Question

Solve the following equations which have equal roots
${x^6} - 3{x^5} + 6{x^3} - 3{x^2} - 3x + 2 = 0$

Answer 1

Hint- In this question, first by trial method we will find one of the roots of the given equation. After this , we will find the derivative of the given function and equate it zero. And again we will find the root of the given equation by trial method. The roots common to both f(x) and f’(x) occur as double roots.

Complete step-by-step solution -
In the question, we have given an equation and we have to find the equal roots of the given equation.
The given equation is:
$f(x) = {x^6} - 3{x^5} + 6{x^3} - 3{x^2} - 3x + 2 = 0$
Let us find the root by trial method.
We will calculate $f(1)$.
$f(1)$=${1^6} - 3 \times {1^5} + 6 \times {1^3} - 3 \times {1^2} - 3 \times 1 + 2$ =$1 -3+6 -3 -3+2 =0$.
Therefore , we can say that $x =1$ is the root of the given equation.
$f(-1)$=${( - 1)^6} - 3 \times {( - 1)^5} + 6 \times {( - 1)^3} - 3 \times {( - 1)^2} - 3 \times - 1 + 2$ =$1 +3-6 -3 +3+2 =0$.
Therefore , we can say that $x = -1$ is also the root of the given equation.
Now let us find the f’(x).
$f'(x) = 6{x^5} - 15{x^4} + 18{x^2} - 6x - 3$
Let us find the roots of f’(x) by trial method.
$f’(1)$= $6 \times {1^5} - 15 \times {1^4} + 18 \times {1^2} - 6 \times 1 - 3$ =$24 – 24 = 0$.
So, $x = 1$ is a root of f’(x).
Now , we calculate at $x = -1$.
$f’(-1)$=$6 \times {( - 1)^5} - 15 \times {( - 1)^4} + 18 \times {( - 1)^2} - 6 \times - 1 - 3$ = $-24 +24 =0$.
Therefore, we can say that $x = 1, -1$ are the roots of both $f(x)$ and $f’(x)$.
So, these roots will occur as double roots or we can say that $x =1$ and $x = -1$ are double or equal roots of the f’(x).
So, ${(x - 1)^2}$ and ${(x + 1)^2}$ are the factors of the function f(x).
$\therefore $ f(x) can be factored as ${(x + 1)^2}{(x - 1)^2}({x^2} - 3x + 2)$ .
So, the function f(x) can be written in factored form as follow:
$f(x) = {(x + 1)^2}{(x - 1)^2}(x - 1)(x - 2) = {(x + 1)^2}{(x - 1)^3}(x - 2)$
Therefore, the roots of the given equation are : 1, 1, 1,-1, -1, 2.

Note- In the question involving finding the roots of an equation having higher powers of ‘x’ , we generally use hit and trial method to find some of its roots . The trial method is done by taking negative and positive values of the factor of the constant term of the given polynomial.