How do you use the quadratic formula to solve the equation, #x^2-x= -1#?

1 Answer
Oct 18, 2016

Answer:

NO ROOTS in #x !in RR#
ROOTS #x in CC#
#x=(1+isqrt3)/2#
OR
#x=(1-isqrt3)/2#

Explanation:

#x^2-x=-1#
#rArrx^2-x+1=0#
We have to factorize
#color(brown)(x^2-x+1)#
Since we can not use polynomial identities so we will calculate #color(blue)(delta)#

#color(blue)(delta=b^2-4ac)#
#delta=(-1)^2-4(1)(1)=-3<0#

NO ROOTS IN #color(red)(x !in RR)# because #color(red)(delta<0)#

But roots exist in #CC#
#color(blue)(delta=3i^2)#

Roots are
#x_1=(-b+sqrtdelta)/(2a)=(1+sqrt(3i^2))/2=(1+isqrt3)/2#
#x_2=(-b-sqrtdelta)/(2a)=(1-sqrt(3i^2))/2=(1-isqrt3)/2#

The equation is:
#x^2-x+1=0#
#rArr(x-(1+isqrt3)/2)(x-(1-isqrt3)/2)=0#
#(x-(1+isqrt3)/2)=0rArrcolor(brown)( x=(1+isqrt3)/2)#
OR
#(x-(1-isqrt3)/2)=0rArrcolor(brown)(x=(1-isqrt3)/2)#

So the roots exist only in #color(red)(x in CC)#