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

Answer:

OPTION E: $\frac{2+\sqrt{14}i}{2}$

Step-by-step explanation:

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

$x=\frac{-b\±\sqrt{b^2-4ac} }{2a}$

In this case you can identify that:

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

Then you can substitute values into the Quadratic formula:

$x=\frac{-(-4)\±\sqrt{(-4)^2-4(2)(9)} }{2(2)}$

$x=\frac{4\±\sqrt{(56} }{4}$

Remember that $\sqrt{-1}=i$, then:

$x=\frac{4\±2\sqrt{14}i}{4}$

Simplifying, you get:

$x=\frac{2(2\±i\sqrt{14}i)}{4}\\\\x=\frac{2\±\sqrt{14}i}{2}\\\\\\x_1=\frac{2+\sqrt{14}i}{2}\\\\x_2=\frac{2-\sqrt{14}i}{2}$