Answer:
Option D
Explanation:
Option A
x² - 3x² + 2x² = 0
Therefore, the given polynomial has no roots.
Option B
x² + 2x + 8
By quadratic formula,
x =
![(-2\pm √(4-32) )/(2)](https://img.qammunity.org/2021/formulas/mathematics/high-school/ffe1kc9gldmcdzsomuxfzqqsqbof1nuafd.png)
=
![(-2\pm √(-28) )/(2)](https://img.qammunity.org/2021/formulas/mathematics/high-school/cfc5my1f7ngnf1x75b6uu2q083i9cdb5zn.png)
Therefore, roots are imaginary and for a polynomial with degree 1 or more than 1, both the roots can't be imaginary.
So the polynomial can't have exactly two roots.
Option C
99x³ - 33x + 1
Since the polynomial is of degree 3, so it will have three roots.
Therefore, the polynomial will not have exactly 2 roots.
Option D
√2x - 3x² + 7√2
By quadratic formula,
x =
![\frac{-√(2)\pm\sqrt{(√(2))^2-4(-3)(7√(2) )}}{2(-3)}](https://img.qammunity.org/2021/formulas/mathematics/high-school/e08yenin2gy101ifzivyzn1ocdz5t91omn.png)
x =
![\frac{-√(2)\pm\sqrt{2+84√(2) )}}{(-6)}](https://img.qammunity.org/2021/formulas/mathematics/high-school/8y2lo2ar4rvxwxbgf31vzwh5ng8sqgdf82.png)
There are exactly two real roots.
Therefore Option D is the answer.
Option E
4x + 11x - 111
Since this polynomial is of a degree 1.
There will be only one root.