How do you find the roots, real and imaginary, of #y=-x^2+5x-9 # using the quadratic formula?

1 Answer
Mar 16, 2016

Answer:

real roots: none
imaginary roots: #x=(5+-sqrt(11)i)/2#

Explanation:

Since the given equation is already in standard form, identify the #color(blue)a,color(darkorange)b,# and #color(violet)c# values. Then plug the values into the quadratic formula to solve for the roots.

#y=color(blue)(-1)x^2# #color(darkorange)(+5)x# #color(violet)(-9)#

#color(blue)(a=-1)color(white)(XXXXX)color(darkorange)(b=5)color(white)(XXXXX)color(violet)(c=-9)#

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#x=(-(color(darkorange)(5))+-sqrt((color(darkorange)(5))^2-4(color(blue)(-1))(color(violet)(-9))))/(2(color(blue)(-1)))#

#x=(-5+-sqrt(25-36))/-2#

#x=(-5+-sqrt(-11))/-2#

#x=(-(-5+sqrt(11)i))/2color(white)(X),color(white)(X)(-(-5-sqrt(11)i))/2#

#x=(5-sqrt(11)i)/2color(white)(X),color(white)(X)(5+sqrt(11)i)/2#

#color(green)(|bar(ul(color(white)(a/a)x=(5+-sqrt(11)i)/2color(white)(a/a)|)))#

#:.#, the imaginary roots are #x=(5+-sqrt(11)i)/2#.