How do you solve #x^2 - 2x + 10 = 0#?

2 Answers
May 14, 2018

Answer:

#x=1+-3i#

Explanation:

#"the equation is in standard form "color(white)(x)ax^2+bx+c=0#

#"with "a=1,b=-2" and "c=10#

#"there are no whole number values which allow us to"#
#"factor the quadratic"#

#"check the value of the "color(blue)"discriminant"#

#•color(white)(x)Delta=b^2-4ac#

#b^2-4ac=(-2)^2-(4xx1xx10)=4-40=-36#

#"since "Delta<0" then roots are complex"#

#"to calculate use the "color(blue)"quadratic formula"#

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

#rArrx=(2+-sqrt(-36))/2=(2+-6i)/2#

#rArrx=1+-3i#

May 14, 2018

Answer:

Solution: # x = 1 +3 i and x = 1- 3 i#

Explanation:

# x^2-2 x +10=0 or x^2-2 x = -10# or

#x^2-2 x +1 = 1-10# or

#(x-1)^2 = -9 or (x-1) = +- sqrt (-9)#

It has complex roots.

#:. (x-1) = +- sqrt (9 i^2) [i^2=-1]# or

#(x-1) = +- 3 i or x = 1+- 3 i#

Solution: # x = 1 +3 i and x = 1- 3 i# [Ans]