How do you find the zeros, real and imaginary, of y=19x^2-4x-3 using the quadratic formula?

1 Answer
The Legend
Mar 27, 2018

x=(2+sqrt(41))/(19) and x=(2-sqrt(41))/(19).

Explanation:

The problem says to use the quadratic formula, so we will use that. The quadratic formula is x=(-b+-sqrt(b^2-4ac))/(2a). a,b, and c come from ax^2+bx+c, the standard form of a quadratic equation.

The equation is already in standard form, so we can simply substitute and simplify!

x=(-b+-sqrt(b^2-4ac))/(2a)
x=(-(-4)+-sqrt((-4)^2-4(19)(-3)))/(2(19))
x=(4+-sqrt(16-(-228)))/(38)
x=(4+-sqrt(244))/(38)
244 is not a perfect square, so we could stop here. However, notice that 244 is divisible by 4, which is a perfect square, so we can simplify sqrt(244)!
x=(4+-sqrt(4*41))/(38)
x=(4+-2sqrt(41))/(38)
We can pull a 2 out of each term and simplify:
x=(cancel2(2+-sqrt(41)))/(cancel2(19))
x=(2+-sqrt(41))/(19)
So our solutions are x=(2+sqrt(41))/(19) and x=(2-sqrt(41))/(19).

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