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:
[tex]x^{3}-4 x^{2}-7 x+28=0[/tex]
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
[tex](x-4) \times\left(x^{2}-7\right)=0[/tex]
We know the formula, [tex]a^{2}-b^{2}=(a+b)(a-b)[/tex], make use of this, we get
[tex](x-4) \times(x-\sqrt{7}) \times(x+\sqrt{7})=0[/tex]
So, [tex]x=4, \sqrt{7},-\sqrt{7}[/tex]
Hence, from the above expression, we have three values of x as x= 4, 2.64 and -2.64
