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

1 Answer
LizZ
Dec 13, 2017

-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!

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