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

1 Answer
Mar 9, 2016

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