# How do you find the roots, real and imaginary, of y=x^2-5x + 6 using the quadratic formula?

Mar 26, 2017

$x \in \left\{2 , 3\right\}$

#### Explanation:

The quadratic formula works for equations of the form:

$y = a {x}^{2} + b x + c$

And states that the solutions for $x$ are:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

In this case, the quadratic equation is $y = {x}^{2} - 5 x + 6$, so:

$a = 1$
$b = - 5$
$c = 6$

So, just plug these values into the quadratic equation and simplify.

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- \left(- 5\right) \pm \sqrt{{\left(- 5\right)}^{2} - 4 \left(1\right) \left(6\right)}}{2 \left(1\right)}$

$x = \frac{5 \pm \sqrt{25 - 24}}{2}$

$x = \frac{5 \pm \sqrt{1}}{2}$

$x = \frac{5 - 1}{2} \textcolor{w h i t e}{\text{XX" or color(white)"XX}} x = \frac{5 + 1}{2}$

$x = 2 \textcolor{w h i t e}{\text{XX" or color(white)"XX}} x = 3$

Therefore, the solutions to this quadratic are $x = 2$ and $x = 3$. You can write this any way that you like (or whichever way your teacher prefers), but a common way to write it is like this, using set notation:

$x \in \left\{2 , 3\right\}$

This just means that $x$ could be any number in the set $\left\{2 , 3\right\}$ to make the equation true.

Mar 26, 2017

$x = 2 \text{ or } x = 3$

#### Explanation:

The standard form of a quadratic function is.

• y=ax^2+bx+c ; a!=0

$\text{For } y = {x}^{2} - 5 x + 6$

$\Rightarrow a = 1 , b = - 5 \text{ and } c = 6$

Using the $\textcolor{b l u e}{\text{quadratic formula}}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\Rightarrow x = \frac{- \left(\textcolor{red}{- 5}\right) \pm \sqrt{25 - \left(4 \times 1 \times 6\right)}}{2}$

$\textcolor{w h i t e}{\Rightarrow x} = \frac{5 \pm \sqrt{1}}{2}$

$\Rightarrow {x}_{1} = \frac{5 + 1}{2} \text{ or } {x}_{2} = \frac{5 - 1}{2}$

$\Rightarrow \text{ roots are " x_1=3" or } {x}_{2} = 2$

Both roots are real. There are no imaginary roots.