Answer:
B. The fundamental theorem of algebra tells you that the equation will have two complex roots since the degree of the polynomial is 2. The roots are x=3±i√11/2 .
Step-by-step explanation:
We know that,
Fundamental theorem of algebra states that ' a polynomial of degree n will have n number of roots'.
Since, the polynomial [tex]y=2x^{2}-6x+10[/tex] has degree 2, so it will have 2 roots.
We know that the polynomial [tex]ax^{2}+bx+c=0[/tex] has roots given by [tex]x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}[/tex].
So, the roots of polynomial [tex]2x^{2}-6x+10=0[/tex] will be,
[tex]x=\frac{6 \pm \sqrt{(-6)^{2}-4 \times 2 \times 10}}{2 \times 2}[/tex]
i.e. [tex]x=\frac{6 \pm \sqrt{36-80}}{4}[/tex]
i.e. [tex]x=\frac{6 \pm \sqrt{-44}}{4}[/tex]
i.e. [tex]x=\frac{6 \pm 2i \sqrt{11}}{4}[/tex]
i.e. [tex]x=3 \pm \frac{\sqrt{11}i}{2}[/tex]
Hence, the roots are given by [tex]3 \pm \frac{\sqrt{11}i}{2}[/tex]
So, option B is correct.
