Question

(a) Use the fundamental theorem of algebra to determine the number of roots for 2x^2+4x+7.

(b) What are the roots of 2x^2+4x+7 ? show work please.

Answer 1

`\textsf{(a)}\quad \textsf{2 roots}`

`\textsf{(b)} \quad x_1=(-2+√(10)\:i)/(2), \quad x_1=(-2-√(10)\:i)/(2)`

Explanation:

Part (a)

The fundamental theorem of algebra states that every non-constant polynomial equation in one variable with complex coefficients has at least one complex root. Furthermore, the number of complex roots (counting multiplicity) is equal to the degree of the polynomial equation.

It is important to note that even when the solutions appear to be real, they can also be expressed as complex numbers, where the imaginary part is zero, effectively making them real numbers.

In this case, we have the polynomial equation 2x² + 4x + 7. Since it is a second-degree polynomial, it will have two complex roots.

`\hrulefill`

Part (b)

To find the roots of 2x² + 4x + 7, we can use the quadratic formula.

`\boxed{\begin{array}{l}\underline{\sf Quadratic\;Formula}\\\\x=(-b \pm √(b^2-4ac))/(2a)\\\\\textsf{when} \;ax^2+bx+c=0 \\\end{array}}`

In this case:

• a = 2
• b = 4
• c = 7

Substitute these values into the quadratic formula and solve for x:

`x=(-(4) \pm √((4)^2-4(2)(7)))/(2(2))`

`x=(-4\pm √(16-56))/(4)`

`x=(-4\pm √(-40))/(4)`

Rewrite -40 as (2² · 10 · (-1)):

`x=(-4\pm √(2^2\cdot 10\cdot (-1)))/(4)`

`\textsf{Apply the radical rule:} \quad √(abc)=\sqrt{\vphantom{b}a}√(b)\sqrt{\vphantom{b}c}`

`x=(-4\pm √(2^2)√(10)√(-1))/(4)`

`\textsf{Apply the imaginary number rule:} \quad √(-1)=i`

`x=(-4\pm 2√(10)\:i)/(4)`

Simplify the expression by dividing the numerator and denominator by the common factor of 2:

`x=(-2\pm √(10)\:i)/(2)`

So, the roots of the equation 2x² + 4x + 7 are:

`x_1=(-2+√(10)\:i)/(2)`

`x_1=(-2-√(10)\:i)/(2)`

These are complex numbers because they involve the imaginary unit
`i`.