Question

Solve the equation:
${x^4} - 10{x^3} + 26{x^2} - 10x + 1 = 0$

Answer 1

Hint: Here the degree of the given expression is fourth power and so first of all we will convert the given expression in the terms of square by diving with the power of “X” and then simplify using the factorization method and find the value for “x”.

Complete step-by-step answer:
Take the given expression: ${x^4} - 10{x^3} + 26{x^2} - 10x + 1 = 0$
Divide the above expression by and simplify the expression by using the negative exponent rule, when bases are same and in division powers are subtracted.
${x^2} - 10x + 26 - \dfrac{{10}}{x} + \dfrac{1}{{{x^2}}} = 0$
The above expression can be re-arranged -
${x^2} + \dfrac{1}{{{x^2}}} - 10x - \dfrac{{10}}{x} + 26 - = 0$
Take common multiple common –
$\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right) - 10\left( {x + \dfrac{1}{x}} \right) + 26 - = 0$ ….. (I)
Now, take $\left( {x + \dfrac{1}{x}} \right) = a$ …. (A)
Squaring both the sides of the above expression –
${\left( {x + \dfrac{1}{x}} \right)^2} = {a^2}$
Expand the above expression using the identity –
$\left( {{x^2} + 2 + \dfrac{1}{{{x^2}}}} \right) = {a^2}$
Make the required term the subject and move other terms on the opposite side –
$\left( {{x^2} + \dfrac{1}{{{x^2}}}} \right) = {a^2} - 2$ …. (B)
Now using equation (A) and (B)
$\left( {{a^2} - 2} \right) - 10\left( a \right) + 26 = 0$
Simplify the above expression –
${a^2} - 2 - 10a + 26 = 0$
Combine the like terms –
${a^2} - 10a + 24 = 0$
Split the middle term –
$\underline {{a^2} - 4a} - \underline {6a + 24} = 0$
Take common multiple common in the above expression –
$
  a(a - 4) - 6(a - 4) = 0 \\
  (a - 6)(a - 4) = 0 \\
  a = 4,6 \;
 $
By using equation (A) –
$x + \dfrac{1}{x} = 4$ or II) $x + \dfrac{1}{x} = 6$
Simplify the above equations -
$
  \dfrac{{{x^2} + 1}}{x} = 4 \\
  {x^2} + 1 = 4x \\
  {x^2} - 4x + 1 = 0 \\
 $
By comparing with the standard equation –
$a{x^2} + bx + c = 0$ and $\Delta = {b^2} - 4ac$
By using the standard quadratic formula, $x = \dfrac{{ - b \pm \sqrt \Delta }}{{2a}}$
$x = \dfrac{{4 \pm \sqrt {16 - 4} }}{2}$
Simplify the above expression –
$
  x = \dfrac{{4 \pm \sqrt {12} }}{2} \\
  x = 2 \pm \sqrt 3 \;
 $

II)
Similarly for the second case $x + \dfrac{1}{x} = 6$
Simplify the above equations -
$
  \dfrac{{{x^2} + 1}}{x} = 6 \\
  {x^2} + 1 = 6x \\
  {x^2} - 6x + 1 = 0 \;
 $
By comparing with the standard equation –
$a{x^2} + bx + c = 0$ and $\Delta = {b^2} - 4ac$
By using the standard quadratic formula, $x = \dfrac{{ - b \pm \sqrt \Delta }}{{2a}}$
$x = \dfrac{{6 \pm \sqrt {36 - 4} }}{2}$
Simplify the above expression –
$
  x = \dfrac{{6 \pm \sqrt {32} }}{2} \\
  x = 3 \pm 2\sqrt 3 \;
 $
Hence, the roots of the equation are $x = 3 \pm 2\sqrt 3 ,2 \pm \sqrt 3 $
So, the correct answer is “ $x = 3 \pm 2\sqrt 3 ,2 \pm \sqrt 3 $”.

Note: Factorization of any expression can be found using the split of the middle term and find the values for “x”. remember the standard quadratic equation and formula for roots, it is useful when there are complex roots of the “x”. be careful of the sign convention.