QUESTION

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

Hint: Here the equation is of degree 4, so we have to convert the equation to quadratic form and solve for x.

Given equation: ${x^4} - 10{x^3} + 26{x^2} - 10x + 1 = 0$ ……… (1)
Now we have to solve equation (1), so
First divide both sides of equation (1) by ${x^2}$ we get,
$\Rightarrow {x^2} - 10x + 26 - \dfrac{{10}}{x} + \dfrac{1}{{{x^2}}} = 0$
Now adding and subtracting 2 in the LHS of the above equation and taking its terms common, we get,
$\Rightarrow \left( {{x^2} + \dfrac{1}{{{x^2}}} + 2} \right) - 2 - 10\left( {x + \dfrac{1}{x}} \right) + 26 = 0$
$\Rightarrow {\left( {x + \dfrac{1}{x}} \right)^2} - 10\left( {x + \dfrac{1}{x}} \right) + 24 = 0$
Let $y = \left( {x + \dfrac{1}{x}} \right)$ , so above equation becomes
$\Rightarrow {y^2} - 10y + 24 = 0$ ……………… (2)
Now splitting the middle term of equation (2), we get,
$\Rightarrow {y^2} - 6y - 4y + 24 = 0$
$\Rightarrow y (y - 6) - 4 (y - 6) = 0$
$\Rightarrow \left( {y - 4} \right)\left( {y - 6} \right) = 0$
Hence, $y = 4$ or $y = 6$.
So now putting value of $y = \left( {x + \dfrac{1}{x}} \right)$, we get
$\left( {x + \dfrac{1}{x}} \right) = 4$ or $\left( {x + \dfrac{1}{x}} \right) = 6$
Above two equations will become,
$\Rightarrow {x^2} - 4x + 1 = 0$ ……………… (3)
$\Rightarrow {x^2} - 6x + 1 = 0$ ……………… (4)
Solving the above equations using the quadratic formula.
Now, after solving the quadratic equation (3) we get $x = 2 \pm \sqrt 3$
Now, after solving the quadratic equation (4) we get $x = 3 \pm 2\sqrt 2$

Hence solutions of the given degree 4 equation in the question are $x = 2 \pm \sqrt 3$ and $x = 3 \pm 2\sqrt 2$.

Note: Whenever we face these types of problems, where the equation given is of degree 4, then first we have to make the given equation into a quadratic equation and then solve the equation using either factorisation method or quadratic formula.