How do you find the zeros, if any, of #y= 2x^2 -12x+ 23 #using the quadratic formula?

1 Answer
Mar 9, 2016

Answer:

#Delta<0#
#:. cancel(EE)x in RR#

The equation has not solutions in #RR#

or, if the excercise allow it

#x_(1,2)=3+-isqrt(10)/2 in CC#

Explanation:

Given

#y=ax^2+bx+c=2x^2-12x+23=0#

The Quadratic Formula tells:

#x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)=(-b+-sqrt(Delta))/(2a)#

where:

#a=2#
#b=-12#
#c=23#

#:. x_(1,2)=(12+-sqrt(144-4*2*23))/4=(12+-sqrt(144-4*2*23))/4=(12+-sqrt(144-184))/4=(12+-sqrt(-40))/4#

Now, #Delta<0# therefore #cancel(EE)x in RR#

You have solutions in #CC#, if the exercise allows them

#x_(1,2)=(12+-sqrt(-1)sqrt(40))/4=(12+-isqrt(4*10))/4=#
#=(cancel(12)^6+-icancel(2)sqrt(10))/cancel(4)^2=3+-isqrt(10)/2#