How do you find the zeros of # y = 2x^2 + 4x -1 # using the quadratic formula?

1 Answer
Dec 13, 2017

Answer:

#-1+-sqrt(6)/2#

Explanation:

This quadratic function is written in "standard form", or #ax^2+bx+c#.
The quadratic formula uses the numbers a, b, and c like this:
#(-b+-sqrt(b^2-4ac))/(2a)#
comparing the original function to the general, standard form function above, we see that #a=2, b=4, &c=-1#.

Plug these in the quadratic formula:
#(-4+-sqrt(4^2-4(2)(-1)))/(2*2)=(-4+-sqrt(16-(-8)))/4#
#(-4+-sqrt(24))/4#
To simplify that #sqrt(24)#, pull out the largest perfect square that goes in it. Since 4 divides evenly into 24, and neither 9 nor 16 do, break it down in terms of 4: #sqrt(24)=sqrt(4*6)#
You can take out #sqrt(4)=2#:

#sqrt(4*6)=2sqrt(6)#
From here, divide through by 4 to get #-1+-sqrt(6)/2#

...and remember, quadratic functions always have two solutions!