How do you find the roots, real and imaginary, of $y=x(2x-1)-(3x-1)^2$ using the quadratic formula?

Question

1485 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-03T17:30:25

$x = 5/14+-(sqrt(3))/14i$

Let's multiply out the brackets and get ourselves some quadratic terms:

$y = 2x^2 - x - 9x^2 + 6x - 1 = -7x^2 +5x -1$

The quadratic formula for polynomial of the form $ax^2+bx+c = 0$ is given by:

$x = (-b+-sqrt(b^2-4ac))/(2a)$

$a = -7, b = 5, c = -1$

$therefore x = (-5+-sqrt(25-4(-7)(-1)))/(2(-7))$

$x= (-5+-sqrt(-3))/(-14)$

Recall that $i^2 = -1$ so we can rewrite $sqrt(-3)$ as $sqrt(3i^2) = sqrt(3)i$

$therefore x = 5/14+-(sqrt(3))/14i$

How do you find the roots, real and imaginary, of y=x(2x-1)-(3x-1)^2 y=x(2x-1)-(3x-1)^2 using the quadratic formula?