Question

Determine the solution to the system of equations given below y=x^2-5x+15

Answer 1

ANSWER

$x = \frac{5}{2} - \frac{ \sqrt{35}i }{2} \: or \: x = \frac{5}{2} + \frac{ \sqrt{35}i }{2}$

EXPLANATION

The given equation is

$y = {x}^{2} - 5x + 15$

To solve this equation, we equate it to zero.

${x}^{2} - 5x + 15 = 0$

Comparing to ax²+bx+c=0, we have a=1, b=-5, c=15.

The solution is given by the quadratic formula,

$x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}$

We plug in the values to obtain,

$x = \frac{ - - 5\pm \sqrt{ {( - 5)}^{2} - 4(1)(15) } }{2(1)}$

$x = \frac{5\pm \sqrt{ 25 -60 }}{2}$

$x = \frac{5\pm \sqrt{ - 35}}{2}$

Recall that,

$\sqrt{-1}=i$

$x = \frac{5\pm \sqrt{ 35}i}{2}$

$x = \frac{5}{2} - \frac{ \sqrt{35}i }{2} \: or \: x = \frac{5}{2} + \frac{ \sqrt{35}i }{2}$

Answer 2

Answer:

x=(5+i*sqrt(35))/2, (5-i*sqrt(35))/2

Step-by-step explanation:

Use the quadratic formula with the following values.

a = 1

b = -5

c = 15

Substitute and simplify.

(5+-sqrt((-5)^2-4*(1*15)))/2*1

x = (5+-i*sqrt(35))/2

When you get a negative number inside the square root, remember that you can pull out i to make the number inside positive.