How do you find the zeros, real and imaginary, of $y=x^2 -34x+1$ using the quadratic formula?

Question

1711 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-16T13:12:58

See a solution process below:

For $color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0$ , the values of $x$ which are the solutions to the equation are given by:

$x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))$

Substituting:

$color(red)(1)$ for $color(red)(a)$

$color(blue)(-34)$ for $color(blue)(b)$

$color(green)(1)$ for $color(green)(c)$ gives:

$x = (-color(blue)(-34) +- sqrt(color(blue)(-34)^2 - (4 * color(red)(1) * color(green)(1))))/(2 * color(red)(1))$

$x = (34 +- sqrt(1156 - 4))/2$

$x = (34 +- sqrt(1152))/2$

$x = (34 - sqrt(576 * 2))/2$ and $x = (34 + sqrt(576 * 2))/2$

$x = (34 - sqrt(576)sqrt(2))/2$ and $x = (34 + sqrt(576)sqrt(2))/2$

$x = (34 - 24sqrt(2))/2$ and $x = (34 + 24sqrt(2))/2$

$x = 17 - 12sqrt(2)$ and $x = 17 + 12sqrt(2)$

How do you find the zeros, real and imaginary, of y=x^2 -34x+1y=x^2 -34x+1 using the quadratic formula?