Question

Solve the following equation:
${x^5} - 6{x^4} - 17{x^3} + 17{x^2} + 6x - 1 = 0$

Answer 1

Hint: Since, the degree (highest power of the variable) of the given equation is 5, therefore, this is a quintic equation. We can find the roots of this equation easily by trial method. We have to put values in this equation (for ex- x = 1, 2, -1, -2, etc.) and check in which it gave zero. And then dividing the equation by that factor. We have to continue this process till we get a quadratic equation (degree 2).

Complete step by step answer:
Now, moving on to the question
The given equation is- ${x^5} - 6{x^4} - 17{x^3} + 17{x^2} + 6x - 1 = 0$
Let us put x = 1 in the equation, we get
0 = 0, hence x = 1 (x-1) is the root of the given equation.
Now, we will divide the given equation with (x – 1) with the help of long division (or synthetic division), we get a new equation of degree 4 which is ${x^4} - 5{x^3} - 22{x^2} - 5x + 1$
Now we will find the roots of the above equation,
Let ${x^4} - 5{x^3} - 22{x^2} - 5x + 1$ = 0
$({x^4} + 1) - 5({x^3} + x) - 22{x^2} = 0$
Dividing the whole equation by ${x^2}$, we get
$({x^2} + {x^{ - 2}}) - 5(x + {x^{ - 1}}) - 22 = 0$
Substituting $x + {x^{ - 1}} = y$, we get
$({y^2} - 2) - 5y - 22 = 0 \Rightarrow {y^2} - 5y - 24 = 0 \Rightarrow (y - 8)(y + 3) = 0$
Therefore, $x + {x^{ - 1}} = 8$ and $x + {x^{ - 1}} = 3$
Or, ${x^2} - 8x + 1 = 0$, ${x^2} - 3x + 1 = 0$
Solving the above quadratic equations with help of formula,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Solving ${x^2} - 8x + 1 = 0$, we get $x = 4 \pm \sqrt {15} $
Solving ${x^2} - 3x + 1 = 0$, we get $x = \dfrac{{3 \pm \sqrt 5 }}{2}$
Therefore the 5 roots of the equation ${x^5} - 6{x^4} - 17{x^3} + 17{x^2} + 6x - 1 = 0$ are
x = 1, $x = 4 \pm \sqrt {15} $, $x = \dfrac{{3 \pm \sqrt 5 }}{2}$

NOTE: Discriminant Formula-
The discriminant formula is employed to seek out the amount of solutions that a quadratic has. In algebra the discriminant is the name given to the expression that appears under the square root sign in the quadratic formula. The discriminant of a polynomial is a function of its coefficients and represented by the ‘D’ or delta $\Delta $ symbol . It shows the nature of the roots of any quadratic equation where a, b, c are rational numbers. The discriminant formula in the quadratic equation $a{x^2} + bx + c$ is
$D = {b^2} - 4ac$.