Question

What is one root of this equation 2x^2-4x+9=0?

A. -2+sqrt14/2i
B. -1+sqtr14/2i
C. 1+sqrt14/3i
D. 1+sqrt14/2i
E. 2+sqrt14/2i

Answer 1

For this case we have the following quadratic equation:

`2x ^ 2-4x + 9 = 0`

The solutions are given by:

`x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}`

We have to:

`a = 2\\b = -4\\c = 9`

Substituting:

`x = \frac {- (- 4) \pm \sqrt {(- 4) ^ 2-4 (2) (9)}} {2 (2)}\\x = \frac {4 \pm \sqrt {16-72}} {4}\\x = \frac {4 \pm \sqrt {-56}} {4}\\x = \frac {4 \pm \sqrt {-1 * 56}} {4}\\x = \frac {4 \pmi \sqrt {2 ^ 2 * 14}} {4}\\x = \frac {4 \pm2i \sqrt {14}} {4}\\x = \frac {2 \pm i\sqrt {14}} {2}`

Answer:

`x_ {1} = \frac {2 + i \sqrt {14}} {2}\\x_ {2} = \frac {2-i \sqrt {14}} {2}`

Answer 2

OPTION E:
`(2+√(14)i)/(2)`

Explanation:

Given the Quadratic equation
`2x^2-4x+9=0`, you can find the roots by applying the Quadratic formula. This is:

`x=(-b\±√(b^2-4ac) )/(2a)`

In this case you can identify that:

`a=2\\b=-4\\c=9`

Then you can substitute values into the Quadratic formula:

`x=(-(-4)\±√((-4)^2-4(2)(9)) )/(2(2))`

`x=(4\±√((56) )/(4)`

Remember that
`√(-1)=i`, then:

`x=(4\±2√(14)i)/(4)`

Simplifying, you get:

`x=(2(2\±i√(14)i))/(4)\\\\x=(2\±√(14)i)/(2)\\\\\\x_1=(2+√(14)i)/(2)\\\\x_2=(2-√(14)i)/(2)`