Question

Using the quadratic formula to solve 4x2 – 3x + 9 = 2x + 1, what are the values of x? StartRoot 1 plus-or-minus StartRoot 159 EndRoot i Over 8 EndFraction StartRoot 5 plus-or-minus StartRoot 153 EndRoot i Over 8 EndFraction StartRoot 5 plus-or-minus StartRoot 103 EndRoot i Over 8 EndFraction StartRoot 1 plus-or-minus StartRoot 153 EndRoot Over 8 EndFraction

Answer 1

For this case we have that a quadratic equation is of the form:

$ax ^ 2 + bx + c = 0$

The roots are given by:

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

We have the following equation:

$4x ^ 2-3x + 9 = 2x + 1\\4x ^ 2-3x-2x + 9-1 = 0\\4x ^ 2-5x + 8 = 0$

We look for the roots:

$x = \frac {- (- 5) \pm \sqrt {(- 5) ^ 2-4 (4) (8)}} {2 (4)}\\x = \frac {5 \pm \sqrt {25-128}} {8}\\x = \frac {5 \pm \sqrt {-103}} {8}$

We have to:

$i ^ 2 = -1$

So:

$x = \frac {5 \pm \sqrt {103i ^ 2}} {8}\\x = \frac {5 \pm i \sqrt {103}} {8}$

We have two imaginary roots:

$x_ {1} = \frac {5 \ + i \sqrt {103}} {8}\\x_ {2} = \frac {5 \ -i \sqrt {103}} {8}$

Answer:

$x_ {1} = \frac {5 \ + i \sqrt {103}} {8}\\x_ {2} = \frac {5 \ -i \sqrt {103}} {8}$

Answer 2

Answer:

C....... :)

Step-by-step explanation:

option C for short cus aint nobody got time for all that..  >_<