Answer:
Option D
Step-by-step explanation:
Option A
x² - 3x² + 2x² = 0
Therefore, the given polynomial has no roots.
Option B
x² + 2x + 8
By quadratic formula,
x = [tex]\frac{-2\pm \sqrt{4-32} }{2}[/tex]
= [tex]\frac{-2\pm \sqrt{-28} }{2}[/tex]
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 = [tex]\frac{-\sqrt{2}\pm\sqrt{(\sqrt{2})^2-4(-3)(7\sqrt{2} )}}{2(-3)}[/tex]
x = [tex]\frac{-\sqrt{2}\pm\sqrt{2+84\sqrt{2} )}}{(-6)}[/tex]
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.
