1 \pm \sqrt 3 i}}{2}} \right\}$$
B.$$\left\{ {\dfrac{{3 \pm 2\sqrt 2 }}{2}, - 1 \pm \sqrt 3 i} \right\}$$
C.$$\left\{ { - 3 \pm 2\sqrt 2 ,\dfrac{{1 \pm \sqrt 3 i}}{2}} \right\}$$
D.$$\left\{ {\dfrac{{ - 3 \pm 2\sqrt 2 }}{2},1 \pm \sqrt 3 i} \right\}$$

Answer 1

Hint: Here given an algebraic equation we need to find the set of solutions. For this, first we have to convert the given equation to the form quadratic equation $$a{x^2} + bx + c = 0$$ by taking some substitution. Later solve the quadratic equation by using the formula $$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$$ on substituting the values we get the required solution.

Complete step-by-step answer:
Consider the given expression:
$$ \Rightarrow \,\,\,{x^4} - 5{x^3} - 4{x^2} - 5x + 1 = 0$$ -----(1)
Divide the whole equation by $${x^2}$$
$$ \Rightarrow \,\,\,{x^2} - 5x - 4 - \dfrac{5}{x} + \dfrac{1}{{{x^2}}} = 0$$
On rearranging, we have
$$ \Rightarrow \,\,\,{x^2} + \dfrac{1}{{{x^2}}} - 5\left( {x + \dfrac{1}{x}} \right) - 4 = 0$$ ----(2)
Let’s put $$\left( {x + \dfrac{1}{x}} \right) = m$$, then
$${\left( {x + \dfrac{1}{x}} \right)^2} = {m^2}$$
$${x^2} + \dfrac{1}{{{x^2}}} + 2\left( x \right)\left( {\dfrac{1}{x}} \right) = {m^2}$$
$${x^2} + \dfrac{1}{{{x^2}}} + 2\left( x \right)\left( {\dfrac{1}{x}} \right) = {m^2}$$
$${x^2} + \dfrac{1}{{{x^2}}} + 2 = {m^2}$$
$${x^2} + \dfrac{1}{{{x^2}}} = {m^2} - 2$$
Then equation (2) becomes
$$ \Rightarrow \,\,\,{m^2} - 2 - 5m - 4 = 0$$
$$ \Rightarrow \,\,\,{m^2} - 5m - 6 = 0$$ ----(3)
The above equation looks similar as quadratic equation in the form of $$a{x^2} + bx + c$$, then the formula of solution is $$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$$.
In equation (3), a=1, b=-5 and c=-6, then on substituting in formula we have
 $$ \Rightarrow \,\,m = \dfrac{{ - \left( { - 5} \right) \pm \sqrt {{{\left( { - 5} \right)}^2} - 4\left( 1 \right)\left( { - 6} \right)} }}{{2\left( 1 \right)}}$$
$$ \Rightarrow \,\,m = \dfrac{{5 \pm \sqrt {25 + 24} }}{2}$$
$$ \Rightarrow \,\,m = \dfrac{{5 \pm \sqrt {49} }}{2}$$
$$ \Rightarrow \,\,m = \dfrac{{5 \pm 7}}{2}$$
$$ \Rightarrow \,\,m = \dfrac{{5 + 7}}{2}$$ and $$\,m = \dfrac{{5 - 7}}{2}$$
$$ \Rightarrow \,\,m = \dfrac{{12}}{2}$$ and $$m = \dfrac{{ - 2}}{2}$$
$$ \Rightarrow \,\,m = 6$$ and $$m = - 1$$
The solution of equation (3) is $$m = 6$$ and $$m = - 1$$.
If $$m = 6$$
Substitute $$m = \left( {x + \dfrac{1}{x}} \right)$$, then
$$ \Rightarrow \,\,\,\,x + \dfrac{1}{x} = 6$$
Take x as LCM in LHS, then
$$ \Rightarrow \,\,\,\,\dfrac{{{x^2} + 1}}{x} = 6$$
Multiply x on both side
$$ \Rightarrow \,\,\,\,{x^2} + 1 = 6x$$
Subtract 6x on both side
$$ \Rightarrow \,\,\,\,{x^2} - 6x + 1 = 0$$
 Then the formula of quadratic equation
$$ \Rightarrow \,\,x = \dfrac{{ - \left( { - 6} \right) \pm \sqrt {{{\left( { - 6} \right)}^2} - 4\left( 1 \right)\left( 1 \right)} }}{{2\left( 1 \right)}}$$
$$ \Rightarrow \,\,x = \dfrac{{6 \pm \sqrt {36 - 4} }}{2}$$
$$ \Rightarrow \,\,x = \dfrac{{6 \pm \sqrt {32} }}{2}$$
$$ \Rightarrow \,\,x = \dfrac{{6 \pm 4\sqrt 2 }}{2}$$
$$ \Rightarrow \,\,x = \dfrac{{2\left( {3 \pm 2\sqrt 2 } \right)}}{2}$$
On simplification, we get
$$\therefore \,\,x = 3 \pm 2\sqrt 2 $$ -----(4)
If $$m = - 1$$
Substitute $$m = \left( {x + \dfrac{1}{x}} \right)$$, then
$$ \Rightarrow \,\,\,\,x + \dfrac{1}{x} = - 1$$
Take x as LCM in LHS, then
$$ \Rightarrow \,\,\,\,\dfrac{{{x^2} + 1}}{x} = - 1$$
Multiply x on both side
$$ \Rightarrow \,\,\,\,{x^2} + 1 = - x$$
Add x on both side
$$ \Rightarrow \,\,\,\,{x^2} + x + 1 = 0$$
 Then the formula of quadratic equation
$$ \Rightarrow \,\,x = \dfrac{{ - \left( 1 \right) \pm \sqrt {{{\left( 1 \right)}^2} - 4\left( 1 \right)\left( 1 \right)} }}{{2\left( 1 \right)}}$$
$$ \Rightarrow \,\,x = \dfrac{{ - 1 \pm \sqrt {1 - 4} }}{2}$$
$$ \Rightarrow \,\,x = \dfrac{{ - 1 \pm \sqrt { - 3} }}{2}$$
$$ \Rightarrow \,\,x = \dfrac{{ - 1 \pm \sqrt 3 \sqrt { - 1} }}{2}$$
As we know the value of $$\sqrt { - 1} = i$$
$$\therefore \,\,\,\,x = \dfrac{{ - 1 \pm \sqrt 3 \,i}}{2}$$ -----(5)
From (4) and (5)
Hence, the required solution is $$\left\{ {\,x = 3 \pm 2\sqrt 2 \,,\,\dfrac{{ - 1 \pm \sqrt 3 }}{2}} \right\}$$.
Therefore, option (A) is the correct answer.
So, the correct answer is “Option A”.

Note: The equation $$a{x^2} + bx + c$$ is the general form of the quadratic equation. The solution of quadratic can be find by using a formula $$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$$ here, a is the coefficient of $${x^2}$$, b is the coefficient of $$x$$ and c is the constant term. Remember the value of the imaginary number $$\sqrt { - 1} = i$$.