Question

What are the approximate values of the non-integral roots of the polynomial equation? –5.57 –1.95 0.21 1.27 4.73

Answer 1

Answer:

D. 1.27

E. 4.73

Answer 2

Answer:

Option 4 and 5.

Step-by-step explanation:

Consider the given polynomial equation is

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

We need to find approximate values of the non-integral roots of the polynomial equation.

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

Find factor form.

$x(x^{3}-4x^{2}-6x^{1}+12)$

For x=-2 the value of parenthesis is 0. It means (x+2) is a factor of parenthesis.

Divide the parenthesis by (x+2). After division remainder is 0 and quotient is $(x^2 - 6 x + 6)$, so the factor form is

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

Equate the factor form equal to 0, to find the roots.

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

$x=0$

$x+2=0\Rightarrow x=-2$

$x^2 - 6 x + 6=0$            .... (1)

Quadratic formula for $ax^2+bx+c=0$ is

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

In (1), a=1, b=-6, c=6. Using quadratic formula we get

$x=\dfrac{-(-6)\pm \sqrt{(-6)^2-4(1)(6)}}{2(1)}$

$x=\dfrac{6\pm \sqrt{12}}{2}$

$x=\dfrac{6\pm 2\sqrt{3}}{2}$

$x=3\pm \sqrt{3}$

$x=3+1.73, 3-1.73$

$x=4.73, 1.27$

The approximate values of the non-integral roots of the polynomial equation are 4.73 and 1.27.

Therefore, the correct options are 4 and 5.