Question

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

Answer 1

`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)`.