Question

What does the fundamental theorem of algebra state about the equation 2x^2−6x+10=0 ?

Answer 1

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 $y=2x^{2}-6x+10$ has degree 2, so it will have 2 roots.

We know that the polynomial $ax^{2}+bx+c=0$ has roots given by $x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$.

So, the roots of polynomial $2x^{2}-6x+10=0$ will be,

$x=\frac{6 \pm \sqrt{(-6)^{2}-4 \times 2 \times 10}}{2 \times 2}$

i.e. $x=\frac{6 \pm \sqrt{36-80}}{4}$

i.e. $x=\frac{6 \pm \sqrt{-44}}{4}$

i.e. $x=\frac{6 \pm 2i \sqrt{11}}{4}$

i.e. $x=3 \pm \frac{\sqrt{11}i}{2}$

Hence, the roots are given by $3 \pm \frac{\sqrt{11}i}{2}$

So, option B is correct.