Question

Given x^4 − 4x^3 = 6x^2 − 12x, what are the approximate values of the non-integral roots of the polynomial equation?

Answer 1

Answer:

the values of the non-integral roots of the polynomial equation are:

4.73 and 1.27.

Step-by-step explanation:

To find the roots of the polynomial equation, we need to factorize the equation:

x^4 − 4x^3 = 6x^2 − 12x ⇒ x^4 − 4x^3 -6x^2 +12x = 0

⇒ x(x+2)(x -3 + sqrt(3))(x -3 - sqrt(3))

Then, the non integral roots are:

x1 = 3 - sqrt(3) = 1.26 ≈ 1.27

x2 = 3 + sqrt(3) =  4.73

Then, the values of the non-integral roots of the polynomial equation are:

4.73 and 1.27

Answer 2

Answer:

The approximate values of the non-integral roots of the polynomial equation are:

                     1.27 and 4.73

Step-by-step explanation:

We are given an algebraic equation as:

$x^4-4x^3=6x^2-12x$

i.e. it could be written as:

$x^4-4x^3-6x^2+12x=0\\\\i.e.\\\\x(x^3-4x^2-6x+12)=0$

Since, we pulled out the like term i.e. "x" from each term.

Now we know that $x=-2$ is a root of the term:

$x^3-4x^2-6x+12$

Hence, we split the term into factors as:

$x^3-4x^2-6x+12=(x-2)(x^2-6x+6)$

Now, finally the equation could be given by:

$x(x-2)(x^2-6x+6)=0$

Hence, we see that:

$x=0,\ x-2=0\ and\ x^2-6x+6=0\\\\i.e.\\\\x=0,\ x=2\ and\ x^2-6x+6=0$

$x=0\ and\ x=2$ are integers roots.

Now, we find the roots with the help of quadratic equation:

$x^2-6x+6=0$

( We know that the solution of the quadratic equation:

$ax^2+bx+c=0$ is given by:

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$ )

Here we have:

$a=1,\ b=-6\ and\ c=6$

Hence, the solution is:

$x=\dfrac{-(-6)\pm \sqrt{(-b)^2-4\times 1\times 6}}{2\times 1}\\\\i.e.\\\\x=\dfrac{6\pm \sqrt{36-24}}{2}\\\\i.e.\\\\x=\dfrac{6\pm \sqrt{12}}{2}\\\\i.e.\\\\x=\dfrac{6}{2}\pm \dfrac{2\sqrt{3}}{2}\\\\i.e.\\\\x=3\pm 3\\\\i.e.\\\\x=3+\sqrt{3},\ x=3-\sqrt{3}$

Now, we put $\sqrt{3}=1.732$

Hence, the approximate value of x is:

$x=3+1.732,\ x=3-1.732\\\\i.e.\\\\x=4.732,\ x=1.268$