Question

HELP!!!! I think its C but I'm not sure!


What does the fundamental theorem of algebra state about the equation 2x2−4x+16=0 ?



A. 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=1±i7√2 .


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=1±i7√ .


C. The fundamental theorem of algebra tells you that the equation will have two complex roots since the leading coefficient of the equation is 2. The roots are ​ x=1±i7√2 .


D. The fundamental theorem of algebra tells you that the equation will have two complex roots since the leading coefficient of the equation is 2. The roots are x=1±i7√ .

Answer 1

Answer:

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 $x=1\pm i\sqrt{7}$.

Step-by-step explanation:

Consider the provided information.

Algebra's fundamental theorem states that: Every polynomial equation of degree n with complex coefficients has n roots in the complex numbers.

Now consider the provided equation.

$2x^2-4x+16=0$

The degree of the polynomial equation is 2, therefore according to Algebra's fundamental theorem the equation have two complex roots.

Now find the root of the equation.

For the quadratic equation of the form $ax^2+bx+c=0$ the solutions are: $x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute $a=2,\:b=-4,\:\ and \ c=16$ in above formula.

$x_{1,\:2}=\frac{-\left(-4\right)\pm \sqrt{\left(-4\right)^2-4\cdot \:2\cdot \:16}}{2\cdot \:2}$

$x_{1,\:2}=\frac{4\pm \sqrt{16-128}}{4}$

$x_{1,\:2}=\frac{4\pm \sqrt{-112}}{4}$

$x_{1,\:2}=\frac{4\pm 4i\sqrt{7}}{4}$

$x_{1,\:2}=1\pm i\sqrt{7}$

Hence, 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 $x=1\pm i\sqrt{7}$.