Question

Solve the following equation: $4{{x}^{4}}-20{{x}^{3}}+33{{x}^{2}}-20x+4=0$

Answer 1

Hint: First, divide both the sides by ${{x}^{2}}$. Rearrange the terms and the substitute $x+{{x}^{-1}}=y$ in the resultant equation. Next, use completing the square method to solve for y. put these values of y in $x+{{x}^{-1}}=y$ and then solve for x.

Complete step-by-step answer:
In this question, we need to solve the equation $4{{x}^{4}}-20{{x}^{3}}+33{{x}^{2}}-20x+4=0$.
Given the equation $4{{x}^{4}}-20{{x}^{3}}+33{{x}^{2}}-20x+4=0$
We will divide both the sides by ${{x}^{2}}$.

On dividing both the sides by ${{x}^{2}}$, we will get the following:
$4{{x}^{2}}-20x+33-20{{x}^{-1}}+4{{x}^{-2}}=0$

On rearranging the terms, we will get the following:
$4\left( {{x}^{2}}+{{x}^{-2}} \right)-20\left( x+{{x}^{-1}} \right)+33=0$
Substituting $x+{{x}^{-1}}=y$ in the above equation, we will get the following:
$4\left( {{y}^{2}}-2 \right)-20y+33=0$
$4{{y}^{2}}-20y+25=0$

Now, we will use completing the square method to solve the above equation.
Let us first define what completing the square method actually is.
In elementary algebra, completing the square is a technique for converting a quadratic polynomial of the form $a{{x}^{2}}+bx+c=0$ to the form $a{{\left( x-h \right)}^{2}}+k=0$ for some h and k.

Its steps are:
1. Transform the equation so that the constant term, c , is alone on the right side.
2. If a, the leading coefficient is not equal to 1, divide both sides by a .
3. Add the square of half the coefficient of the x -term, ${{\left( \dfrac{b}{2a} \right)}^{2}}$ to both sides of the equation.
4. Factor the left side as the square of a binomial.
5. Take the square root of both sides
5. Solve for x.

Using the above steps, we convert $4{{y}^{2}}-20y+25=0$ to the following:
${{\left( 2y-5 \right)}^{2}}=0$
For this equation, we have equal roots, y = $\dfrac{5}{2}$.

Putting this in the initial substitution $x+{{x}^{-1}}=y$, we will get the following:
$x+{{x}^{-1}}=\dfrac{5}{2}$
$2{{x}^{2}}-5x+2=0$
$\left( 2x-1 \right)\left( x-2 \right)=0$
$x=\dfrac{1}{2},2$
So, the roots of the given equation: $4{{x}^{4}}-20{{x}^{3}}+33{{x}^{2}}-20x+4=0$ are $\dfrac{1}{2},\dfrac{1}{2},2,2$

Note: In this question, it is very important to know/do the following: divide both the sides by ${{x}^{2}}$, substituting $x+{{x}^{-1}}=y$, and that completing the square is a technique for converting a quadratic polynomial of the form $a{{x}^{2}}+bx+c=0$ to the form $a{{\left( x-h \right)}^{2}}+k=0$ for some h and k.