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

Jun 10, 2016

Zeros of $x \left(2 x - 1\right) - {x}^{2} - \frac{5}{2} x + \frac{1}{2}$ are

$\frac{7}{4} - \frac{\sqrt{41}}{4}$ and $\frac{7}{4} + \frac{\sqrt{41}}{4}$

#### Explanation:

Quadratic formula gives the roots or zeros of general form of quadratic equation $a {x}^{2} + b x + c$ as $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Hence we should first reduce $y = x \left(2 x - 1\right) - {x}^{2} - \frac{5}{2} x + \frac{1}{2}$ to general form.

$x \left(2 x - 1\right) - {x}^{2} - \frac{5}{2} x + \frac{1}{2}$

= $2 {x}^{2} - x - {x}^{2} - \frac{5}{2} x + \frac{1}{2}$

= ${x}^{2} - \frac{7}{2} x + \frac{1}{2}$

Hence zeros of $x \left(2 x - 1\right) - {x}^{2} - \frac{5}{2} x + \frac{1}{2}$ are

$\frac{- \left(- \frac{7}{2}\right) \pm \sqrt{{\left(- \frac{7}{2}\right)}^{2} - 4 \cdot 1 \cdot \frac{1}{2}}}{2 \cdot 1}$ or

$\frac{\frac{7}{2} \pm \sqrt{\frac{49}{4} - 2}}{2}$ or

$\frac{\frac{7}{2} \pm \sqrt{\frac{41}{4}}}{2}$ or

$\frac{7}{4} \pm \frac{\sqrt{41}}{4}$

i.e. $\frac{7}{4} - \frac{\sqrt{41}}{4}$ and $\frac{7}{4} + \frac{\sqrt{41}}{4}$