How do you solve using the quadratic formula for #x^2 + x + 5 = 0#?

2 Answers
Apr 13, 2018

Answer:

The answer is #(-1+-isqrt(19))/2#.

Explanation:

The quadratic formula is #x=(-b+-sqrt(b^2-4ac))/(2a# for the equation #ax^2+bx+c#.

In this case, #a=1#, #b=1#, and #c=5#.

You can therefore substitute in those values to get:

#(-1+-sqrt(1^2-4(1)(5)))/(2(1)#.

Simplify to get #(-1+-sqrt(-19))/2#.

Because #sqrt(-19)# is not a real number, we have to stick to imaginary solutions. (If this problem asks for real number solutions, there are none.)

The imaginary number #i# equals #sqrt(-1)#, therefore we can substitute it in:

#(-1+-sqrt(-1*19))/2 rarr (-1+-sqrt(-1)*sqrt(19))/2 rarr (-1+-isqrt(19))/2#, the final answer.

Hope this helps!

Apr 13, 2018

Answer:

See application of the quadratic formula below in obtaining the result:
#color(white)("XXX")x=-1/2+-sqrt(19)i#

Explanation:

#x^2+x+5=0# is equivalent to #color(red)1x^2+color(blue)1x+color(magenta)5=0#

Applying the general quadratic formula #x=(-color(blue)b+-sqrt(color(blue)b^2-4color(red)acolor(magenta)c))/(2color(red)a#
for #color(red)ax^2+color(blue)bx+color(magenta)c=0#

to this specific case, we have
#color(white)("XXX")x=(-color(blue)1+-sqrt(color(blue)1^2-4 * color(red)1 * color(magenta)5))/(2 *color(red)1)#

#color(white)("XXXXX")=(-1+-sqrt(-19))/2#

There are no Real solutions, but as Complex values:
#color(white)("XXX")x=-1/2+sqrt(19)icolor(white)("XXX")"or"color(white)("XXX")x=-1/2-sqrt(19)i#