How do you find the roots, real and imaginary, of $y=x(2x-1)-x^2 - 5/2x + 1/2$ using the quadratic formula?

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Answer 1 · 2016-06-10T13:00:42

Zeros of $x(2x-1)-x^2-5/2x+1/2$ are

$7/4-sqrt41/4$ and $7/4+sqrt41/4$

Quadratic formula gives the roots or zeros of general form of quadratic equation $ax^2+bx+c$ as $(-b+-sqrt(b^2-4ac))/(2a)$

Hence we should first reduce $y=x(2x-1)-x^2-5/2x+1/2$ to general form.

$x(2x-1)-x^2-5/2x+1/2$

= $2x^2-x-x^2-5/2x+1/2$

= $x^2-7/2x+1/2$

Hence zeros of $x(2x-1)-x^2-5/2x+1/2$ are

$(-(-7/2)+-sqrt((-7/2)^2-4*1*1/2))/(2*1)$ or

$(7/2+-sqrt(49/4-2))/2$ or

$(7/2+-sqrt(41/4))/2$ or

$7/4+-sqrt41/4$

i.e. $7/4-sqrt41/4$ and $7/4+sqrt41/4$

How do you find the roots, real and imaginary, of y=x(2x-1)-x^2 - 5/2x + 1/2 y=x(2x-1)-x^2 - 5/2x + 1/2 using the quadratic formula?