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

##### 2 Answers

$x = \frac{5 \setminus \pm \setminus \sqrt{17}}{2}$

#### Explanation:

Given quadratic function:

$y = {x}^{2} - \left(x - 2\right) \left(2 x - 1\right)$

$y = {x}^{2} - 2 {x}^{2} + 5 x - 2$

$y = - {x}^{2} + 5 x - 2$

The roots of given function will be at $y = 0$ i.e.

$- {x}^{2} + 5 x - 2 = 0$

${x}^{2} - 5 x + 2 = 0$

Now, using quadratic formula, the roots of above quadratic equation are given as

${x}_{1 , 2} = \setminus \frac{- \left(- 5\right) \setminus \pm \setminus \sqrt{{\left(- 5\right)}^{2} - 4 \left(1\right) \left(2\right)}}{2 \left(1\right)}$

$= \setminus \frac{5 \setminus \pm \setminus \sqrt{17}}{2}$

Thus, the roots of given quadratic function are real & distinct given as

$x = \frac{5 \setminus \pm \setminus \sqrt{17}}{2}$

Jul 24, 2018

The solutions are $S = \left\{0.44 , 4.56\right\}$

#### Explanation:

The quadratic equation is

$y = {x}^{2} - \left(x - 2\right) \left(2 x - 1\right)$

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

$= {x}^{2} - 2 {x}^{2} + 5 x - 2$

$\implies$, $- {x}^{2} + 5 x - 2 = 0$

$a = - 1$

$b = 5$

$c = - 2$

The discriminant is

$\Delta = {b}^{2} - 4 a c = {\left(5\right)}^{2} - 4 \left(- 1\right) \left(- 2\right) = 25 - 8 = 17$

As $\Delta > 0$, there are $2$ real roots

The solution is

$x = \frac{- b \pm \sqrt{\Delta}}{2 a}$

$= \frac{- 5 \pm \sqrt{17}}{- 2}$

${x}_{1} = 0.438$

${x}_{2} = 4.56$

graph{-x^2+5x-2 [-10, 10, -5, 5]}