Question

2 Points

Answer 1

Two or zero expresses the possible number of positive real  solutions for the given polynomial equation.

Answer: Option D

Step-by-step explanation:

Given equation:

        $x^{3}-4 x^{2}-7 x+28=0$

First, we put hit and trial method to find out the one solution. So, if we put x=4 then the above expression will become zero.  We can also write the above expression as  

      $(x-4) \times\left(x^{2}-7\right)=0$

We know the formula, $a^{2}-b^{2}=(a+b)(a-b)$, make use of this, we get

   $(x-4) \times(x-\sqrt{7}) \times(x+\sqrt{7})=0$

So, $x=4, \sqrt{7},-\sqrt{7}$

Hence, from the above expression, we have three values of x as x= 4, 2.64 and -2.64